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

^{-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 + H_{2}O <----------> ClO^{-} +
H_{3}O^{+}

Let X be [H_{3}O^{+}] and ClO^{-}

At equilibrium,

K_{a} = X^{2} / (0.05515-X) =
3*10^{-8}

X^{2} = 3*10^{-8} [ 0.05515-X]

X^{2} - 1.6545*10^{-9} + 3*10^{-8} X =
0

Solve for X ,

X = 4.06605*10^{-5} M = [H_{3}O^{+}]

PH = -log [H_{3}O^{+}] = -log
[4.06605*10^{-5}] = 4.3908 ~ 4.4

Therefore PH = 4.4

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).

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