Calculate the pH of the solution when the equivalence point has been reached in the titration of 0.0450 M KOH against 25.00 mL of 0.233 M HOCl. pKa = 7.45.

Answer 1

Let's look at the balanced equation for this reaction:

KOH(aq) + HOCl(aq) ==> KClO(aq) + H₂O(l)

At the equivalence point, moles KOH = moles HOCl

moles HOCl = 25.00 ml x 1 L/1000 ml x 0.233 mol/L = 0.005825 moles HOCl

Since there will also be 0.005825 mol KOH, we can now find the volume of KOH...

volume KOH: (x L)(0.0450 mol/L) = 0.005825 mol and x = 0.129 L = 129 ml

TOTAL VOLUME at equivalence = 129 ml + 25.0 ml = 154 mls = 0.154 L

Since the titration is between a strong base (KOH) and a weak acid (HOCl), the salt that is produced (KClO) will produce an alkaline pH when dissolved in water. The FINAL [ClO-] = .005825 mol/0.154 L = 0.962 M

To find the pH, we need to look at the hydrolysis of KClO or simply of ClO-....

ClO- + H2O ==> HClO + OH-

From pKa for HClO we can find the pKb for ClO- and then Kb as follows:

pKb = 14 - pKa = 14 - 7.45 = 6.55 and Kb = 1x10^-6.55 = 3.55x10^-7

Then we use the expression:

Kb = [HClO][OH-]/[ClO^-]

3.55x10^-7 = (x)(x)/0.962

x² = 3.41x10^-7

x = 5.84x10^-4 = [OH^-]

pOH = -log 5.84x10^-4 = 3.23

pH = 10.8