Question

A) A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 20.0 mL of KOH.

B) A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 28.0 mL of HNO3.

C) A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 15.0 mL of NaOH.

Answer 1

A)

millimole of HBr = 50 x 0.15 = 7.5

millimoles of KOH = 0.25 x 20 = 5

acid millimoles > base millimoles

[H+] = 7.5 - 5 / (50 + 20) = 0.0357 M

pH = -log [H+] = -log (0.0375)

pH = 1.45

B)

millimoles of base NH3 = 75 x 0.2 = 15

millimoles of HNO3 =28 x 0.5 = 14

NH3 + HCl ------------------> NH4Cl

15         14                               0

1           0                                 14

pOH = pKb + log (NH4Cl / NH3)

pOH = 4.75 + log (14/1)

pOH = 5.89

pH + pOH = 14

pH = 8.11

C)

millimoles of CH3COOH = 52 x 0.35 = 18.2

millimoles of NaOH = 0.40 x 14 = 6

CH3COOH + NaOH ------------------> CH3COONa + H2O

18.2                  6                                       0                0

12.2                 0                                        6

pH = pKa + log [CH3COONa / CH3COOH]

pH = 4.74 + log (6 /12.2)

pH = 4.43