Question

A 100 mL volume of a 0.100 M weak base (pKb = 5.00) was titrated with 1.00 M HClO4. Find the pH at the following volumes of added acid: 0.00, 1.00, 5.00, 10.00, and 10.10 mL.

Answer 1

ANSWER:-

first, estimate the equivalence volume with the equation

                    V(weak base)*M(weak base) = V(HClO₄)*M[HClO₄]

100 x 0.100M = V(HClO₄) x 1.00

V(HClO₄) = 10 ml

at 0.00 ml

0.00 mL added [OH- ] = 0.100 M

[H3O]+ = 1.00*10-13

pH = 13.00

at 1.00 ml

[OH- ] = [(mmol OH initial) – (1.00 mL HClO₄)(1.00 M H+ )]/Vt = [(10.0 – 1.00)]/101.00] = 0.089 M

[H3O]+ = 1.1*10-13

pH = 12.95

at 5.00 ml

[OH- ] = [(mmol OH initial) – (5.00 mL HClO₄)(1.00 M H+ )]/Vt = [(10.0 – 5.00)]/105.00] = 0.048 M

[H3O]+ = 2.1 *10-13

pH = 12.68

at 10.00 ml

equiv. pt

[OH- ] = [(mmol OH initial) – (10.00 mL HClO₄)(1.00 M H+ )]/Vt = [(10.0 – 1.00)]/110.00] = 0 (= 10-7 M from Kw rxn)

[H3O]+ = 10-7

pH = 7.00

at 10.10 ml

After the equiv. point the excess H+ is what controls the pH.

[H+ ] = [(10.10 mL HClO₄)(1.00 M H+ )-(mmol OH initial)]/Vt = [10.10 - 10.0) / (110.10) = 9 *10-4 M

pH = 3.0