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Problems · Problem 6.20

Q.Calculate the pH of 0.08M solution of hypochlorous acid, HOCl. The ionization constant of the acid is 2.5 × 10⁻⁵. Determine the percent dissociation of HOCl.

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HOCl is a weak acid: [H+]=KaC=1.41×10−3[\text{H}^+] = \sqrt{K_a C} = 1.41\times10^{-3} M, giving pH=2.85\text{pH} = 2.85 and a percent dissociation of 1.76%1.76\%.

HOCl⇌H++OCl−,Ka=2.5×10−5,C=0.08 M\text{HOCl} \rightleftharpoons \text{H}^+ + \text{OCl}^-, \qquad K_a = 2.5\times10^{-5}, \quad C = 0.08 \text{ M}

1. Equilibrium expression. With x=[H+]=[OCl−]x = [\text{H}^+] = [\text{OCl}^-]:

Ka=x20.08−xK_a = \frac{x^2}{0.08 - x}

2. Small-xx approximation (KaK_a is small, so 0.08−x≈0.080.08 - x \approx 0.08):

x2≈KaC=(2.5×10−5)(0.08)=2.0×10−6x^2 \approx K_a C = (2.5\times10^{-5})(0.08) = 2.0\times10^{-6}

x=2.0×10−6=1.41×10−3 Mx = \sqrt{2.0\times10^{-6}} = 1.41\times10^{-3} \text{ M}

3. pH. …

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