Question

1. Calculate the pH during the titration of 20.00 mL of 0.1000 M C₆H₅COOH(aq) with 0.1000 M NaOH(aq) after 19 mL of the base have been added. K_a of benzoic acid = 6.5 x 10^-5.

2. Determine the volume in mL of 0.15 M KOH(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 35 mL of 0.18 M HNO₂(aq). The K_a of nitrous acid is 7.1 x 10^-4.

Answer 1

1)

Given:

M(C6H5COOH) = 0.1 M

V(C6H5COOH) = 20 mL

M(KOH) = 0.1 M

V(KOH) = 19 mL

mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)

mol(C6H5COOH) = 0.1 M * 20 mL = 2 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 19 mL = 1.9 mmol

We have:

mol(C6H5COOH) = 2 mmol

mol(KOH) = 1.9 mmol

1.9 mmol of both will react

excess C6H5COOH remaining = 0.1 mmol

Volume of Solution = 20 + 19 = 39 mL

[C6H5COOH] = 0.1 mmol/39 mL = 0.0026M

[C6H5COO-] = 1.9/39 = 0.0487M

They form acidic buffer

acid is C6H5COOH

conjugate base is C6H5COO-

Ka = 6.5*10^-5

pKa = - log (Ka)

= - log(6.5*10^-5)

= 4.1871

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.1871+ log {0.0487/0.0026}

= 5.4658

pH = 5.47

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