Question

Determine the pH change when 0.110 mol KOH is added to 1.00 L of a buffer solution that is 0.418 M in HNO2 and 0.375 M in NO2-.

Answer 1

Initial pH of buffer,

pH = pKa + log(base/acid)

     = 3.15 + log(0.375/0.418)

     = 3.10

After addition of KOH

New [HNO2] = (0.418 M x 1 L - 0.110)/1 L = 0.308 M

New [NO2-] = (0.375 M x 1 L + 0.110)/1 L = 0.485 M

New pH of buffer,

pH = pKa + log(base/acid)

     = 3.15 + log(0.485/0.308)

     = 3.35

change in pH = (final - initial) pH

                      = 3.35 - 3.10

                      = 0.25