Acids and Bases. Chapter 10. Solutions for Practice Problems. Student Textbook page 378

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1 Chapter 10 Acids and Bases Solutions for Practice Problems Student Textbook page Problem Hydrogen cyanide is a poisonous gas at room temperature. When this gas dissolved in water, the following reaction occurs: HCN (aq) + H2O (l) H 3 O + (aq) + CN (aq) Identify the conjugate acid-base pairs. You have to identity the conjugate acid-base pairs in the hydrogen cyanide reaction. The balanced chemical equation is given. Step 1 Identity the proton donor on the left side of the equation. This is the acid. The other reactant is the proton receiver, which is the base. Step 2 Identity the proton receiver as a product on the right side of the equation. It has the same basic formula as the base, plus one extra proton. This is the conjugate acid of the base reactant. Step 3 The other product is, by default, the conjugate base of the acid reactant. It has one less proton than the acid reactant. Conjugate Acid-Base Pairs: HCN (aq) /CN (aq) H 2 O (l) /H 3 O + (aq) The formulas of the conjugate acid-base pairs differ by only one proton, as expected. 2. Problem Sodium acetate is a good electrolyte. In water, the acetate ion reacts as follows: CH 3 COO (aq) + H 2 O (l) CH 3 COOH (aq) + OH (aq) Identify the conjugate acid-base pairs. You have to identity the conjugate acid-base pairs in the sodium acetate reaction. The balanced chemical equation is given. Step 1 Identity the proton donor on the left side of the equation. This is the acid. The other reactant is the proton receiver, which is the base. Step 2 Identity the proton receiver as a product on the right side of the equation. It has the same basic formula as the base, plus one extra proton. This is the conjugate acid of the base reactant. 165

2 Step 3 The other product is, by default, the conjugate base of the acid reactant. It has one less proton than the acid reactant. Conjugate Acid-Base Pairs: CH 3 COOH (aq) /CH 3 COO (aq) H 3 O + (aq) /H 2 O (l) The formulas of the conjugate acid-base pairs differ by only one proton, as expected. 3. Problem Write equations to show how the hydrogen sulfide ion, HS, can react with water. First show the ion acting as an acid, then show the ion acting as a base. You have to write balanced equations for the reaction of the hydrogen sulfide ion with water. The reactants are HS and H 2 O. HS can be both an acid and a base. HS as a base: The water is, by default, the acid. Step 1 Write the formula for the conjugate acid. This will be the same formula as the base HS ion, plus one proton, making it H 2 S. Step 2 Write the formula for the conjugate base. This will be the same formula as the acid H 2 O, minus one proton, making it the negative OH ion. Step 3 Identify the states of the reactants and products. Water is liquid and the HS ion is aqueous. The products will be aqueous too. Step 4 Put the equation together and check that the number of atoms on the left equal the atoms on the right. HS as an acid: The water is, by default, the base. Step 1 Write the formula for the conjugate base. This will be the same formula as the acid HS ion, minus one proton, making it S 2. Step 2 Write the formula for the conjugate acid. This will be the same formula as the base H 2 O, plus one proton, making it the positive H 3 O + ion. Repeat Steps 3 and 4 above. As a base: HS (aq) + H 2 O (l) H 2 S (aq) + OH (aq) As an acid: HS (aq) + H 2 O (l) S 2 (aq) + H 3 O + (aq) The formulas of the conjugate acid-base pairs differ by only one proton, as expected. The number of H, S, and O atoms on the left equal their numbers on the right. Solutions for Practice Problems Student Textbook page Problem (a) Write the chemical formula for hydrobromic acid. Then write the name and formula for the anion that it forms. (b) Hydrosulfuric acid, H 2 S, forms two anions. Name them and write their formulas. 166

3 (a) You need to give the chemical formula of hydrobromic acid and the name and formula of its anion. (b) You need to give the chemical formula of hydrosulfuric acid and the name and formula of its two anions. The formula for hydrosulfuric acid is H 2 S. Hydrobromic acid has one anion, while hydrosulfuric acid has two anions. To determine the chemical formula, we use the naming rules for acids and binary compounds. Hydrobromic acid is hydrogen bromide, while hydrosulfuric acid is hydrogen disulfide. We can apply the same steps as used for Practice Problems in Chapter 3 to determine the chemical formula. The anion of the acid is essentially its conjugate base in an acid-base reaction. That is, it has the acid formula minus one proton, and acquires a negative charge. If the conjugate base is able to dissociate further, it becomes an acid in the next reaction; then its own conjugate base is its same formula minus one proton, plus an extra negative charge. The charge of a conjugate base becomes one negative charge for every proton that is lost. (a) HBr (aq) gives the Bromide ion, Br (b) H 2 S (aq) gives the hydrogen sulfide ion, HS (aq) and the sulfide ion, S 2 (aq) The formulas of the ions (conjugate bases) differ by only one proton and one negative charge to their parent acid formula, as expected. 5. Problem Write the chemical formula for the following acids. Then name and write the formulas for the oxoanions that form from each acid. Refer to Chapter 3, Table 3.5, Names and Valences of Some Common Polyatomic Ions, as necessary. (a) nitric acid (b) nitrous acid (c) hyponitrous acid (d) phosphoric acid (e) phosphorus acid (f) periodic acid You need to give the chemical formula of the acids listed and then name and write the formula of its oxoanion. The names of the acids are given. Chapter 3, Table 3.5, Names and Valences of Some Common Polyatomic Ions, is given as a reference guide. To determine the chemical formula, we use the naming rules for acids and binary compounds with polyatomic ions. We can apply the same steps as used for Practice Problems in Chapter 3. The oxoanion of the acid is essentially its conjugate base in an acid-base reaction. That is, it has the acid formula minus one proton, and acquires a negative charge. If the conjugate base is able to dissociate further, it becomes an acid in the next reaction; then its own conjugate base is its same formula minus one proton, plus an extra 167

4 negative charge. The charge of a conjugate base becomes one negative charge for every proton that is lost from its parent acid. Table 3.5 in Chapter 3 can then be used to identify the name of the oxoanion created. (a) HNO 3(aq) ; nitrate NO 3 (b) HNO 2(aq) ; nitrite NO 2 (c) HNO (aq) ; hyponitrite NO (d) H 3 PO 4(aq) ; dihydrogen phosphate H 2 PO 4 ; hydrogen phosphate HPO 2 4 ; 3 phosphate PO 4 (e) H 3 PO 3(aq) ; dihydrogen phosphite H 2 PO 3 ; hydrogen phosphite HPO 2 3 ; 3 phosphite PO 3 (f) HIO 4(aq) periodate IO 4 The formulas of the oxoanions (conjugate bases) differ by only one proton and one negative charge to their parent acid formula, as expected. You can search other reference sources or the Internet to reaffirm the names and formulas of the oxoanions. Solutions for Practice Problems Student Textbook page Problem Calculate the ph of each solution, given the hydronium ion concentration. (a) [H 3 O + ] = mol/l (b) [H 3 O + ] = mol/l (c) [H 3 O + ] = mol/l (d) [H 3 O + ] = mol/l You need to calculate the solution ph for the hydronium ion concentrations listed. The [H 3 O + ] is given for each solution. Apply the equation: ph = log [H 3 O + ] (a) ph = log = 2.57 (b) ph = log = 7.14 (c) ph = log = 4.01 (d) ph = log = (a) [H 3 O + ] > mol/l, therefore the ph should be less than 7 and the solution should be acidic, which it is at (b) [H 3 O + ] is slightly lower than mol/l, therefore the ph of the solution should be just above neutral, which it is at (c) [H 3 O + ] > mol/l, therefore the ph should be less than 7 and the solution should be acidic, which it is at (d) [H 3 O + ] < mol/l, therefore ph should be more than 7 and the solution should be basic, which it is at

5 7. Problem [H 3 O + ] in a cola drink is about mol/l. Calculate the ph of the drink. State whether the drink is acidic or basic. You have to calculate the ph of a cola drink. [H 3 O + ] = mol/l Use the equation: ph = log [H 3 O + ] ph = log = Therefore, the cola drink is acidic. [H 3 O + ] > mol/l, therefore the ph should be less than 7 and the cola drink should be acidic, which it is at Problem A glass of orange juice has [H 3 O + ] of mol/l. Calculate the ph of the juice. State whether the result is acidic or basic. You have to calculate the ph of a sample of orange juice. [H 3 O + ] = mol/l Use the equation: ph = log [H 3 O + ] ph = log = Therefore, the orange juice is acidic. Check on Your Solution [H 3 O + ] > mol/l, therefore the ph should be less than 7 and the juice should be acidic, which it is at Problem (a) [H 3 O + ] in a dilute solution of nitric acid, HNO 3, is mol/l. Calculate the ph of the solution. (b) [H 3 O + ] of a solution of sodium hydroxide is mol/l. Calculate the ph of the solution. You have to calculate the ph of the nitric acid and sodium hydroxide solutions. [H 3 O + ] = mol/l for HNO 3 [H 3 O + ] = mol/l for NaOH Use the equation: ph = log [H 3 O + ] (a) ph = log = 2.2, therefore, the solution is acidic. (b) ph = log = 9.18, therefore, the solution is basic. 169

6 (a) [H 3 O + ] > mol/l, therefore the ph should be less than 7 and the solution should be acidic, as would be expected for nitric acid. (b) [H 3 O + ] < mol/l, therefore the ph should be more than 7 and the solution should be basic, as would be expected for NaOH. Solutions for Practice Problems Student Textbook page Problem ml of nitric acid neutralizes ml of mol/l NaOH (aq). What is the concentration of the nitric acid? You have to find the concentration of nitric acid needed to neutralize the given amount of NaOH. Volume of HNO 3 = ml = L Volume of NaOH = ml = L Concentration of NaOH = mol/l Step 1 Write the balanced equation for the reaction. Step 2 Calculate the number of moles of NaOH used by multiplying its concentration by the given volume. Step 3 From the mole ratio of acid:base in the balanced equation and the number of moles of NaOH from step 2, equate and solve for the number of moles of HNO 3 reacted. Step 4 Divide the number of moles of HNO 3 by the given volume to get its concentration. Step 1 HNO 3(aq) + NaOH (aq) NaNO 3(aq) + H 2 O (l) Step 2 Number of moles of NaOH = 0.15 mol/l L = mol Step 3 HNO 3 reacts with NaOH in a 1:1 ratio so there must be mol HNO 3 Step 4 [HNO 3 ] = mol / L = mol/l Therefore, the concentration of HNO 3 is mol/l. The concentration of nitric acid is greater than that of sodium hydroxide, so one would expect the volume of acid needed for the neutralization to be less than the volume of the base. This is the case with the volumes given in the question. 11. Problem What volume of mol/l magnesium hydroxide is needed to neutralize 40.0 ml of 1.60 mol/l hydrochloric acid? You have to find the volume of magnesium hydroxide needed to neutralize the given amounts of hydrochloric acid. Volume of hydrochloric acid = 40.0 ml = 0.04 L Concentration of hydrochloric acid = 1.60 mol/l 170

7 Concentration of magnesium chloride = mol/l Step 1 Write the balanced equation for the reaction. Step 2 Calculate the number of moles of hydrochloric acid used by multiplying its concentration by the given volume. Step 3 From the mole ratio of acid:base in the balanced equation and the number of moles of HCl from step 2, equate and solve for the number of moles of magnesium hydroxide reacted. Step 4 Divide the number of moles of magnesium hydroxide by the given concentration to get its volume in L. Step 1 2HCl (aq) + Mg(OH) 2(aq) MgCl 2(aq) + 2H 2 O (l) Step 2 Number of moles of HCl = 1.6 mol/l 0.04 L = mol Step 3 HCl reacts with Mg(OH) 2 in a 2:1 ratio. Therefore, the number of moles of Mg(OH) 2 needed is 1 mol Mg(OH) 2 2 mol HCl mol HCl = mol Mg(OH) mol Step 4 Volume of Mg(OH) 2 = = L = 31.5 ml mol/l Therefore, the volume of magnesium hydroxide used is 31.5 ml. The mole ratio of HCl to Mg(OH) 2 is 2:1, meaning twice as much acid is needed to neutralize the base. Although the given concentration of HCl is higher than Mg(OH) 2, it is still nowhere near twice that of the base, so one would expect the volume of acid needed for the neutralization to still be more than the volume of the base. This is the case with the volume of Mg(OH) 2 calculated. Your answer is reasonable. 12. Problem What volume of mol/l hydrochloric acid is needed to neutralize each solution below? (a) 25.0 ml of mol/l sodium hydroxide (b) 20.0 ml of mol/l ammonia solution (c) 80 ml of mol/l calcium hydroxide You need to calculate the volume of hydrochloric acid needed to neutralize the given amounts of bases. Concentration of hydrochloric acid = mol/l (a) Volume of sodium hydroxide = 25.0 ml = L; concentration = mol/l (b) Volume of ammonia solution = 20.0 ml = L; concentration = mol/l (c) Volume of calcium hydroxide = 80 ml = L; concentration = mol/l For each case, do the following steps. Step 1 Write the balanced equation for the reaction. Step 2 Calculate the number of moles of base used by multiplying its concentration by the given volume. 171

8 Step 3 From the mole ratio of acid:base in the balanced equation and the number of moles of base from step 2, equate and solve for the number of moles of hydrochloric acid reacted. Step 4 Divide the number of moles of hydrochloric acid by the given concentration to get its volume in L. Note: An aqueous ammonia solution, i.e., NH 3 + H 2 O, is represented as NH 4 (OH) (aq). (a) Step 1 HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O (l) Step 2 Number of moles of NaOH = mol/l L = mol Step 3 NaOH reacts with HCl in a 1:1 ratio. Therefore, mol NaOH reacts with mol HCl Step 4 Volume ofhcl = mol mol/l = L = 22.5 ml (b) Step 1 HCl (aq) + NH 4 OH (aq) NH 4 Cl (aq) + H 2 O (l) Step 2 Number of moles of NH 4 OH = mol/l L = mol Step 3 NH 4 OH reacts with HCl in a 1:1 ratio. Therefore, mol NH 4 OH reacts with mol HCl mol Step 4 Volume of HCl = = L = 24.7 ml mol/l (c) Step 1 2HCl (aq) + Ca(OH) 2(aq) CaCl 2(aq) + 2H 2 O (l) Step 2 Number of moles of Ca(OH) 2 = mol/l L = mol Step 3 Ca(OH) 2 reacts with HCl in a 1:2 ratio. Therefore, mol Ca(OH) 2 reacts with mol HCl Step 4 Volume of HCl = mol mol/l = L = 4.8 ml Check your Solution (a) The mole ratio of HCl to NaOH is 1:1. The given concentration of HCl is higher than NaOH, so its volume should be less than that of NaOH. This is the case with the volume of 22.5 ml calculated. Your answer is reasonable. (b) The mole ratio of HCl to NH 4 (OH) is 1:1. The given concentration of HCl is lower than NH 4 (OH), so its volume should be more than that of NH 4 (OH). This is the case with the volume of 24.7 ml calculated. Your answer is reasonable. (c) The mole ratio of HCl to Ca(OH) 2 is 2:1 meaning twice as much acid is needed to neutralize the base. However, the given concentration of HCl is already almost three times that of Ca(OH) 2, so its volume should be less than that of Ca(OH) 2. This is the case with the volume of 4.8 ml calculated. Your answer is reasonable. 13. Problem What concentration of sodium hydroxide solution is needed for each neutralization reaction? (a) ml of sodium hydroxide neutralizes ml of mol/l hydrofluoric acid. (b) ml of sodium hydroxide neutralizes ml of mol/l sulfuric acid. (c) ml of sodium hydroxide neutralizes ml of mol/l phosphoric acid. You need to calculate the concentration of sodium hydroxide used to neutralize the given amounts of acids. (a) Volume of sodium hydroxide = ml = L [hydrofluoric acid] = mol/l; volume = ml = L 172

9 (b) Volume of sodium hydroxide = ml = L [sulfuric acid] = mol/l; volume = ml = L (c) Volume of sodium hydroxide = ml = L [phosphoric acid] = mol/l; volume = ml = L For each case, do the following steps. Step 1 Write the balanced equation for the reaction. Step 2 Calculate the number of moles of acid used by multiplying its concentration by the given volume. Step 3 From the mole ratio of acid:base in the balanced equation and the number of moles of acid from step 2, equate and solve for the number of moles of sodium hydroxide used. Step 4 Divide the number of moles of sodium hydroxide by the given volume to get its concentration in mol/l. (a) Step 1 HF (aq) + NaOH (aq) NaF (aq) + H 2 O (l) Step 2 Number of moles of HF = mol/l L = mol Step 3 HF reacts with NaOH in a 1:1 ratio so there must be mol NaOH Step 4 Concentration of NaOH = mol = mol/l L (b) Step 1 H 2 SO 4(aq) + 2NaOH (aq) Na 2 SO 4(aq) + 2H 2 O (l) Step 2 Number of moles of H 2 SO 4 = mol/l 0.02 L = mol Step 3 H 2 SO 4 reacts with NaOH in a 1:2 ratio so there must be mol NaOH Step 4 Concentration of NaOH = mol = mol/l L (c) Step 1 H 3 PO 4(aq) + 3NaOH (aq) Na 3 PO 4(aq) + 3H 2 O (l) Step 2 Number of moles of H 3 PO 4 = mol/l L = mol Step 3 H 3 PO 4 reacts with NaOH in a 1:3 ratio so there must be mol NaOH Step 4 Concentratio n of NaOH = mol = mol/l L Check your Solution (a) The mole ratio of HF to NaOH is 1:1. The given volume of HF is lower than that of NaOH, so its given concentration must be more than that of NaOH. This is the case with the [NaOH] of mol/l calculated. Your answer is reasonable. (b) The mole ratio of H 2 SO 4 to NaOH is 1:2, meaning twice as much base is needed to neutralize the acid. The given volume of H 2 SO 4 is only slightly lower than that of NaOH, so its given concentration should be almost half of that of NaOH. This is the case with the [NaOH] of mol/l calculated. Your answer is reasonable. (c) The mole ratio of H 3 PO 4 to NaOH is 1:3, meaning three times as much base is needed to neutralize the acid. However, the given volume of NaOH is only a little over one half that of H 2 PO 4, so its concentration should be very much higher than that of the acid. This is the case with the [NaOH] of mol/l calculated. Your answer is reasonable. 173

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