Answer:
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
