Question

. A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1532 M solution of NaOH at 25 °C. The Ka value for HF is 6.8 x 10-4 . (a) What is the pH of the HF solution before titrant is added? (b) How many milliliters of titrant are required to reach the equivalence point? (c) What is the pH at 0.50 mL before the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH at 0.50 mL after the equivalence point?

Answer 1

a) Through the dissociation reaction, we calculate the hydronium concentration:

HF + H2O = H3O + + F-

Ka = [H3O +] * [F-] / [HF]

on balance:

6.8 x10 ^ -4 = x ^ 2 / 0.25-x

Dependent on:

x ^ 2 + 6.8x10 ^ -4 * x - 1.7x10 ^ -4 = 0

using the equation of the second degree:

x = [H3O +] = 0.013 M

pH = -Log (0.013) = 1.87

b) Neutralization ratio is used for the concentration and volume of acid:

Ca * Va = Cb * Vb

Volume of NaOH required:

Vb = Ca * Va / Cb = 0.25 M * 35 mL / 0.1532 M = 57.11 mL

c) V NaOH added = 57.11 - 0.5 = 56.61 mL = 0.05661 L

We calculate the moles of NaOH and HF:

n NaOH = 0.1535 mol / L * 0.05661 L = 8.67x10 ^ -3 mol NaOH

n HF = 0.25 mol / L * 0.035 L = 8.75 x10 ^ -3 mol HF = mol H3O +

We calculate the moles of H3O + that remain:

n HF = 8.75 x10 ^ -3 mol - 8.67x10 ^ -3 mol = 8x10 ^ -5 mol remaining

We calculate the final NaOH and HF concentrations:

[HF] = 8x10 ^ -5 mol / 0.09161 L = 8.73x10 ^ -4 M

[NaOH] = [F-] = 8.67x10 ^ -3 mol / 0.09161 L = 0.0946 M

pKa = - Log Ka = - Log (6.8 x10 ^ -4) = 3.17

We calculate pH:

pH = pKa + Log [F-] / [HF] = 3.17 + Log (0.0946 / 8.73x10 ^ -4) = 5.20

d) According to the reaction:

F- + H2O = HF + OH-

Calculate Kb:

Kb = Kw / Ka = 10 ^ -14 / 6.8 x10 ^ -4 = 1.47x10 ^ -11

Kb = [HF] * [OH-] / [F-] = x ^ 2 / 0.0946 -x = 1.47x10 ^ -11

clearing to x we ​​have:

X ^ 2 + 1.47x10 ^ -11 * X - 1.40x10 ^ -12 = 0

by equation of the second degree, we have:

x = [OH-] = 1.18x10 ^ -6

pOH = - Log (1.18x10 ^ -6) = 5.93

pH = 14 - 5.93 = 8.07

e) Total V = 57.11 mL + 0.5 mL = 57.61 mL = 0.05761 L

nNaOH aggregate = 0.1532 mol / L * 0.05761 L = 8.83x10 ^ -3 moles

nHF = 8.75 x10 ^ -3 moles

nNaOH final = 8.83x10 ^ -3 moles - 8.75 x10 ^ -3 moles = 8x10 ^ -5 moles

[NaOH] final = [OH-] = 8x10 ^ -5 moles / 0.05761 L = 1.39x10 ^ -3 M

pOH = - Log (1.39x10 ^ -3) = 2.86

pH = 14 - 2.86 = 11.14.