104. A chemist titrated 35.00 mL of a 0.150 mol/L solution of hypobromous acid, HBrO(aq), Ka = 2.8 ´ 10–9. Calculate the resulting pH after the addition of 15.00 mL of 0.1000 mol/L sodium hydroxide solution, NaOH(aq). What Is Required? You need to find the pH at a certain point in the titration using hypobromous acid and potassium hydroxide solutions. What Is Given? You know the volume of the hypobromous acid: VHBrO = 35.00 mL You know the concentration of the hypobromous acid: cHBrO = 0.1500 mol/L You know the volume of the sodium hydroxide: VNaOH = 15.00 mL You know the concentration of the sodium hydroxide: cNaOH = 0.1000 mol/L You are given the acid dissociation constant for hypobromous acid: Ka = 2.8 × 10‒9. Plan Your Strategy Write the balanced equation for the reaction between HBrO (aq) and NaOH(aq).
Act on Your Strategy Because NaOH is a strong base, the reaction essentially goes to completion. HBrO(aq) (aq) + NaOH(aq) → NaBrO(aq) + H2O(ℓ)
Use the formula n = cV to calculate the amount in moles, n, of HBrO(aq) present before any base was added. Use the formula n = cV to calculate the amount in moles, n, of NaOH(aq).
Calculate the amount in moles, n, of BrO‒(aq) present after the base was added.
nHBrO = cV æ 1 L ö æ mol ö = ç 0.1500 ÷ ÷ 35.00 mL ç L ø è è 1000 mL ø = 5.250 ´ 10 –3 mol nNaOH = cV æ 1 L ö æ mol ö = ç 0.1000 ÷ ÷ 15.00 mL ç L ø è è 1000 mL ø = 1.500 ´ 10 –3 mol Because the reaction goes to completion and because the ratio of BrO‒ to NaOH is 1:1, for every mole of NaOH added, one mole of HBrO will be converted into BrO‒. Therefore: nBrO- = nNaOH
(
)
(
)
= 1.500 ´ 10-3 mol Calculate the amount in moles of unreacted HBrO(aq).
n( excess HBrO) = n(i HBrO) - n BrO- formed
(
= 5.250 ´ 10-3 mol - 1.500 ´ 10-3 mol = 3.750 ´ 10-3 mol
264 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
)
Determine the final volume of the solution.
V = VHBrO + VNaOH = 35.00 mL + 15.00 mL = 50.00 mL ´
n to determine the V concentrations of HBrO(aq) and BrO‒(aq) in mol/L. Use c =
Write equations for any equilibrium reactions occurring and determine whether the solution has buffering capacity.
1L 1000 mL
= 0.05000 L n cHBrO = V 3.750 ´ 10-3 mol = 0.0500 L = 7.500 ´ 10-2 mol/L n cBrO- = V 1.500 ´ 10-3 mol = 0.0500 L = 3.000 ´ 10-2 mol/L HBrO(aq) + H2O(ℓ) BrO‒(aq) + H3O+(aq)
BrO‒(aq) + H2O(ℓ)
HBrO (aq) + OH‒(aq)
The solution has buffering capacity. The reactions of HBrO and BrO‒ reacting with water balance each other and very little change actually occurs. Therefore, you can use the amounts of HBrO and BrO‒ present after the reaction with NaOH and the equilibrium constant to find the concentration of hydronium ion, H3O+ present. Write the equilibrium expression, substitute in the calculated concentrations and solve for the concentration of the hydronium ion.
[BrO - ][H 3 O + ] [HBrO] (3.000 ´ 10-2 )( x) 2.8 ´ 10 –9 = 7.500 ´ 10-2 (2.8 ´ 10-9 )(7.500 ´ 10-2 ) x= 3.000 ´ 10-2 = 7.000 ´ 10 –9 Ka =
[H3O+] = x = 7.000 ×10–9 mol/L
Unit 4 Part B ● MHR 265
Calculate pH using: pH = –log [H3O+].
pH = –log [H3O+] = –log (7.000 × 10–9) = 8.1549 = 8.15
Check Your Solution The number of significant digits to the right of the decimal is the same as the number of significant digits in the given Ka.
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105. A solution of 50.00 mL of a 0.120 mol/L nitrous acid, HNO2(aq), is titrated with 0.1000 mol/L of potassium hydroxide, KOH(aq). Determine the resulting pH of the solution after 11.25 mL of the base has been added. For the acid, Ka = 5.6 ´ 10–4. What Is Required? You need to find the pH at a certain point in the titration using nitrous acid and potassium hydroxide solutions. What Is Given? You know the volume of the nitrous acid: VHNO2 = 50.00 mL You know the concentration of the nitrous acid: cHNO2 = 0.120 mol/L You know the volume of the potassium hydroxide: VKOH = 11.50 mL You know the concentration of the potassium hydroxide: cKOH = 0.1000 mol/L You know the acid dissociation constant: Ka = 5.6 × 10‒4 Plan Your Strategy Write the balanced equation for the reaction between HNO2(aq) and KOH(aq).
Act on Your Strategy Because KOH is a strong base, the reaction essentially goes to completion. HNO2(aq) + KOH(aq) →KNO2(aq) + H2O(ℓ)
Use the formula n = cV to calculate the amount in moles, n, of HNO2(aq) present before any base was added. Use the formula n = cV to calculate the amount in moles, n, of KOH(aq).
Calculate the amount in moles, n, of NO2‒(aq) present after the base was added.
nHNO2 = cV æ 1 L ö æ mol ö = ç 0.120 ÷ ÷ 50.00 mL ç L ø è è 1000 mL ø = 6.00 ´ 10 –3 mol nKOH = cV æ 1 L ö æ mol ö = ç 0.100 ÷ ÷ 11.25 mL ç L ø è è 1000 mL ø = 1.125 ´ 10 –3 mol Because the reaction goes to completion and because the ratio of NO2‒ to KOH is 1:1, for every mole of KOH added, one mole of HNO2 will be converted into NO2‒. Therefore: nNO - = nKOH
(
(
)
)
2
Calculate the amount in moles of unreacted HNO2(aq).
= 1.125 ´ 10-3 mol n( excess HNO2 ) = n(i HNO2 ) - n NO - formed ( 2 ) = 6.000 ´ 10-3 mol - 1.125 ´ 10-3 mol = 4.875 ´ 10-3 mol
Unit 4 Part B ● MHR 267
Determine the final volume of the solution.
V = VHNO2 + VKOH
n to determine the V concentrations of HNO2(aq) and NO2‒(aq) in mol/L.
n V 4.875 ´ 10-3 mol = 0.06125 L = 7.9592 ´ 10-2 mol/L n cNO - = 2 V 1.125 ´ 10-3 mol = 0.06125 L = 1.8367 ´ 10-2 mol/L HNO2(aq) + H2O(ℓ) NO2‒(aq) + H3O+(aq)
Use c =
Write equations for any equilibrium reactions occurring and determine whether the solution has buffering capacity.
= 50.00 mL + 11.25 mL 1L = 61.25 mL ´ 1000 mL = 0.06125 L
cHNO2 =
NO2‒(aq) + H2O(ℓ)
HNO2 (aq) + OH‒(aq)
The solution has buffering capacity. The reactions of HNO2 and NO2‒ reacting with water balance each other and very little change actually occurs. Therefore, you can use the amounts of HNO2 and NO2‒ present after the reaction with KOH and the equilibrium constant to find the concentration of hydronium ion, H3O+ present. Write the equilibrium expression, substitute in the calculated concentrations and solve for the concentration of the hydronium ion.
[NO 2 - ][H 3 O + ] [HNO 2 ] (1.8367 ´ 10-2 )( x) 5.6 ´ 10 –4 = 7.9592 ´ 10-2 (5.6 ´ 10 –4 )(7.9592 ´ 10-2 ) x= 1.8367 ´ 10-2 = 2.43267 ´ 10-3 Ka =
[H3O+] = x = 2.43267 × 10-3 mol/L
268 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
Calculate pH using: pH = –log [H3O+].
pH = –log [H3O+] = –log (2.43267 × 10-3) = 2.615 = 2.61
Check Your Solution The pH is consistent with having unreacted HNO2 present. The number of significant digits to the right of the decimal is the same as the number of significant digits in the given Ka. The answer seems reasonable.
Unit 4 Part B ● MHR 269
106. If 100.00 mL of a 0.400 mol/L hydrofluoric acid solution, HF(aq), is titrated with a 0.2000 mol/L sodium hydroxide solution, NaOH(aq), determine the pH that results when 20.00 mL of the base is added. For the hydrofluoric acid solution, Ka = 6.3 ´ 10–4. What Is Required? You need to find the pH at a certain point in the titration of hydrofluoric acid with sodium hydroxide solution. What Is Given? You know the volume of the hydrofluoric acid: VHF = 100.00 mL You know the concentration of the hydrofluoric acid: cHF = 0.4000 mol/L You know the volume of the sodium hydroxide: VNaOH = 20.00 mL You know the concentration of the sodium hydroxide: cNaOH = 0.2000 mol/L You are given the acid dissociation constant for hydrofluoric acid: Ka = 6.3 × 10‒4 Plan Your Strategy Write the balanced equation for the reaction between HF (aq) and NaOH(aq).
Act on Your Strategy Because NaOH is a strong base, the reaction essentially goes to completion. HF(aq) (aq) + NaOH(aq) → NaF(aq) + H2O(ℓ)
Use the formula n = cV to calculate the amount in moles, n, of HF(aq) present before any base was added. Use the formula n = cV to calculate the amount in moles, n, of NaOH(aq).
Calculate the amount in moles, n, of F‒(aq) present after the base was added.
nHF = cV æ 1 L ö æ mol ö = ç 0.4000 ÷ ÷ 100.00 mL ç L ø è è 1000 mL ø = 4.000 ´ 10-2 mol nNaOH = cV æ 1 L ö = 0.2000 mol/ L 20.00 mL ç ÷ è 1000 mL ø = 4.000 ´ 10 –3 mol Because the reaction goes to completion and because the ratio of F‒ to NaOH is 1:1, for every mole of NaOH added, one mole of HF will be converted into F‒. Therefore: nF- = nNaOH
(
(
)
)(
)
= 4.000 ´ 10-3 mol Calculate the amount in moles of unreacted HF(aq).
n( excess HF) = n(i HF) - n F- formed
(
= 4.000 ´ 10
-2
)
mol - 4.000 ´ 10-3 mol
= 3.600 ´ 10-2 mol
270 MHR ● Chemistry 12 Solutions Manual 978-0-07-106042-4
Determine the final volume of the solution.
V = VHF + VNaOH = 100.00 mL + 20.00 mL = 120.00 mL ´
n to determine the V concentrations of HF(aq) and F‒(aq) in mol/L. Use c =
Write equations for any equilibrium reactions occurring and determine whether the solution has buffering capacity.
1L 1000 mL
= 0.1200 L n cHF = V 3.600 ´ 10-2 mol = 0.1200 L = 0.300 ´ 10-1 mol/L n cF- = V 4.000 ´ 10-3 mol = 0.1200 L = 3.333 ´ 10-2 mol/L HF(aq) + H2O(ℓ) F‒(aq) + H3O+(aq)
F‒(aq) + H2O(ℓ)
HF(aq) + OH‒(aq)
The solution has buffering capacity. The reactions of HF and F‒ reacting with water balance each other and very little change actually occurs. Therefore, you can use the amounts of HF and F‒ present after the reaction with NaOH and the equilibrium constant to find the concentration of hydronium ion, H3O+ present. Write the equilibrium expression, substitute in the calculated concentrations and solve for the concentration of the hydronium ion.
[F- ][H 3 O + ] [HF] (3.333 ´ 10-2 )( x) 6.3 ´ 10 –4 = 3.000 ´ 10-1 (6.3 ´ 10 –4 )(3.000 ´ 10 -1 ) x= 3.333 ´ 10-2 = 5.6706 ´ 10 –3 Ka =
[H3O+] = x = 5.6706 ×10–3 mol/L
Unit 4 Part B ● MHR 271
Calculate pH using: pH = –log [H3O+].
pH = –log [H3O+] = –log (5.6706 ×10–3) = 2.2464 = 2.25
Check Your Solution The pH is consistent with having unreacted HF present. The number of significant digits to the right of the decimal is the same as the number of significant digits in the given Ka. The answer seems reasonable.
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