Question

1. Calculate the pH during the titration of 20.00 mL of 0.1000 M C₆H₅SH(aq) with 0.1000 M KOH(aq) after 11 mL of the base have been added. K_a of thiophenol = 3.2 x 10^-7.

2. Calculate the pH during the titration of 20.00 mL of 0.1000 M trimethylamine, (CH₃)₃N(aq), with 0.2000 M HCl(aq) after 7 mL of the acid have been added. K_b of trimethylamine = 6.5 x 10^-5.

Answer 1

1)

Given:

M(C6H5SH) = 0.1 M

V(C6H5SH) = 20 mL

M(KOH) = 0.1 M

V(KOH) = 11 mL

mol(C6H5SH) = M(C6H5SH) * V(C6H5SH)

mol(C6H5SH) = 0.1 M * 20 mL = 2 mmol

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

mol(KOH) = 0.1 M * 11 mL = 1.1 mmol

We have:

mol(C6H5SH) = 2 mmol

mol(KOH) = 1.1 mmol

1.1 mmol of both will react

excess C6H5SH remaining = 0.9 mmol

Volume of Solution = 20 + 11 = 31 mL

[C6H5SH] = 0.9 mmol/31 mL = 0.029M

[C6H5S-] = 1.1/31 = 0.0355M

They form acidic buffer

acid is C6H5SH

conjugate base is C6H5S-

Ka = 3.2*10^-7

pKa = - log (Ka)

= - log(3.2*10^-7)

= 6.4949

use:

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

= 6.4949+ log {0.0355/0.029}

= 6.582

pH = 6.582

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