Calculate the ph for each case in the titration of 50.0 ml of 0.150 m hclo (aq) with 0.150 m koh (aq). use the ionization constant for hclo

Answer;

pH = 7.58

Explanation;

The KOH will react with the HClO to produce KClO. You will have a solution containing unreacted HClO and KOH. This is a buffer solution.

Equation:

HClO + KOH → KClO + H2O

HClO reacts with KOH in 1:1 molar ratio

Mol HClO in 50mL of 0.150M solution = 50/1000*0.150 = 0.0075 mol HClO

Mol KOH in 30mL of 0.150M solution = 30/1000*0.150 = 0.0045 mol KOH

These react to produce 0.0045 mol KClO and there is 0.0030 mol HClO unreacted

Volume of solution = 50mL + 30mL = 80mL = 0.080L

Molarity of HClO in solution = 0.0030/0.080 = 0.0375M

Molarity of KClO in solution = 0.0045/0.080 = 0.0562M

Using the Henderson-Hasselbalch equation we can calculate the pH;

pKa HClO = - log (4.0*10^-8) = 7.40

pH = pKa + log ([KClO]/[HClO])

pH = 7.40 + log (0.0562/0.0375)

pH = 7.40 + log 1.50

pH = 7.40 + 0.18

pH = 7.58