Blood, ph and Logarithms Unit (Level IV Graduate Math) Draft (NSSAL) C. David Pilmer 0 (Last Updated: June 0) Important Note: This unit is only available to learners who are taking the ALP Graduate Level IV Math and ACC Health Math Dual Accreditation program.
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Table of Contents Introduction ii Negotiated Completion Date. ii The Big Picture.. iii Course Timelines... iv How are these Terms Related?. Introduction to Eponents. More Eponents... Scientific Notation... More Scientific Notation.. 0 Logarithms... Logarithms and ph.. Logarithms and poh More poh. Putting It Together 0 Post-Unit Reflections Answers. NSSAL i Draft 0 C. D. Pilmer
Introduction Normally in Level IV Graduate Math, learners must complete five required units and choose two of four optional units (Linear Functions and Systems of Equations Unit, Trigonometry Unit, Statistics Unit). There are, however, learners who wish to pursue careers in the health sciences and will also need to take the NSCC's ACC Health Math courses. There is significant overlap between ALP Graduate Math and ACC Health Math courses, so the Department of Labour and Advanced Education has worked collaboratively with the college to create a pathway between the two courses so learners can move through the two programs as quickly as possible and as seamlessly as possible. For this reason, we have created this particular unit. Learners interested in the health sciences will now take the original five required units for Graduate Math, the ALP Statistics Unit, the ALP Blood, ph, and Logarithm Unit, and the ACC Dosages, Solutions, and IV Therapy Unit. By doing so, the learner will receive the ALP Graduate Math credit which can be applied to their diploma, and the ACC Health Math credits that will officially be recognized following the learner's graduation from ALP. We stress that the ACC credits will not be applied to the learner's high school graduation diploma. Negotiated Completion Date After working for a few days on this unit, sit down with your instructor and negotiate a completion date for this unit. Start Date: Completion Date: Instructor Signature: Student Signature: NSSAL ii Draft 0 C. D. Pilmer
The Big Picture This flow chart only applies to learners who are enrolled in the ALP Graduate Level IV Math and ACC Health Math dual credit program. Math in the Real World Unit Solving Equations Unit Consumer Finance Unit Graphs and Functions Unit Measurement Unit Statistics Unit Blood, ph and Logarithms Unit Completion of ALP Graduate Math Dosages, Solutions, and IV Therapy Unit NSSAL iii Draft 0 C. D. Pilmer
How are these Terms Related? You are probably looking at the three terms in the unit title, The Blood, ph and Logarithms Unit, and asking, "How are these terms related?" How are Blood and ph related? Health care workers monitor a host of patient vitals which include respiration, heart rate, and blood pressure. Many do not know they also monitor patients' blood ph levels. The kidneys and lungs maintain the proper balance of chemicals, called acids and bases, in the body. If the body correctly maintains this balance then the blood ph range for an adult should be between. and.. For infants and children ( yrs.), the blood ph levels should be between. and.. What is ph? When some substances are dissolved in water, they alter the hydrogen ion (H + ) concentration. Acidic solutions (e.g. lemon juice) increase the hydrogen ion concentration. The ph is, however, low for these solutions. Alkaline or basic solutions (e.g. Soapy Water) decrease the hydrogen ion concentration. The ph is, however, high for these alkaline or basic solutions. Acidic Solutions Increase the H + Concentration Low ph Alkaline Solutions Decrease the H + Concentration High ph In conclusion, ph tells you whether are dealing with an acidic solutions, basic solutions, or something in between (i.e. a neutral solution). How do our bodies regulate ph? There are five mechanisms that the body uses to regulate ph levels.. The lungs vent carbon dioide (CO ) from our systems. This carbon dioide is a waste product of the metabolism of oygen. This waste product gets ecreted into our blood, then moved to our lungs, and finally ehaled. When still in the blood, this carbon dioide can react with the water in our blood to create carbonic acid. It is important that the lungs function properly to ensure that CO can be removed in a timely fashion so that carbonic acid does not rise to dangerous levels.. The kidneys filter blood and remove ecess acids and bases. They do have the ability to alter the amount of acid or base that is ecreted, but these adjustments occur slowly generally taking several days.. The lower intestines epel ecess acid or base in feces. NSSAL Draft 0 C. D. Pilmer
. The skin ecretes a mild acid through perspiration. This acid also maintains the strength and integrity of the skin and prevents infection by inhibiting the growth of bacteria. A buffer is a solution that can maintain a nearly constant ph if it is diluted, or if strong acids or bases are added. The body uses chemical buffers in fluids and bones to balance ph levels. The buffers are released to neutralize ecess acid or base. When blood ph levels are off If the blood ph level for an infant, child, or adult falls below. then the patient is said to have acid-base disorder acidosis. The condition is considered etreme if the ph drops to. or lower. When blood ph levels drop this low, coma and even death can occur. If the blood ph level for an adult rises above. or the blood ph level for an enfant or child rises above., then the patient is said to have the acid-base disorder alkalosis. This condition is considered etreme if it rises to. or higher in adults and. or higher in infants or children. When these elevated ph levels occur, the patients will likely convulse and these convulsions may ultimately lead to death if the condition is not addressed. Etreme Acidosis Normal Range Etreme Alkalosis Adult.. to.. Infant or Child.. to.. It is important to note that these two acid-base disorders (acidosis and alkalosis) are not diseases themselves; rather they are signs of an underlying disease or condition. Both acidosis and alkalosis are found in respiratory and metabolic forms. Respiratory acidosis or alkalosis is caused by various malfunctions of the lungs. Metabolic acidosis or alkalosis is caused by various metabolic disorders which result in an ecessive build up or loss of acids or bases. Metabolic disorders are problems associated with the various processes within the body that convert food and other substances into energy and other metabolic byproducts used by the body. Respiratory Acidosis Respiratory acidosis results from a build-up of carbon dioide (CO ) due to hypoventilation (i.e. slow shallow breathing). This depressed breathing can be caused by drugs, central nervous system trauma, or lung diseases (e.g. pneumonia). Carbon dioide reacts with the water in our blood and forms carbonic acid (H CO ). Ecess levels of CO, result in higher levels of the acid H CO, and correspondingly lower blood ph readings. Metabolic Acidosis Metabolic acidosis results from an increased production of acid or from the decreased availability of bicarbonate (HCO, a base) due to diabetes, kidney disease, poisoning (aspirin, ethylene glycol, or methanol), or severe diarrhea. NSSAL Draft 0 C. D. Pilmer
Respiratory Alkalosis Respiratory alkalosis results from decreased CO levels due to hyperventilation (i.e. rapid deep breathing). This over breathing can be caused by aniety, pain, or fever. Metabolic Alkalosis Metabolic alkalosis results when the body loses too much acid or from the increased availability of bicarbonate (HCO, a base). Prolonged vomiting is one way a body can lose ecessive amounts of acid. In some cases (e.g. poisoning), hospital staff have to suction out stomach contents, reducing acid levels which can lead to alkalosis. In addition, metabolic alkalosis can occur if sodium or potassium levels fall too low. When this happens, the kidneys are unable to maintain the proper acid base balance. Important Note: You will have to be familiar with these four terms and the blood ph ranges in later sections of this unit. How are logarithms and ph related? We have learned that ph is a way of distinguishing acidic solutions from basic solutions. The net obvious question is why does the ph rules seem to be backwards. For eample, if a solution has a high concentration of H +, it has a low ph. Similarly, if the solution has a low concentration of H +, the ph is high. The answer to this apparent backwardness is how ph is described using logarithms. You may have also noticed that the range of blood ph levels from dangerously low to normal to dangerously high seems quite small. For eample the difference between an etreme low (.) to the low normal (.) is only 0.. Similarly the difference between a high normal (.) in adults and an etreme high (.) in adults is only 0.0. These differences seem very small but the implications for the patients are huge. Why are these differences so small? Again, the answer is logarithms. Now we cannot just dive into the topic of logarithms because there is some fundamental work with eponents and scientific notation that must first be completed. In the net few sections we will work on these fundamental concepts with the understanding that they will ultimately lead to logarithms and further discussions of blood ph. NSSAL Draft 0 C. D. Pilmer
Introduction to Eponents Before you start this unit, it is epected that you are familiar with powers of,,, and. You should know that: and conversely and conversely and conversely and conversely and conversely and conversely and conversely and conversely and conversely and conversely and conversely Make sure you are familiar with these before proceeding. The Laws of Eponents When given the epression base. n a, we say that n a is the power, n is the eponent, and a is the Therefore when given the power, we state that the base is and the eponent is. This power can be written as. The Law of Products of Powers To multiply two powers that have the same base, keep the base and add the eponents. m n mn a a a Eamples: r r r r The Law of Quotients of Powers To divide two powers that have the same base, keep the base and subtract the eponents. m n mn a a a Eamples: r r r NSSAL Draft 0 C. D. Pilmer
NSSAL Draft 0 C. D. Pilmer Eamples: (a) (b) b a b a b a (c) k k k (d) d c cd d c (e) (f) n m n m n m (g) v u v u v u (h) z y z y z y (i) 0 0 y y y y y (j) 0 a ab b a ab b a b a The Law of Powers of Powers To raise a power to a power, keep the same base and multiply the eponents. n m n m a a Eamples: h h The Law of Powers of Products If a product in parentheses is raised to an eponent, the parentheses indicate that each factor must be raised to that eponent. n n n b a ab Eamples: y y y y r r r The Law of Power of Quotients If a quotient in parentheses is raised to an eponent, the parentheses indicate that both the numerator and denominator must be raised to that power. n n n b a b a where 0 b Eamples: y y b a b a 0 0 y y y
Questions. Evaluate. (a) (b) (c) (e) (g) (i) (d) (f) 00 (h) (j). Simplify each of the following epressions. (a) (c) (e) (g) (b) a b a b (d) c c d y y (f) p q p q h h (h) y y (i) (k) a (j) c d k (l) a b (m) c d (n) y (o) y (p) ab c (q) c d c d (r) pq pq p (s) y (t) a b NSSAL Draft 0 C. D. Pilmer
(The remaining questions will require two steps. Show your work.) (u) y y y y (v) y 0 0 p q (w) p q () c d cd c d (y) a b a b ab y y y (z). Determine. No work needs to be shown. (a) = (b) = (c) = (d) = (e) = (f) = (g) = (h) = NSSAL Draft 0 C. D. Pilmer
More Eponents Zero Eponents n n. Consider We know that when we take one quantity and divide it by the same quantity, the quotient is always. For eample:,,, and. Therefore we can n conclude that. The only eception to this rule is if the two quantities are equal to n 0 zero. 0 Based on the Law of Quotients of Powers, we know that to divide two powers that have the same base, we keep the base and subtract the eponents. Therefore our epression n can now be epressed in the following manner. n n nn 0 n n n 0 In the first bullet we stated that. In the second bullet we stated that. n n Therefore we can conclude 0 (as long as does not equal zero). Eamples: 0 0 0 0 0 (Note: In this question only, not -, is raised to the eponent 0.) 0. 0 (Note: In this question only 0., not -0., is raised to the eponent 0.) NSSAL Draft 0 C. D. Pilmer
Negative Eponents Consider. The epression can be simplified by dividing through (i.e. canceling out) like quantities. Using the Law of Quotients of Powers, our epression can now be epressed in the following manner. In the first bullet we stated that. In the second bullet we stated that. Therefore we can conclude n n or. If we research this further we could generalize that where is not equal to zero. Therefore the negative eponent just means that we deal with the reciprocal of the base. Eamples: NSSAL Draft 0 C. D. Pilmer
Fractional Eponents Part : The Fractional Eponent Suppose we were asked to find two identical numbers that multiplied to give us the product. Most people would quickly respond because. This mathematical statement could also be written as the following. Now let s work with the variable. If we needed to find two identical terms that multiply to give us the product, we would end up with the following statement. Again we are going to find two identical terms that multiply to give us the product but we are not going to rely on radicals (i.e. square roots, cube roots, fourth roots, ). Instead we are going to rely on our understanding of the Law of Products of Powers. We need to find two powers with the base of and the same eponent whose product is.?? The only eponent that makes sense is one half. or In the first bullet we stated that. In the second bullet we stated that Part : The Fractional Eponent. Therefore we can conclude that. Suppose we were asked to find three identical numbers that multiplied to give us the product. Most people would quickly respond because. This mathematical statement could also be written as the following. Now let s work with the variable. If we needed to find three identical terms that multiply to give us the product, we would end up with the following statement. Again we are going to find three identical terms that multiply to give us the product but we are not going to rely on radicals (i.e. square roots, cube roots, fourth roots, ). Instead we are going to rely on our understanding of the Law of Products of Powers. We need to find three powers with the base of and the same eponent whose product is.??? The only eponent that makes sense is one third. or In the first bullet we stated that. In the second bullet we stated that. Therefore we can conclude that. NSSAL 0 Draft 0 C. D. Pilmer
Conclusion: If and, its fair to conclude that n n. Note: We can only take the square roots and fouth roots of positive numbers. 00 0 no real solution no real solution We can take the cube roots and fifth roots of both positive and negative numbers. Eamples: no real solution (We cannot take the square root of a negative number.) To Sum It Up: Zero Eponents: 0 where 0 Eample: 0 n n Negative Eponents: where 0 Eample: Fractional Eponents: n n Eample: Operations with Zero, Negative, and Fractional Eponents The laws of eponents that we learned in the previous section still apply when dealing with powers with zero, negative, and fractional eponents Eamples: 0 c c c c 0 p q p q p q NSSAL Draft 0 C. D. Pilmer
NSSAL Draft 0 C. D. Pilmer d d d d q p q p q p f f f 0 0 z y z y z y Questions:. Evaluate each of the following. Do not use a calculator. (a) (b) 0 (c) (d) (e) (f) 0 (g) (h) (i) 0 (j) (k) (l) (m) (n) 0 (o) (p) (q) (r) (s) (t)
. Epress each as a power. (a) p (b) b (c) h (d) d. Determine. No work needs to be shown. (a) = (b) = (c) = (d) 0 0 = (e) 0 = (f) = (g) = (h) = (i) = (j) =. Simplify each of the following epressions. (a) (b) (d) h h (c) (e) 0 k b b b (f) k a (h) c c (g) b (i) (k) p q (j) a b y y (l) c NSSAL Draft 0 C. D. Pilmer
NSSAL Draft 0 C. D. Pilmer (m) d c d c (n) 0 t s r (o) gh h g (p) pq q p
Scientific Notation You have probably done some calculations on your calculator and ended up with pretty weird looking answers. The screen shot shown on the right contains two such calculations. In one case we took 000 and raised it to the eponent, and in the other case we took 0.0 and divided it by 0 000. You might not have realized that your calculator is epressing the answer in scientific notation. 000. 0 000 000 000 000 (This is a huge number!) 0.0.0 0.00000 0000 (This is a tiny number!) When scientists (and people in health care professions) work with very large or very small numbers, they often epress these numbers using scientific notation. A positive number is in scientific notation when it is in the form. a is a number greater than or equal to and less than 0, and. n is an integer ( -, -, -, 0,,,, ) n a 0, where: Before we can start, we have to look at powers with a base of 0 and eponents that are integers. In last section project we learned that negative eponents meant we were dealing with n n reciprocals (i.e. where 0 ). If we apply this to powers with a base of 0, we get the following. 0 0 0 0 0 0 0 0 000 00 0 0 0 0 00 0 000 This knowledge can now be used to convert numbers from standard form to scientific notation and vise versa. For eample, 0 means 000 or 000. NSSAL Draft 0 C. D. Pilmer
Eample Convert the following numbers in standard form to scientific notation. (a) 00 (b) 000 (c) 0.00 (d) 0.0000 Answers: We will show you two very similar methods for tackling these questions. You choose the one you prefer. Method Method (a) 00 =. 000 We change the given number to a number greater =. 0 than or equal to and less than 0. Therefore we change 00 to.. In doing this, we moved the decimal point three places to the left. So now we have the following but we do not know the eponent for the power.? 00 =. 0 Since the decimal point was moved three places to the left, we know that the eponent for the power must be. 00 =. 0 (b) 000 =. 00 000 =. 0 (c) 0.00 =. 0.00 =. 000 =. 0 We change 000 to.. In doing this, we moved the decimal point five places to the left. Now we have the following.? 000 =. 0 Since the decimal point was moved five places to the left, we know that the eponent for the power must be. 000 =. 0 We change the given number to a number greater than or equal to and less than 0. Therefore we change 0.00 to.. In doing this, we moved the decimal point three places to the right. Now we have the following.? 0.00. 0 Since the decimal point was moved three places to the right, we know that the eponent for the power must be -. 0.00. 0 NSSAL Draft 0 C. D. Pilmer
(d) 0.0000 = 0.0000 = 00000 = 0 We change 0.0000 to. In doing this, we moved the decimal point five places to the right. Now we have the following.? 0.0000 0 Since the decimal point was moved five places to the right, we know that the eponent for the power must be -. 0.0000 0 Eample Convert the following numbers in scientific notation to standard form. (a). 0 (b). 0 (c). 0 (d) 0 Answers: Again we will show you two very similar methods; choose the one you prefer. Method Method (a). 0.0000 Since the eponent of our power is, we move = 000 the decimal point, for the number., four places to the right and add zeros as placeholders.. 0 = 000 (b). 0. 000000 = 0 000 Since the eponent of our power is, we move the decimal point, for the number., si places to the right and add zeros as placeholders.. 0 = 0 000 (c). 0. 00. 0.0 0.0 Since the eponent of our power is -, we move the decimal point, for the number., two places to the left and add zeros as placeholders.. 0 0.0 (d) 0 0000 0.000 0.000 Since the eponent of our power is -, we move the decimal point, for the number, four places to the left and add zeros as placeholders. 0 0.000 NSSAL Draft 0 C. D. Pilmer
Questions:. Convert the following numbers in standard form to scientific notation. (a) 0 000 (b) 000 000 (c) 00 000 000 (d) 0.0 (e) 0.000 (f) 0.00000 (g) 0 000 (h) 0.00 (i) 00 000 (j) 0.0000. Convert the following numbers in scientific notation to standard form. (a) 0 (b) 0 (c) 0 (d) 0 (e). 0 (f). 0 (g). 0 (h).0 0 (i). 0 (j). 0 NSSAL Draft 0 C. D. Pilmer
. A certain thyroid tablet contains 0.000000 g of medication. Epress this amount in scientific notation.. Circle the appropriate response. When a number is epressed in scientific notation and the eponent of the 0 is a negative, then the number is (less than or greater than) one.. Order these numbers from smallest to largest. (a) 00, 0.000,., 0.00 (b) 0.000,.00,.0, 0.00 (c). 0,. 0, 0 (d). 0, 0,. 0 (e) 0,. 0,. 0 (f). 0, 0,. 0,. 0 (g). 0,. 0, 0,. 0 (h). 0, 0,. 0,. 0 NSSAL Draft 0 C. D. Pilmer
More Scientific Notation In this section we will learn how to complete operations (addition, subtraction, multiplication, and division) with numbers in scientific notation. Adding and Subtracting Numbers in Scientific Notation Numbers in scientific notation can only be added if they have the same eponent on the power of base 0. We add or subtract the decimal part and leave the eponent unchanged. Eample Complete the operation. (a). 0.0 (b) (c) 0 0 (d) (e) 0 0 (f). 0.0. 0.0. 0.0 Answers: The first four questions are a little easier because the eponents of the powers are the same in each case. (a). 0.0. 0 (b).0.0.0 (c) 0 0 0 (d).0.0.0 Our answer is not in the proper scientific notation form because the first number () is not greater than or equal to and less than 0. We have to do more work. 0.0.0 0 As with the last question, our answer has to be changed to the proper scientific notation form..0.0.0 0 With the two remaining questions, the eponents are different. This means that we must change one of the numbers (the smaller number) such that the two numbers end up with the same eponent. NSSAL 0 Draft 0 C. D. Pilmer
(e) 0 0 Change the smaller number, 0. (f). 0. 0 Change the smaller number,. 0. 0 0. 0 0. 0 0 Now we can do the addition. 0 0.0. 0. 0 0.0 0.0 0 Now we can do the subtraction.. 0 0.0.0 Multiplying Numbers in Scientific Notation One can multiply any two numbers written in scientific notation. Unlike addition and subtraction, we do not need the same eponent on the powers with a base of 0. To multiply numbers in scientific notation we merely multiply the numbers to the left of the 0 notation and then multiply the powers with a base of 0 (by adding the eponents). We also have to make sure that the final answer is in the proper scientific notation form. Eample Complete the operation. Write your answer in scientific notation. 0 0 0 0 (a) (b) (c). 0. 0 (d). 0. 0 Answers: 0 (a) 0 0 0 0 0 (c). 0. 0..0 0. 0. 0 Not in the proper form. (b) 0 0 0 0 0 0 (d). 0. 0..0 0. 0. 0 Not in the proper form..0.0.0 0.0.0.0 0 NSSAL Draft 0 C. D. Pilmer
Dividing Numbers in Scientific Notation Similar to multiplication, one can divide any two numbers written in scientific notation. We merely divide the numbers to the left of the 0 notation and then divide the powers with a base of 0 (by subtracting the eponents). Again one has to make sure the final answer is in the proper scientific notation form. Eample Complete the operation. Write your answer in scientific notation. 0 0. 0.0 (a) (b) (c). 0. 0 (d). 0.0 Answers: 0 (a) 0 ( ) 0 0 0 0 (c). 0. 0..0 0 0. 0 0. 0 0 Not in the proper form. 0.0. 0. 0 0 0 0 (b). 0. 0..0 0. 0. 0 (d).0.0..0 0 0.0 0.0 Not in the proper form 0.0.0.0 0 Questions. Complete the operations. Write your answer in scientific notation. (a). 0. 0 (b). 0.0 (c) 0. 0 (d). 0. 0 NSSAL Draft 0 C. D. Pilmer
(e). 0.0 (f) 0. 0 (g) 0. 0 (h).. 0. 0. Complete the operations. Write your answer in scientific notation. 0 0. 0. 0 (a) (b) (c). 0. 0 (d) 0 0 (e). 0. 0 (f).00. 0. Complete the operations. Write your answer in scientific notation. 0 0 0 0 (a) (b) NSSAL Draft 0 C. D. Pilmer
(c). 0. 0 (d). 0.0 (e) 0 0 (f). 0. 0. Complete the operations. Write your answer in scientific notation. (a). 0.0. 0. 0 (b) (c). 0. 0 (d). 0. 0 (e). 0.0 (f). 0. 0 (g). 0. 0 (h). 0. 0 NSSAL Draft 0 C. D. Pilmer
(i). 0. 0 (j). 0. 0 (k). 0 0 (l). 0. 0 0 0 (m) 0. 0. 0 (n) 0.0.0 (o) 0. 0.0 (p). 0 NSSAL Draft 0 C. D. Pilmer
Logarithms When we first learned how to solve simple linear equations like, we also learned about inverse operations. Consider the eamples below. 0 0 The and are being added together on the left hand side of the equation. In order to solve for, we had to undo that addition. This was accomplished by subtracting from both sides of the equation. Subtraction is the inverse operation to addition. The subtraction reverses the work already done by the addition. We can conclude that: 0 is equivalent to 0 The and are being multiplied together. In order to reverse this process of multiplication we used the inverse operation division. We can conclude that: is equivalent to In the two eamples above, we were able to use some of the basic algebraic operations (i.e. addition, subtraction, multiplication, division, and roots) to solve for. Unfortunately these same basic algebraic operations can not be used to solve eponential equations. Mathematicians had to create a new inverse operation to handle eponential equations. Complete the following using a calculator (i.e. Use the 0 button.). (a) Evaluate (b) Evaluate (c) Evaluate. 0, then find the log of that solution. Your Answer: 0. 0, then find the log of that solution. Your Answer:. 0, then find the log of that solution. Your Answer: (d) Did the log button undo the work done by the eponent button? Here are the equivalent forms of equations we previously learned. Eample: Yes or No a b is equivalent to b a 0 is equivalent to 0 a b is equivalent to b a is equivalent to a b is equivalent to b is equivalent to a b is equivalent to a b a is equivalent to n a is equivalent to n a is equivalent to NSSAL Draft 0 C. D. Pilmer
Now that we know that logs undo the work done by eponents, we can now add a new equivalent form of equations to our list. Eample: a b is equivalent to log b 0 is equivalent to log 0 a Eample Change the following from eponential to logarithmic form. (a) (b) (c) Answers: (a) log (b) log We would say, Log base of is. (c) log Eample Change the following from logarithmic to eponential form. (a) log (b) log (c) log Answers: (a) (b) (c) Eample Evaluate each of the following logarithms. (a) log (b) log (c) log Answers: Do not use your log button on your calculator to solve any of these problems. The log button works only with logarithms with a base of 0. Our questions have the bases, and. The easiest approach is to make each of these epressions equal to, change from logarithmic form to eponential form, and then evaluate. NSSAL Draft 0 C. D. Pilmer
(a) log log (b) log log (c) log log Eample Evaluate each of the following logarithms. (a) log 0 (b) log 0.0 (c) log (-00) Answers: When the base is not supplied on a logarithm, one is to assume that the base is 0. Therefore: log log0 Since the base is 0, we can use the log button on our calculator to evaluate each of these. (a) log 0 =. This could be restated as 0. 0. The answer. seems reasonable because 0 00 and 0 000. Since 0 is between 00 and 000, then we would epect our answer to be between and. (b) log 0.0 = -. This could be restated as 0. 0. 0. The answer -. seems reasonable because 0 or 0., and 0 or 0.0. Since 0.0 is between 0. and 0.0, then we 0 00 would epect our answer to be between - and -. (c) log (-00) no solution The calculator is showing an error because this question can be restated as 0 00. It is impossible to take 0 and raise it to an eponent such that a negative number is generated. NSSAL Draft 0 C. D. Pilmer
Eample Solve for. (a) log (b) log (c) log (d) log (e) log. (f) log. (g) log (h) log Answers: With all these questions, we will start by changing the equation from its logarithmic form to its eponential form. Once this is done, the first four questions (a to d) can be quickly solved in your head. The last two questions (e and f) will require one to use a calculator's 0 button. (a) log (b) log (c) log (d) log log 0 0 00 (e) log. log. 0. 0 (calculator) 0. (f) log. log0.. 0 (calculator) 0.0 (g) log (h) log NSSAL Draft 0 C. D. Pilmer
Questions:. Change the following from eponential to logarithmic form. (a) (b) (c) (d) (e) (f) 0. Change the following from logarithmic to eponential form. (a) log (b) log (c) log (d) log (e) log (f) log 0. 00 (g) log (h) log0000. Evaluate each of the following logarithms. (a) log (b) log (c) log (d) log NSSAL 0 Draft 0 C. D. Pilmer
(e) log (f) log 000 (g) log (h) log log (j) log (i) 0. (k) log (l) log (m) log (n) log. Solve for. (a) log (b) log (c) log (d) log NSSAL Draft 0 C. D. Pilmer
(e) log (f) log (g) log (h) log (i) log (j) log 0. (k) log (l). log (m) log (n) log (o) log (p) log. NSSAL Draft 0 C. D. Pilmer
Logarithms and ph When an acid is added to water, it releases hydrogen ions (H + ) which logically drives up the concentration of hydrogen ions (H + ). This resulting acidic solution is said to have a ph lower than. For eample, black coffee (a weak acidic solution) has a ph of while stomach acid (a strong acidic solution) has a ph of. When a base is added to water, it releases hydroide ions (OH - ) which in turns drives down the concentration of hydrogen ions. This resulting basic (or alkaline) solution is said to have a ph greater than. For eample, milk of magnesia (a weak basic solution) has a ph of 0 while bleach (a strong basic solution) has a ph of. This ph system was developed in 0 by the Swedish chemist S.P.L. Sorensen. He defined ph as the logarithm (base 0) of the reciprocal of the concentration of the hydrogen ions, measured in moles per litre, in a solution. ph log Note: [H + ] represents the concentration of hydrogen ions [H ] From chemistry you have learned that one mole of a substance represents.0 0 representative particles (atoms, molecules or ions) of that substance. The concentration measured in moles per litre (mol/l) is referred to as molarity. Note: We will talk more about hydroide ions later in the unit and eplain why it affects hydrogen ion concentrations. Eample If the concentration of hydrogen ions in a solution is 0.00 moles per litre, then what is the corresponding ph of that solution? Answer: ph log [H ] ph log 0.00 ph log. ph. Please Note: Many chemistry tetbooks define ph as the following. ph log [H ] This is equivalent to the formula that we are using. To prove this, we can answer eample using this alternate formula and show that we get the same answer. ph log [H ] ph log0.00 ph. ph. NSSAL Draft 0 C. D. Pilmer
Eample If the ph of a solution is., then what is the concentration of hydrogen ions in moles per litre? Answer: ph log [H ]. log [H ]. log0 [H ] It was changed from logarithmic to eponential form. 0. [H ]. [H ] 0 [ H ]. 0 [H ].00 moles per litre (It can also be written as 0.000000000 mol/l.) Eample The blood for one adult patient is showing hydrogen ion concentrations of. 0 moles per litre. (a) Determine the ph for this patient's blood. (b) Is this patient's blood in the normal ph range or is he/she eperiencing acidosis or alkalosis? Answer: (a) ph log [H ] ph log -.0 ph log ph. (b) The normal blood ph range for adults is between. and. (See pages and.). In this case the blood ph is below this range, therefore the patient is eperiencing acidosis. NSSAL Draft 0 C. D. Pilmer
Eample The blood ph for a year old child is.. The child has also been vomiting for the last two days and as a result can not keep down food or fluids. (a) Based on the blood ph levels and the patient's symptom, what can one conclude? (b) Determine the concentration of hydrogen ions in moles per litre in this patient's blood. Answer: (a) The normal blood ph range for children (< years) is between. and. (See pages and.). In this case, the ph reading is higher than this meaning the patient is eperiencing alkalosis. The prolonged vomiting indicates that we are dealing with metabolic alkalosis, rather than respiratory alkalosis (Again see pages and.). (b) ph log [H ]. log [H ]. log0 [H ] 0. [H ]. [H ] 0 [ H ]. 0 [H ].00 moles per litre (It can also be written as 0.00000000 mol/l.) Questions:. The hydrogen ion concentrations for different solutions have been provided below. Determine the corresponding ph for each. (a) vinegar:. 0 mol/l (b) window cleaner:. 0 mol/l NSSAL Draft 0 C. D. Pilmer
(c) cider:.0 0 mol/l (d) urine:. 0 mol/l. The ph readings of various solutions have been provided below. Determine the corresponding hydrogen ion concentrations in moles per litre. (a) saliva: ph. (b) sea water: ph. (c) buttermilk: ph. (d) wine: ph NSSAL Draft 0 C. D. Pilmer
. The blood ph for an adult male is.0. The patient's breathing is slow and shallow.. (a) Based on the blood ph levels and the patient's symptom, what can one conclude? (b) Determine the concentration of hydrogen ions in moles per litre in this patient's blood.. The blood for one child patient (< years) is showing hydrogen ion concentrations of. 0 moles per litre. (a) Determine the ph for this patient's blood. (b) Is this patient's blood in the normal ph range or is he/she eperiencing acidosis or alkalosis?. The blood ph for a year old female is.. The child has also been having severe diarrhea over the last three days. (a) Based on the blood ph levels and the patient's symptom, what can one conclude? (b) Determine the concentration of hydrogen ions in moles per litre in this patient's blood. NSSAL Draft 0 C. D. Pilmer
. The blood for an adult female is showing hydrogen ion concentrations of. 0 moles per litre. (a) Determine the ph for this patient's blood. (b) Based on the blood ph readings and that the patient suffers from aniety attacks that result in prolonged bouts of rapid deep breathing, what can one conclude?. Complete the following ph Table (i.e. fill in the hydrogen ion concentrations). Initially you might think that you have to do lots of calculations to complete this table, but after you have done a few, you should be able to see a pattern that will allow you to quickly complete the table. ph Hydrogen Ion Eamples Concentration (in moles per litre) Liquid Drain Cleaner Bleach Soapy Water Ammonia 0 Milk of Magnesia Baking Soda Egg Whites Pure Water Milk Black Coffee Tomato Juice Orange Juice Lemon Juice Stomach Acid 0 Battery Acid. Look at your completed ph table from the previous question and answer the following. (a) As the ph increases by one, the hydrogen ion concentration decreases by a factor of. (b) As the ph decreases by one, the hydrogen ion concentration increases by a factor of. NSSAL Draft 0 C. D. Pilmer
Logarithms and poh Up to this point, we have talked about using the concentration of hydrogen ions as a means of describing the acidity or alkalinity of a solution. This, however, is not the complete story. There is another ion, the hydroide ion, which we must consider. The obvious question is what is the relationship between the two ions (hydrogen ions and hydroide ions)? If you were to eamine pure water, you would probably think that is totally comprised of molecules of H O. However, approimately two out of every billion water molecules ionize to form hydrogen ions (H+) and hydroide ions (OH-). H O (l) H + (aq) + OH - (aq) So for pure water, it is predominately comprised of H O molecules but there is a very small concentration ( 0 moles per litre) of H +, and a very small concentration ( 0 moles per litre) of OH -. Notice that for pure water the concentration of H + and OH - are equal and that when multiplied give 0 moles per litre. This special relationship leads to what is referred to as the ionization constant for water, K w. K w = [H + ][OH - ] K w = 0 0 K w = 0 moles per litre If you were asked to find the ph of pure water, you would calculate the following. ph log [H ] ph log - 0 ph log 0 ph poh is defined as the logarithm (base 0) of the reciprocal of the concentration of the hydroide ions, measured in moles per litre, in a solution. We can use this formula to calculate poh of pure water. poh log Note: [OH - ] represents the concentration of hydroide ions [OH ] Please Note: poh log - 0 Many chemistry tetbooks define poh using the following poh log 0 formula. poh poh log [OH ] This formula is equivalent to the one we will be using. Please note that for pure water the ph + poh =. NSSAL Draft 0 C. D. Pilmer
Add an Acidic Solution to Pure Water So what happens when an acidic solution is added to pure water? For eample, when hydrochloric acid (HCL) is added to pure water, it increases the amount of H + in the new solution while the amount of OH - remains the same. However, the concentration of H + multiplied by the concentration of the OH - remains constant at 0 moles per litre. [H + ][OH - ] = 0 Initially, one might think that these two statements do not seem to align. How can one increase the amount of one ion (H + ), keep the amount of the other ion (OH - ) the same, yet end up with the same constant 0 when the concentrations are multiplied? The key to understanding this is recognizing the difference between the amount of an ion (measured in moles) and the concentration of an ion (measured in moles per litre). Consider the following situation. Let's say that we have one litre of pure water. This water would have 0 moles of hydrogen ions and 0 moles of hydroide ions. 0 moles of H + litre of pure water 0 moles of OH - Suppose we add one litre of a specific acidic solution to the pure water such that the amount of hydrogen ions will change from 0 moles to 0 moles. Adding an acidic solution to the water will not change the amount of hydroide ions, therefore they will remain at 0 moles. 0 moles of H + litres of solution 0 moles of OH - For this new two litre solution, let's look at the ions as concentrations (moles per litre) rather than amounts (moles) For the hydrogen ions (H+) we have: 0 moles per litres NSSAL 0 Draft 0 C. D. Pilmer
This can be simplified to: For the hydroide ions (OH - ) we have: This can be simplified to: 0 moles per litres 0 moles per litres 0. 0 moles per litre (or 0 moles per litre) Now let's see if the products of the two ion concentrations equal 0 as stated earlier. [H + ][OH - ] 0 0.0 0 It works! Now let's eamine the ph and poh for this litre solution. ph log poh log [H ] [OH ] ph log - poh log - 0 0.0 ph log000000 poh log0000000 ph. poh. Notice that the sum of the ph and poh is. (i.e.. +. = ) Add a Basic Solution to Pure Water So what happens when a basic solution is added to pure water? For eample, when sodium hydroide (NaOH) is added to pure water, it increases the amount of OH - in the new solution while the amount of H + remains the same. However, the concentration of H + multiplied by the concentration of the OH - remains constant at 0 moles per litre. [H + ][OH - ] = 0 The logic behind this is very similar to that discussed when we looked at adding an acidic solution to pure water. Similarly there is a relationship between the ph and poh ph + poh = NSSAL Draft 0 C. D. Pilmer
Questions. Complete the following table (i.e. complete the last two columns). ph Hydrogen Ion Concentration (mol/l) [H + ] 0 0 Eamples Hydroide Ion Concentration (mol/l) [OH - ] 0 0 0 0 Liquid Drain Cleaner or 0 Bleach 0 Soapy Water 0 0 Ammonia 0 0 Milk of Magnesia 0 0 Baking Soda 0 0 Egg Whites 0 0 Pure Water 0 0 Milk 0 0 Black Coffee 0 0 Tomato Juice 0 0 0 Orange Juice 0 0 Lemon Juice 0 0 Stomach Acid 0 0 0 or 0 Battery Acid 0 poh [H + ][OH - ]. Look at your completed table from the previous question and answer the following. (a) As the poh increases by one, the hydroide ion concentration decreases by a factor of. (b) As the poh decreases by one, the hydroide ion concentration increases by a factor of. (c) What can you say about the ph and poh of the same solutions? (d) What can you say about the hydrogen ion concentration and hydroide ion concentration for the same solution? NSSAL Draft 0 C. D. Pilmer
More poh In this section, we will apply the four formulas we have learned. ph log poh log [H + ][OH - ] = 0 ph + poh = [H ] [OH ] Eample What is the ph of a solution that has a poh of.? Are we dealing with an acidic, basic or neutral solution? Answer: ph + poh = ph +. = ph = -. ph =. When the ph is greater than (or the corresponding poh is less than ), then we are dealing with a basic solution. Eample What is the hydrogen ion concentration for a solution that has a hydroide ion concentration of. 0 moles per litre? Answer: [H + ][OH - ] = 0-0 H - OH - 0 H.0 H.0 moles per litre Eample What is the poh of a solution with a hydroide ion concentration of Answer: poh log [OH ] poh log -.0 poh log poh.. 0 moles per litre? NSSAL Draft 0 C. D. Pilmer
Eample What is the hydroide ion concentration in a solution having a poh of.? Answer: poh log [OH ]. log [OH ]. log0 [OH ] It was changed from logarithmic to eponential form. 0. [OH ]. [OH ] 0 [ OH ]. 0 0 [OH ].0 moles per litre Eample What is the poh of a solution having [H + ] =. Answer: Method A: First find the ph and then find the poh. ph log [H ] ph log ph + poh = -.0. + poh = ph log0 poh = -. poh =. ph. Method B: First find the [OH - ] and then find the poh. 0 moles per litre? [H + ][OH - ] = 0 - - 0 OH H - - 0 OH.0 - OH.0 poh log [OH ] poh log -.0 poh log0000 poh. NSSAL Draft 0 C. D. Pilmer
Eample : If the ph of a solution is., determine the concentration of hydroide ions in moles per litre. Answer: Method A: First find the poh and then find [OH - ] ph + poh =. + poh = poh = -. poh =. poh log [OH ]. log [OH ]. log0 [OH ] It was changed from logarithmic to eponential form. 0. [OH ]. [OH ] 0 [ OH ]. 0 [OH ].0 moles per litre Method B: First find [H + ] and then find [OH - ]. ph log [H ]. log0 [H ] It was changed from logarithmic to eponential form. 0. [H ]. [H ] 0 [ H ]. 0 0 [H ].0 moles per litre [H + ][OH - ] = 0 - - 0 OH H NSSAL Draft 0 C. D. Pilmer
- - 0 OH 0.0 - OH.0 moles per litre Questions. If the ph of saliva is., find its poh.. What is the poh of a solution which has a hydroide ion concentration of. 0 moles per litre?. If the hydrogen ion concentration of milk is. 0 moles per litre, determine the concentration of hydroide ions in moles per litre.. What is the ph of a solution which has a hydrogen ion concentration of litre?. 0 moles per NSSAL Draft 0 C. D. Pilmer
. A certain hair conditioner has a poh of.. What is the hydroide ion concentration in that conditioner?. If the poh of a glass of lemonade is., what is the corresponding ph?. If the hydroide ion concentration of a solution is concentration of hydrogen ions in moles per litre.. 0 moles per litre, determine the. A solution has a ph of.. What is the hydroide ion concentration in that solution? NSSAL Draft 0 C. D. Pilmer
. If the poh of a solution is., determine the concentration of hydrogen ions in moles per litre. 0. What is the ph of a solution having [OH - ] =. 0 moles per litre?. Determine the concentration of hydroide ions in a solution that has a ph of.. NSSAL Draft 0 C. D. Pilmer
. Determine the poh of a solution that has a hydrogen ion concentration of per litre? Are we dealing with an acidic, basic, or neutral solution?. 0 moles. The hydrogen ion concentration in an adult blood sample in. 0 moles per litre. (a) What is the ph of this blood sample? (b) Is this patient's blood in the normal ph range or is he/she eperiencing acidosis or alkalosis? (c) What is the poh of this blood sample? (d) What is the hydroide ion concentration in this blood sample? NSSAL Draft 0 C. D. Pilmer
Putting It Together Questions. Simplify each of the following epressions. (a) (b) b a b a (c) 0 n (d) c d c d 0 n (e) h (f) p q (g) c d (h) y z m n (i) m n 0 a b a b (j) ab. Evaluate each of the following. Do not use a calculator. (a) (c) (b) 0 (d) 00 (e) 0 (f) (g) (h) (i) 0 (j) (k) (l) NSSAL 0 Draft 0 C. D. Pilmer
(m) (n) (o) (p) (q) (r). Simplify each of the following epressions. (a) (b) a b a b c m n (c) (d) c m n (e) p (f) y (g) p q (h) ab c. Convert the following numbers in scientific notation to standard form. (a). 0 (b). 0 (c). 0 (d) 0. Convert the following numbers in standard form to scientific notation. (a) 00 000 (b) 00 (c) 0.000 (d) 0.00 NSSAL Draft 0 C. D. Pilmer
. Order these numbers from smallest to largest. (a). 0, 0, 0,. 0 (b). 0, 0,. 0,.0 0. Complete the operations. Write your answer in scientific notation. (a). 0.0. 0. 0 (b) (c). 0. 0 (d). 0. 0 (e). 0. 0 (f).0. 0 (g). 0. 0 (h). 0. 0 NSSAL Draft 0 C. D. Pilmer
0 0 (i) 0 0. 0 (j) 0. Change the following from eponential to logarithmic form. (a) (b). Change the following from logarithmic to eponential form. (a) log (b) log 0. Evaluate each of the following logarithms. (a) log (b) log (c) log (d) log log (f) log 0000 (e) NSSAL Draft 0 C. D. Pilmer
log (h) log (g). Solve for. (a) log (b) log (c) log (d) log (e) log (f) log (g) log (h) log (i). log (j) log NSSAL Draft 0 C. D. Pilmer
. The blood for a female child ( years old) is showing hydrogen ion concentrations of. 0 moles per litre. (a) Determine the ph for this patient's blood. (b) Based on the blood ph readings and that the patient has been vomiting repeatedly, what can one conclude?. The blood ph for a year old male is.0. The adult's breathing has been slow and shallow for the last five hours. (a) Based on the blood ph levels and the patient's symptom, what can one conclude? (b) Determine the concentration of hydrogen ions in moles per litre in this patient's blood.. If the poh of a solution is., determine the concentration of hydrogen ions in moles per litre. NSSAL Draft 0 C. D. Pilmer
. What is the ph of a solution having [OH - ] =. 0 moles per litre?. The hydroide ion concentration in an adult blood sample in. 0 moles per litre. (a) What is the poh of this blood sample? (b) What is the ph of this blood sample? (c) What is the hydrogen ion concentration in this blood sample? (d) Is this patient's blood in the normal ph range or is he/she eperiencing acidosis or alkalosis? NSSAL Draft 0 C. D. Pilmer
Post-Unit Reflections What is the most valuable or important thing you learned in this unit? What part did you find most interesting or enjoyable? What was the most challenging part, and how did you respond to this challenge? How did you feel about this math topic when you started this unit? How do you feel about this math topic now? Of the skills you used in this unit, which is your strongest skill? What skill(s) do you feel you need to improve, and how will you improve them? How does what you learned in this unit fit with your personal goals? NSSAL Draft 0 C. D. Pilmer
Answers Introduction to Eponents (pages to ). (a) (b) (c) (d) (e) (f) (g) 0 (h) (i) (j). (a) (b) a b (c) (d) c d (e) y (f) p q (g) h (h) y (i) a (j) c 0 d (k) k (l) a b c (m) (n) d y a b (o) (p) y c (q) c d (r) p q y (s) (t) a b (u) y 0 (v) y (w) p q () d (y) a b (z) y. (a) (b) (c) (d) (e) (f) (g) (h) More Eponents (pages to ). (a) (b) (c) (d) (e) (f) (g) (h) NSSAL Draft 0 C. D. Pilmer
(i) (k) (m) (j) (l) (n) - (o) (p) (q) (s). (a) p (c) h (r) (t) (b) (d) b d. (a) (b) 0 (c) - (d) (e) (f) (g) (h). (a) (c) (e) (g) (i) (i) (j) - (b) 0 (d) h b (f) k a b (h) c p q (j) a b y (l) (k) 0 c 0 r s (m) c d (n) 0 t (o) g h (p) p q r t or 0 Scientific Notation (pages to ). (a) (c) (e) 0 (b) 0 (d) 0 (f) 0 0 0 NSSAL Draft 0 C. D. Pilmer
(g) (i). 0 (h). 0 (j). 0. 0. (a) 0 000 (b) 0 000 000 (c) 0.00 (d) 0.00000 (e) 00 (f) 0.0000 (g) 0. (h) 0 000 (i) 0 000 000 (j) 0.0..0 g. Less than. (a) 0.000, 0.00,., 00 (b) 0.00, 0.000,.00,.0 (c). 0, 0,. 0 (d) 0,. 0,. 0 (e). 0,. 0, 0 (f). 0, 0,. 0,. 0 (g). 0, 0,. 0,. 0 (h). 0,. 0, 0,. 0 More Scientific Notation (pages 0 to ) Many of the answers have been rounded off.. (a) (c) (e) (g). 0 (b). 0 (d).0 0 (f). 0 (h). 0. 0. 0.0 0. (a) (c) (e). (a) (c) (e). (a) (c) (e) (g) 0 (b). 0 (d). 0 (f) 0 (b). 0 (d). 0 (f). 0 (b). 0 (d). 0 (f). 0 (h). 0. 0.0 0 0. 0. 0. 0. 0. 0. 0 NSSAL 0 Draft 0 C. D. Pilmer
(i). 0 (j) (k). 0 (l) (m) 0 (n) (o). 0 (p) 0. 0. 0. 0 Logarithms (pages to ). (a) log (b) log (c) log (d) log (e) log (f) log 0. (a) (b) (c) (d) (e) (f) 0 0. 00 (g) (h) 0 0000. (a) (b) (c) (d) - (e) - (f) (g) (h) (i) -0. (k) (m) (j) (l).0 (n) no solution. (a) (b) (c) (d) (e) (f) (g) (h) (i) 00 000 (j) 0. (k) (l). (m)- (n) NSSAL Draft 0 C. D. Pilmer
(o) (p) 0.000 Logarithms and ph (pages to ). (a). (b) 0. (c). (d).. (a) (c) 0 moles per litre (b). 0 moles per litre (d) 0 moles per litre. 0 moles per litre. (a) respiratory acidosis (b). 0 moles per litre. (a). (b) normal blood ph range. (a) metabolic acidosis (b). 0 moles per litre. (a). (b) respiratory alkalosis. ph Hydrogen Ion Eamples Concentration (in moles per litre) 0 Liquid Drain Cleaner 0 Bleach 0 Soapy Water 0 Ammonia 0 0 0 Milk of Magnesia 0 Baking Soda 0 Egg Whites 0 Pure Water 0 Milk 0 Black Coffee 0 Tomato Juice 0 Orange Juice 0 Lemon Juice 0 Stomach Acid 0 0 or 0 Battery Acid. (a) 0 (b) 0 NSSAL Draft 0 C. D. Pilmer
Logarithms and poh (pages to ). ph Hydrogen Ion Concentration (mol/l) [H + ] 0 0 Eamples Hydroide Ion Concentration (mol/l) [OH - ] poh [H + ][OH - ] 0 0 0 0 0 Liquid Drain Cleaner or 0 Bleach 0 0 0 Soapy Water 0 0 0 Ammonia 0 0 0 Milk of Magnesia 0 0 0 Baking Soda 0 0 0 Egg Whites 0 0 0 Pure Water 0 0 0 Milk 0 0 0 Black Coffee 0 0 0 Tomato Juice 0 0 0 0 0 Orange Juice 0 0 0 Lemon Juice 0 0 0 Stomach Acid 0 0 0 0 or 0 Battery Acid 0 0. (a) 0 (b) 0 (c) When we add the ph and poh for the same solution we get. For eample, Tomato Juice: ph =, poh = 0, and ph + poh = (d) The product of the hydrogen ion concentration and the hydroide ion concentration for the same solution is 0 moles per litre. For eample, Tomato Juice: [H + ] = 0 moles per litre [OH - 0 ] = 0 moles per litre [H + ][OH - ] = 0 moles per litre More poh (pages to )...... 0 moles per litre. 0. NSSAL Draft 0 C. D. Pilmer
.. 0 moles per litre........0 0 moles per litre 0 moles per litre 0 moles per litre 0.... 0 moles per litre.., basic solution. (a). (b) normal (c). (d). 0 moles per litre Putting It Together (pages 0 to ). (a) (c) (e) a b (b) 0n (d) c d h 0 (f) p q c y (g) (h) 0 d z (i) mn (j) a b. (a) (c) (b) (d) 0 (e) (f) (g) (h) (i) (j) (k) (l) NSSAL Draft 0 C. D. Pilmer
(m) (o) (n) (p) (q) (r). (a) (b) a (c) c (d) m n (e) p (f) 0 y p a b (g) (h) 0 q c b. (a) 00 000 (b) 0.00 (c) 00 (d) 0.0000. (a) (c). (a) (b). (a) (c) (e) (g) (i). 0 (b). 0 (d). 0, 0,. 0, 0. 0, 0,. 0,.0 0. 0 (b). 0 (d). 0 (f). 0 (h) 0 (j). 0 0. 0. 0. 0. 0 0. (a) log (b) log. (a) (b) 0. (a) (b) (c) - (d) (e) (f) (g) (h) -. (a) (b) NSSAL Draft 0 C. D. Pilmer
(c) 00 (e) (g) (d) (f) 0 (h). (i). (j). (a). (b) metabolic alkalosis. (a) respiratory acidosis (b). 0 moles per litre..0 0 moles per litre... (a). (b). (c). 0 moles per litre (d) alkalosis NSSAL Draft 0 C. D. Pilmer