Calculate the pH for each case in the titration of 50.0 mL of 0.220 M HClO (aq) with 0.220 M KOH (aq). Use the ionization constant for HClO. What is the pH before addition of any KOH?

Before adding any KOH, the pH is 4.03

Explanation:

Step 1: Data given

Volume of a 0.220 M HClO = 50.0 mL = 0.050 L

Molarity of KOH = 0.220 M

The ionization constant for HClO is 4.0*10^-8

Step 2: The balanced equation

HClO + KOH → KClO + H2O

Step 3: pH before any addition of KOH

When no KOH is added, we only have HClO, a weak acid.

To calculate the pH of a weak acid, we need the Ka

Ka = [H+] / [acid]

4.0*10^-8 = [H+]² / 0.220

[H+]² = (4.0*10^-8) * 0.220

[H+]² = 8.8*10^-9

[H+] = √ (8.8*10^-9)

[H+] = 9.38*10^-5 M

pH = - log [H+]

pH = - log (9.38*10^-5)

pH = 4.03

Before adding any KOH, the pH is 4.03