An industrial chemist studying bleaching and sterilizing prepares a hypochlorite buffer using 0.350 M HClO and 0.350 M NaClO. (Ka for HClO = 2.9 * 10-8) Find the pH of 1.00 L of the solution after 0.030 mol of NaOH has been added.

pH = 7.45

Explanation:

This is a buffer solution and we can solve it by using the Henderson-Hasselbalch equation:

pH = pKa + log ((A⁻) / (HA))

Here we will first have to calculate the A⁻ formed in the 1. 0 L solution which is formed by the reaction of HClO with the strong base NaOH and add it to the original mol of NaClO

mol NaClO = mole NaCLO originally present in the 1L of M solution + 0.030 mol produced in the reaction of HCLO with NaOH

0.350 mol +.030 mol = 0.380 mol

New concentrations:

HClO = 0. 350 mol-0.030 mol = 0.320 M (have to sustract the 0.030 mol reacted with NaOH)

NaClO = 0.380 mol / 1 L = 0.380 M

Now we have all the values required and we can plug them into the equation

pH = - log (2.9 x 10^-8) + log (0.380/.320) = 7.45