Question

Consider the titration of a 25.0 −mL sample of 0.110 M HC7H5O2 with 0.120 M KOH. Determine each of the following.

The Ka of HC7H5O2 is 6.5 × 10−5.

Part A: the pH at 5.00 mL of added base pH= ?

Answer 1

Given:

M(HC7H5O2) = 0.11 M

V(HC7H5O2) = 25 mL

M(KOH) = 0.12 M

V(KOH) = 5 mL

mol(HC7H5O2) = M(HC7H5O2) * V(HC7H5O2)

mol(HC7H5O2) = 0.11 M * 25 mL = 2.75 mmol

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

mol(KOH) = 0.12 M * 5 mL = 0.6 mmol

We have:

mol(HC7H5O2) = 2.75 mmol

mol(KOH) = 0.6 mmol

0.6 mmol of both will react

excess HC7H5O2 remaining = 2.15 mmol

Volume of Solution = 25 + 5 = 30 mL

[HC7H5O2] = 2.15 mmol/30 mL = 0.0717M

[C7H5O2-] = 0.6/30 = 0.02M

They form acidic buffer

acid is HC7H5O2

conjugate base is C7H5O2-

Ka = 6.5*10^-5

pKa = - log (Ka)

= - log(6.5*10^-5)

= 4.187

use:

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

= 4.187+ log {2*10^-2/7.167*10^-2}

= 3.633

Answer: 3.63