Question

Determine the pH change when 0.050 mol HNO3 is added to 1.00 L of a buffer solution that is 0.464 M in HClO and 0.226 M in ClO-.

Answer 1

Given that volume of buffer = 1.0 L

molarity of ClO- = 0.226 M

molarity of HClO = 0.464 M

[ClO-] = molarity x volume = 0.226 M x 1.0 L = 0.226 mol

    [HClO] = molarity x volume = 0.464 M x 1.0 L = 0.464 mol

[HNO3] = 0.05 mol

Original pH = pKa + log  [ClO-]/ [HOCl] ----------(1)

pH after addition of HNO3:

ClO- + HNO3 ---------> HOCl + NO3-

Hence,

new pH = pKa + log {[ClO-] - [HNO3]/[HOCl] +[HNO3]} --------(2)

Since HNO3 is a strong caid, pH will be decreased compared to original pH.

To get change in pH, subtract (2) from (1).

Then,

change in pH =  pKa + log  [ClO-]/ [HOCl] -  pKa + log {[ClO-] - [HNO3]/[HOCl]+[HNO3]}

=   log  [ClO-]/ [HOCl]  - log {[ClO-] - [HNO3]/[HOCl]+[HNO3]}

= log [ 0.226/ 0.464] - log [ 0.226 - 0.05 ] / [ 0.464+0.05]

  = 0.46

Therefore, change in pH = 0.46