Topic 18 Acids and Bases Exercises

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1 Topic 18 Acids and Bases 18.1 Exercises 1. Define: (a) ph The negative log of the hydrogen ion concentration in a solution. i.e. ph = log[h 3 O + ] (b) poh The negative log of hydroxide ion concentration in a solution. i.e. poh = log[oh ] (c) K w The ion-product constant for water, i.e. K w = [H + ][OH ] = (d) pk w The negative log of the ion-product constant for water, i.e. pk w = 14 (e) K a The ionisation constant for a weak acid, K a = _ [H + ] [A ] _ [HA] (f) pk a The negative log of the ionisation constant for a weak acid, i.e. pk a = log K a (g) K b The ionisation constant for a weak base, K b for NH 3 for example =_ [NH + 4 ][OH ]_ [NH 3 ] (h) pk b The negative log of the ionisation constant for a weak base i.e. pk b = log K b 2. Why is the concentration of water omitted from the expression for K w? Water ionises only very slightly: H 2 O H + + OH. At 29 8 K the [H + ] = [OH ] = 10-7 mol dm -3. The concentration of H 2 O before ionisation equals 55.6 mol dm -3. (1 dm 3 H 2 O has a mass of 1000 g and the Mr mass H 2 O = 18 g / 18 = 55.6.) After ionisation [H 2 O] = = 55.6 mol dm -3. During a reaction, the change in concentration of water is negligible compared to the other reacting species and is hence effectively constant. Therefore it is not necessary to include. Some texts give a value of 1 to [H 2 O] in dilute solutions signifying that it has no effect on the calculations used in determining K and pk values 1

2 3. A student gave this expression for Ka for the weak acid HA: K a = [H + ] 2 / [HA] Discuss the correctness or otherwise of this expression. This expression is only true for monoprotic acid dissociation examples, i.e. when [H + ] = [A - ] 4. Enter the missing numbers and words: Pure water has a ph = 7 at 298 K. The ionization of water is an endothermic reaction so increasing the temperature will increase K w, in accordance with Le Chatelier s principle. The ph of pure water will decrease as it is heated. 5. Calculate the ph values at 298 K of the following solutions: (a) mol dm -3 hydrochloric acid HCl H + + [Cl ], strong acid, therefore fully ioised. Therefore, [HCl] = [H + ] = 1 x 10 2 M Therefore, ph = log 10 2 = 2 (b) 0.10 mol HC1 in 250 cm 3 of aqueous solution [HCl] = 0.10/0.25 = 0.40 mol dm 3 Therefore [H + ] = 0.40 Therefore ph = log 4.0 x 10 1 = = 0.40 = 4.0 x 10-1 (c) 1.0 x 10-1 mol dm -3 ethanoic acid, Ka = 1.8 x 10-5 CH 3 COOH + H 2 O H 3 O + + CH 3 COO Ka = _[H 3 O + ][CH 3 COO ]_ = x 2-5 = 1.8 x 10 [CH 3 COOH] 1.0 x 10 1 Therefore x 2 = 1.8 x 10 6 Therefore x = 1.8 x 10 3 = 1.3 x 10 3 ph = log(1.3 x 10 3 ) = 2.89 = 2.9 (d) M NaOH NaOH strong base, therefore, [OH ] = 2.0 x 10 3 M Therefore poh = log (2 x 10 3 ) = 2.7 Therefore ph = =

3 (e) 0.10 mol HCO - 3 acid in 100 cm 3 of solution, K a = HCO 3 H + + CO 2 3 K a = _[H + ][CO 2 3 ]_ [HCO 3 ] Therefore, x 2 = 10 1 x = Therefore, x = Therefore, ph = log = 5.6 (f) M Ca(OH) 2 Ca(OH) 2 Ca OH Therefore, 0.050M Ca(OH) 2 gives 0.10M of OH Therefore, [OH ] = 10 1 Therefore, poh = 1 Therefore, ph = 13 (g) 1.0 M of HA, Ka = 10-5 Ka = _[H + ][A ]_ = _x 2 _ = 10 6 [HA] 1 Therefore, x 2 = 10 6 Therefore, x = 10 3 Therefore, ph = log10 3 = 3 6. The first ionization of H 2 SO 4 is essentially complete in dilute aqueous solution, but the - second ionisation of HSO 4 to release the second proton is very weak. (a) Account for this difference. The first ionisation is for a strong acid, therefore complete (almost). H 2 O + H 2 SO 4 H + + HSO 4. The equilibrium favours the weaker side of the equilibrium H 2 O + HSO 4 H + (aq) + SO 4 (aq). H + is far stronger an acid than is HSO 4 so the reverse reaction is favoured. - (b) Write an expression for Ka for HSO 4. K a = _ [H + ][SO 2 4 ]_ [HSO 4 ] (c) Suggest possible Ka values for each ionization: (i) for H 2 SO 4 Very high (strong acid) say

4 - (ii) for HSO 4 Very low (weak acid) say The acid constant, Ka, for a weak monoprotic acid, HA, is 1 x (a) What is meant by monoprotic? Only one proton from each molecule is donated during protolysis, e.g. HA + H 2 O H + (aq) + A (aq) (b) What is the ph of a 0.05 M solution of HA? K a = _ [H + ][A ]_ = x 2 = 1 x 10-8 [HA] 5 x 10 2 Therefore, x 2 = 5 x Therefore, x = 5 x 10 5 = 2.2 x 10 5 = log(2.24 x 10 5 ) = = (a) Given the Ka for acid A, ethanoic acid, is 1.8 x 10-5 mol dm -3 calculate the ph of a 0.20 M solution. Say HA H + + A, therefore, 1.8 x 10 5 = _ [H + ][A ]_ HA Since [H + ] = [A ] let [H + ] = x which means 1.8 x 10 5 = x 2 2 x 10 1 Therefore, x 2 = 2 x 1.8 x 10 6 Therefore, x = 1.8 x 10 3 = 1.34 x 10 3 Therefore, ph = (log1.34 x 10 3 ) = 2.9 (b) A 0.10 M solution of the weak acid B, HA, has a ph of 3.5. Calculate the value for the Ka of this acid. Say HB H + + B, therefore, K a = _ [H + ][B ]_ [HB] = _ x _ 1 x 10 1 K a = 1 x 10 6 mol dm 3 4

5 (c) A solution of acid C with a Ka value of 5 x 10-5 mol dm -3 has a ph of 4.0. What is the concentration of this acid? Say HC H + + C, therefore, Ka = _ [H + ][C ]_ [HC] For ph 4.0, [H + ] = Therefore, 5 x 10 5 = (10 4 x 10 4 ) / [HC] Therefore, [HC] = 10 8 /5 x 10 5 = 2 x 10 4 M (d) Arrange acids A, B and C above in increasing order of acid strength, and explain how you arrived at your answer. B < A < C Ka gives acid strength, not ph. 1 x 10 6 < 1.8 x 10 5 < 5 x In a solution that has a concentration of H + (aq) = 2 x 10-4 calculate: (a) The OH - concentration [H + ][OH ] = 10 14, therefore, [OH ] = _ _ 2 x 10 4 = 5 x mol dm 3 (b) The ph ph = log(2 x 10 4 ) = A saturated solution of barium hydroxide contains 40 g/dm 3 of Ba(OH) 2 8H 2 O. Calculate the ph of this solution. Convert the concentration given into mol dm -3 by dividing 40 g by the molar mass which is (the 8 H 2 O must be included), i.e g/mol. The concentration of the Ba(OH) 2 solution is M. Assuming complete dissociation and considering the equation: Ba(OH) 2 8H 2 O Ba OH - + 8H 2 O The concentrations are [Ba 2+ ] = M and because it is 2OH -, [OH - ] = M. So poh = - log [OH - ] = - log 2.54 x10-1 = ph + poh = 14, therefore ph = = An acid found in coconut oil has the formula HC 6 H 11 O 2. The ph of a solution of this acid, of concentration of 11.0 g/dm 3 is What is its Ka value? State any assumptions made in your calculation. 5

6 Convert the concentration (which we will assume is the initial concentration) given into mol dm -3 By dividing the mass, 11.0 g, by the molar mass, which is 116 g/mol. This gives the initial concentration of the coconut oil as M. Calculate the concentration of hydrogen ions by considering the ph. i.e. [H + ] = 10 -ph = = x 10-3 M. This H + came from the dissociation of the coconut oil. The concentration from the ionisation of water can be ignored. Therefore, the concentration of the coconut oil at equilibrium is x 10-3 = M. Another assumption could be to ignore this change in [HC 6 H 11 O 2 ], leaving it as By considering the equation: HC 6 H 11 O 2 H + + C 6 H 11 O 2 we see that [H + ] = [C 6 H 11 O 2 ] and, therefore, K a = [H + ] 2 [HC 6 H 11 O 2 ] Which gives K a = 1.40 x 10-5 Acids and bases are to be found in so many of the things we use and eat. Rain water, car batteries, our stomachs soft drinks and alcoholic beverages contain acids, while washing powders, dish washing liquids, oven cleaners, paint strippers contain bases. Even examiners enjoy writing questions about acid-base equilibria! 6

7 18.2 Exercises 1. Supply the missing words: Buffers are solutions containing a weak acid and a salt of this acid. Such a solution is able to resist changes to its ph. They contain H + (aq) in an equilibrium situation and if more H + (aq) is added the equilibrium will move to offset this change. If H + (aq) is removed by adding an alkali, more H + (aq) forms to offset this change. This is an example of whose principle? Le Chatelier s. 2. How are buffer solutions prepared? Give an example of a buffer resistant to changes in ph greater than 7. By mixing a solution of a weak acid, or a weak base, with one of its salts. The acid (or base) concentrations must be reasonably high. The mixture NH 3 (aq) with NH 4 Cl(aq) will resist changes to increase its ph. The added OH - + reacts with the NH 4 from the NH 4 Cl to form NH 3 (aq) + H 2 O. Buffers maximise their action when the concentrations of the weak acid and its conjugate base are kept large and approximately equal. The buffer range, i.e. the ph range over which a buffer is effective, can be calculated from this equation: ph = pka + log {[conjugate base] / [acid]} As [conjugate base] = [acid] in a buffer these cancel out in the above equation so making the effective range for buffers dependent on their pka, or pkb, values. For ethanoic acid / sodium ethanoate buffers the effective range is between ph 3.7 to 5.7 as pka for ethanoic acid = 4.7. For a weak base / conjugate acid buffer, e.g. NH 3 / NH 4 Cl, pka applies the conjugate acid of the weak base, here the NH + 4. pka for NH + 4 = 9.2, therefore best operating range will be between ph 8.2 to Discuss with your teacher whether or not this extension should be studied. 3. An aqueous solution, 0.10 M with respect to both CH 3 COOH and NaCH 3 COO, is called a buffer solution. (a) Explain what is meant by a buffer solution. A solution that is able to resist any changes to their ph. (b) Explain the effect on the ph of this solution by adding to it small quantities of HCl (aq) or NaOH (aq). The addition of HCl will combine with the CH 3 COO and Na + to form CH 3 COOH and NaCl. The ph 7

8 is not significantly changed. The added NaOH will react with the acid, CH 3 COOH, resulting in the formation of CH 3 COO. In both cases the concentration of H+ and OH - remains relatively stable. (c) Given Ka for ethanoic acid = 1.7 x 10-5 calculate the ph of the above solution. [H + ] = K a x [HA] / [A - ] [H + ] = 1.7 x 10 5 [CH 3 COOH]/ [CH 3 COO ] = 1.7 x 10 5 Therefore, ph = log (1.7 x 10 5 ) = Blood is an example of a buffered solution. Both the H 2 CO 3 /HCO - 3 and the H 2 PO /HPO 4 buffers help to maintain the ph of blood at about (a) Write the Ka expression for each. H 2 CO 3 H + + HCO 3 K a = [H + ][HCO 3 ] / [H 2 CO 3 ] H 2 PO 4 H + + HPO 2 4 Therefore, K a = [H + ][HPO 2 4 ] / [H 2 PO 4 ] (b) Calculate the [H + (aq)] of blood. ph = 7.4, therefore [H + ] = moldm 3 = 3.98 x 10 8 = 4.0 x 10 8 M (2 sig fiqs) (c) Acidosis is the medical term to describe a decrease in the ph of blood. If blood ph falls below 6.8 the result could be fatal. Calculate the [H + (aq)] in blood with a ph of 6.8. How many times more concentrated with H + (aq) is this blood compared to normal blood? ph = 6.8, therefore, [H + ] = = 1.58 x 10 7 = 1.6 x 10 7 M 5. The weak monoprotic acid, HA, has a Ka value of 1 x (a) Calculate the ph of a 0.01 M solution of this acid. Ka = [H + ][A ] / HA = 1 x 10 6 = x 2 / 1 x 10 2 Therefore, x 2 = 10 8, therefore, x = 10 4 Therefore, ph = 4 (one sig. fig) (b) What would be the ph of a solution 0.01 M with respect to this acid, and also 0.01 M with respect to the sodium salt of this acid, NaA? 8

9 [H + ] = K a x [HA] / [A ] = 1 x 10 6 [0.01] / [0.01] = 1 x 10 6 Therefore, ph = 6 (one siq. fig) (c) What assumptions are made when calculating for answers in (a) and (b)? You assume the concentrations given are initial concentrations and are not significantly changed during protolysis. A reasonable assumption because they are weak acids 6. Diluting a buffer solution with water will not change its ph. Why is this? Consider the equation [H + ] = K a x [HA] / [A ] K a remains a constant and the concentration of both [HA] and [A ] reduce. However since they both reduce by the same amount the ratio of the two remains constant, hence [H + ] remains constant and therefore the ph does not change. 9

10 18.3 Exercises 1. Explain why an aqueous solution of Na 2 CO 3 is alkaline. CO 2 3 is a weak base. It can attract a few H + from the weak acid, water, producing some OH, hence forming a slightly alkaline solution. CO H 2 O HCO 3 + OH 2. Use the example in question 1 above to explain the following terms: (a) salt hydrolysis A salt, e.g. Na 2 CO 3, reacting with water (hydrolysis) to produce a change in ph. In this case it is alkaline due to OH present. (b) acid/conjugate base In the reaction: CO 2 3 (aq) + H 2 O HCO 3 (aq) + OH H 2 O is the acid and its conjugate base, OH, forms from the loss of a proton. An acid / conjugate base differ by a proton. (c) base/conjugate acid CO 2 3 accepts a proton (it is a base) to form its conjugate acid HCO 3. A base / conjugate acid differ by a proton 2- (d) Write an expression for Kb for CO 3 CO 3 2 K b + H 2 O HCO 3 + OH = _ [HCO 3 ][OH ] [CO 2 3 ] 3. Explain why a solution of NH 4 Cl (aq) is acidic. In the reaction: NH + 4 (aq) + H 2 O NH 3 (aq) + H 3 O + (aq). The weak acid NH + 4 (aq) can donate a proton, hence the [H 3 O + ] increases resulting in an acidic solution. 4. Write an expression for Ka for: + (a) NH 4 (aq) NH H 2 O NH 3 H 3 O + 10

11 Ka = [NH 3 ][H 3 O + ] [NH + 4 ] (b) [Fe (H 2 O) 6 ] 3+ (aq) Fe 3+ (aq), like Al 3+ (aq), reacts as follows ; [Fe(H 2 O 6 )] 3+ + H 2 O [Fe OH(H 2 O) 5 ] 2+ + H 3 O + Therefore, Ka = [(Fe OH (H 2 O) 5 ) 2+ ][H 3 O + ] [(Fe OH(H 2 O) 5 ) 3+ ] 5. Solutions of salts that have highly charged ions, e.g. Al 3+ (aq), are acidic. Explain, with an equation, why this is so. They are acidic because of hydrolysis. The high charge on Al 3+, Fe 3+ and Cr 3+ ions attracts electrons from the surrounding water molecules, making it easier for a water molecule to lose a proton, H +. Protons are readily accepted by H 2 O. The reaction is: Al(H 2 O) H 2 O Al(H 2 O) 5 OH 2+ + H 3 O Would you expect solutions of Cr 3+ (aq) to be neutral, acidic or basic? Explain your answer, using an equation. Acidic, as the Cr 3+ ion attracts electrons making it easier for a bonded H 2 O molecule to lose a proton. 3+ Cr(H 2 O) 6 + H 2 O Cr(H 2 O) 5 OH 2+ + H 3 O + 7. When a solution of aluminium chloride is added to a solution of sodium hydrogencarbonate a vigorous reaction occurs. Explain, with the help of equations, what is happening. AlCl 3 contains Al (H 2 O) This undergoes salt hydrolysis producing H 3 O Al(H 2 O) 6 + H 2 O Al(H 2 O) 5 OH 2+ + H 3 O +. The acid hydrogen carbonate reaction produces CO 2 (g). HCO 3 (aq) + H 3 O + (aq) 2H 2 O + CO 2 (g) Think of hydrolysis as water splitting. The word comes from the Greek, hydro, water, and lusis, loosen. 11

12 18.4 Exercises 1. Some revision first. In volumetric analysis the following terms are used. Explain each. (a) First reading The reading from your burette at the beginning of a titration. (b) Final reading The reading taken at the end-point, i.e. when the indicator changes colour. (c) Titre The difference between the final reading and the first reading. (d) Titrant The solution in the burette. The substance you determine a titre for. (e) End-point The point when the indicator colour changes by the addition of 1 drop. (f) Equivalence point The point when the reactants are in stoichiometric proportions. (g) Indicator Are weak acids or bases used in titrations to determine end-points. How they work is conveyed in the next topic. (h) Standard solution a solution of accurately known concentration. (i) Meniscus The curved surface of a liquid in a tube. When titrating the bottom of the meniscus determines reading. (j) Neutralisation point In acid-base reactions when neither the acid nor base are in excess. (k) Inflexion point The point on the graph which has the maximum gradient. 2. The apparatus used in titration work includes a chemical balance, analytical grade reagents (AR), a volumetric flask, a volumetric pipette and a burette. For each explain their use and, for the glassware include how they are cleaned, and the points to note in their use. (a) The balance Determining mass accurately. Cleanliness essential. 12

13 (b) An analytical reagent Designated AR. A reagent of high purity. Used to prepare standard solutions. (c) Volumetric flask Use to measure an exact known volume of a standard solution. Rinse after use and before use with distilled water. (d) Volumetric pipette Used to accurately deliver a volume of a solution. Rinse with distilled water and drain, then rinse with solution to be used. Drain. Fill and deliver with tip against the flask. (e) Burette To deliver a measured volume of solution during a titration. Rinse with distilled water, then solution to be used. Ensure tip contains no air bubbles. 3. Draw and explain the titration curve showing ph against volume of titrant (in this case the alkali) for a titration using 0.10M HCl (aq) and 0.10M NaOH (aq). Indicate the equivalence point Strong acid strong base. Very little change occurs until close to equivalence point. Then 1 more drop of alkali can change ph from 3(+) to 12(+) shown by the inflexion point. This point of inflexion is the equivalence point. 4. Draw and explain the titration curve showing ph against volume of titrant (in this case the acid) for a titration using 0.10M HCl (aq) and 0.10M Na 2 CO 3 (aq). Indicate the equivalence point. 13

14 Strong acid, weak base. Gradual change as HCl(aq) added. At equivalence point rapid change from 9(-) to (a) Draw and explain the titration curve showing the ph against volume of alkaline titrant for a titration using 0.10 M CH 3 COOH and 0.10 NaOH. Indicate the equivalence point. Weak acid, strong base. Initial ph of acid 4, gradual change then at equivalence point rapid change over ph band 7(+) to 12(+). (b) Mark in the maximum buffering point on this graph and explain why you have chosen this point. Best buffering effect is when [CH 3 COO ] = [CH 3 COOH], i.e. halfway to neutralization. There is sufficient acid, CH 3 COOH, to offset addition of OH, and sufficient conjugate base, CH 3 COO, to offset addition of H +. (c) At what point will the pk a for ethanoic acid be equal to the ph of the mixture? Explain your reasoning. Consider [H + ] = K a x [CH 3 COO ] / [CH 3 COOH] If we multiply both sides by log we get -log [H + ] = log K a log [CH 3 COO ] / [CH 3 COOH] Which is equivalent to: 14

15 ph = pk a log [CH 3 COO ] / [CH 3 COOH] When [CH 3 COO ] = [CH 3 COOH], [CH 3 COO ] / [CH 3 COOH] = 1, therefore log [CH 3 COO ] / [CH 3 COOH] = 0, therefore ph = pk a The equation ph = pk a log [base] / [acid] is called the Henderson-Hasselbalch equation. Two chemists, with surnames beginning with H, became very interested in relationships involving ph! In writing ionization equations for weak acids and weak bases, e.g. HA(aq) H + (aq) + A (aq), the use of the is essential. Using will cost you marks. An experimental error? 15

16 18.5 Exercises 1. Explain how acid base indicators work. Indicators are weak organic acids or bases (Hln) with their ionised from (In ) being a different colour to that of the unionised form. In an acidic solution the Hln form (colour 1) dominates in alkalis the In (colour 2) dominates. Le Chatelier s principle. 2. Explain how the ph range for the colour change of an indicator relates to its pka value. Ka for indicator Hln = [H + ][In colour 2] [Hln colour 1] Mid-point colour occurs when [colour 2] = [colour 1]. At this point they cancel out in the expression, so making Ka = [H + ]. Therefore, pka = ph at the mid-point colour Increase ph, decease colour 2; decrease ph, increase colour The indicator, HIn, changes colour over the ph range of In which two of the following titrations would you use this indicator? A. strong acid strong base B. strong acid weak base C. weak acid strong base D. weak acid weak base Option A and C. If you had a problem with 3 above look at the appropriate titration curves. The equivalence point must cover, as a straight line, ph range < 7.9 to > A strong base must be involved for this, therefore A and C. 16

17 4. Do you recognise this compound? Yes, it is phenolphthalein. Circle the species present in an alkaline solution, and explain your choice. Alkaline solutions contain OH. This attracts an H + from an -OH group in the unionised form, now making it negative. This molecule, in an alkaline solution, would be red. HIn (colourless) + OH (aq) H 2 O + In (colour red) 5. The range and colour change for some common indicators is given. Indicator ph range Colour change Alizarin yellow yellow red orange Phenolphthalein colourless pink Bromothymol blue yellow blue Bromophenol blue yellow violet Methyl orange orange red orange yellow Use this information to draw titration curves for: (a) 0.10 M NaOH (aq) added to 25 cm M HCl (aq) (b) 0.10 M NaOH (aq) added to 25 cm M CH 3 COOH (aq) 17

18 (c) 0.10 M NH 3 (aq) added to 25 cm M HCl (aq) (d) 0.10 M NH 3 (aq) added to 25 cm M CH 3 COOH (aq) On your graph show the ph at equivalence point, a suitable indicator for the determination of the end point, an indicator not suitable for the determination of end point, and the volume of alkali added at equivalence point. (a) ph at equivalence point: range 3(-) 11(+) Volume of NaOH added 25.0cm 3 Suitable indicators: All indicators (b) 18

19 ph at equivalence point: range 8(-) 11(+) Volume of NaOH added 25.0cm 3 Suitable indicators: alizarin yellow or phenolphthalein Indicators not suitable: methyl orange, bromothymol blue, bromophenol blue (c) ph at equivalence point: range 4(-) 8(+) Volume of NH 3 added 25.0cm 3 Suitable indicators: Methyl orange, bromophenol blue, bromothymol blue Indicators not suitable: alizarin yellow, phenolphthalein (d) ph at equivalence point: range 6(-) 8(+) Suitable indicators: none 19

20 6. A ph titration was performed with the aid of a ph meter. 25 cm 3 of M HNO 3 (aq) was titrated against NaOH (aq). The titration curve obtained is shown. (a) Calculate the concentration of the NaOH (aq) HNO 3(aq) + NaOH (aq) NaHNO 3(aq) +H 2 O mol ratio = 1:1 Therefore, 25 x = 20.5 x? Therefore,? = 25 x 0.050/20.5 = M = 6.1 x 10 2 mol dm 3 (b) Suggest a suitable indicator that could be used for a visual end point determination. Methyl orange. 7. This photograph shows from left to right three different indicators in an acid, neutral and alkaline solution. Name each indicator From left to right its: Methyl Orange, Bromothymol blue, Phenolphtalein. Suggest the ph for the neutral solution of each indicator above. Now explain to a friend how you arrived at your answers. 20

21 Your Exam Questions and Answers 21

22 22

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