Question

What will be the pH at the equivalence point in the titration of 45.00 mL of 0.115 M NaClO with 0.106 M HCl?

ClO^–(aq) + HCl(aq) → HClO(aq) + Cl^–(aq) Ka(HClO)=3.0·10^-8  

You only need to provide the final answer (pH).

Don't forget significant figures!

Answer 1

moles of NaClO = 0.115 mol/L * (45/1000)L = 5.175*10^-3 mol
mol of HCl required to reach the equivalence point = 5.175*10^-3 mol
volume of HCl = 5.175*10^-3 mol / 0.106mol/L = 0.04882
total volume = 0.04882L + 0.045 = 0.09382 L
ClO^- = (5.175*10^-3 mol / 0.09382 L) = 0.05515 M
HClO + H₂O <----------> ClO^- + H₃O⁺
Let X be [H₃O⁺] and ClO^-
At equilibrium,
K_a = X² / (0.05515-X) = 3*10^-8
X² = 3*10^-8 [ 0.05515-X]
X² - 1.6545*10^-9 + 3*10^-8 X = 0
Solve for X ,
X = 4.06605*10^-5 M = [H₃O⁺]
PH = -log [H₃O⁺] = -log [4.06605*10^-5] = 4.3908 ~ 4.4
Therefore PH = 4.4