Question

Calculate the pH at the equivalence point in titrating 0.120 M solutions of each of the following with 9.0×10−2 MNaOH.

Part A

hydrobromic acid (HBr)

Part B

chlorous acid (HClO2)

Part C

benzoic acid (C6H5COOH)

Answer 1

Part A.Since HBr and NaOH are a strong acid and a strong base, the net ionic equation for the reaction is just:

H+ + OH- --> H2O. So, the pH of that solution at the equivalence point will be pH = 7.00.

Part B. Chlorous acid, HClO₂, K_a = 1.1 x 10^-2

The volumes are not given but this should still be solvable. At equivalence point, mmoles of acid equal mmoles of base. That is,

M_base x V_base = M_acid x V_acid

thus V_acid = (M_base x V_base) / M_acid

What we need to solve this problem is the total volume at the equivalence point and the number of mmoles of the salt that is formed:

[ClO₂^-] = (M_base x V_base )/(V_base + V_acid)

= (M_base x V_base) / (V_base + (M_base x V_base) / M_acid)

The remaining V term, V_base will cancel out:

[ClO₂^-] = M_base /(1 + (M_base/M_acid)) = M_baseM_acid/(M_base+M_acid)
[ClO₂^-] = (0.120)(9 * 10^-2)/0.21 = 0.051

ClO₂^-(aq) + H₂O(l) <====> HClO₂ (aq) + OH^-(aq)
K_b = K_w/K_a= (1.0 x 10^-14)/(1.1 x 10^-2) = [HC₃H₅O₃][OH^-]/[ C₃H₅O₃^-]

9.1 x 10^-13 = [OH^-]²/0.051
[OH^-] = 2.2 x 10^-7, pOH = 6.7, pH = 7.3

Part C. Benzoic acid, HC₇H₅O₂, K_a = 6.46 x 10^-5

The volumes are not given but this should still be solvable. At equivalence point, mmoles of acid equal mmoles of base. That is,

M_base x V_base = M_acid x V_acid

thus V_acid = (M_base x V_base) / M_acid

What we need to solve this problem is the total volume at the equivalence point and the number of mmoles of the salt that is formed:

[C₇H₅O₂^-] = (M_base x V_base )/(V_base + V_acid)

= (M_base x V_base) / (V_base + (M_base x V_base) / M_acid)

The remaining V term, V_base will cancel out:

[C₇H₅O₂^-] = M_base /(1 + (M_base/M_acid)) = M_baseM_acid/(M_base+M_acid)
[C₇H₅O₂^-] = (0.120)(9 * 10^-2)/0.21 = 0.051

C₇H₅O₂^-(aq) + H₂O(l) <====> HC₇H₅O₂ (aq) + OH^-(aq)
K_b = K_w/K_a= (1.0 x 10^-14)/(6.46 x 10^-5) = [HC₇H₅O₂][OH^-]/[ C₇H₅O₂^-]

1.55 x 10^-10 = [OH^-]²/0.051
[OH^-] = 2.8 x 10^-6, pOH = 5.5, pH = 8.5