A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calcula

A 30.0 mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added:

A) 30.0 mL

B) 34.0 mL

C) 36.0 mL

D) 37.0 mL

E) 43.0 mL

30.0 mL x 0.150 M KOH = 4.50 millimoles KOH (a strong base)
0.125 M HClO4 (a strong acid)
NaClO4 will be neutral

(A) 30.0 mL x 0.125 M HClO4 = 3.75 millimoles HClO4
4.50 - 3.75 = 0.75 millimoles KOH remain in 60 mL solution
0.75 mmole/60 mL = 0.0125 M KOH
pOH = 1.90; pH = 12.10

(B) 34.0 mL x 0.125 M HClO4 = 4.25 millimoles HClO4
4.50 - 4.25 = 0.25 millimoles KOH remain in 64 mL solution
0.25 mmole/64 mL = 0.00391 M KOH
pOH = 2.41; pH = 11.59

(C) 36.0 mL x 0.125 M HClO4 = 4.50 millimoles HClO4
4.50 - 4.50 = 0.00 millimoles remain in 65 mL solution
pOH = 7.00; pH = 7.00

(D) 37.0 mL x 0.125 M HClO4 = 4.625 millimoles HClO4
4.625 - 4.50 = 0.125 millimoles HClO4 remain in 66 mL solution
0.125 mmole/66 mL = 0.00189 M HClO4
pH = 2.72

(E) 43.0 mL x 0.125 M HClO4 = 5.375 millimoles HClO4
5.375 - 4.50 = 0.875 millimoles HClO4 remain in 73 mL solution
0.875 mmole/73 mL = 0.0120 M HClO4
pH = 1.92