Question

A 45.9 mL sample of a 0.503 M aqueous hypochlorous acid solution is titrated with a 0.207 M aqueous barium hydroxide solution. What is the pH after 21.2 mL of base have been added?

pH =

A buffer solution contains 0.294 M C₆H₅NH₃Cl and 0.463 M C₆H₅NH₂ (aniline). Determine the pH change when 0.122 mol HI is added to 1.00 L of the buffer.

pH after addition − pH before addition = pH change =

Determine the pH change when 0.096 mol HCl is added to 1.00 L of a buffer solution that is 0.366 M in HCN and 0.352 M in CN^-.

pH after addition − pH before addition = pH change =

Answer 1

1)

mmoles of HClO = 45.9 x 0.503 = 23.09

mmoles of Ba(OH)2 = 21.2 x 0.207 = 4.3884

mmoles of OH- = 8.778

HClO (aq) +   OH- (aq)   -----------> ClO- (aq) + H2O

23.09             8.778                              0                  0

14.313              0                               8.778

pH = pKa + log [salt / acid]

    = 7.46 + log [8.778 / 14.313]

pH = 7.25

2)

A buffer solution contains 0.294 M C₆H₅NH₃Cl and 0.463 M C₆H₅NH₂ (aniline). Determine the pH change when 0.122 mol HI is added to 1.00 L of the buffer.

pOH = pKb + log [salt / base]

        = 9.13 + log [0.294 / 0.463]

      = 8.93

initial pH = 5.07

when HI is added :

pOH = pKb + log [salt + C / base - C]

      = 9.13 + log [0.294 + 0.122 / 0.463 - 0.122]

     = 9.22

final pH = 4.78

pH change = 4.78 - 5.07 = - 0.29

pH change = -0.29

3)

pKa = 9.40

pH change = - 0.24