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

the answer is the first option

explanation:

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