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Problems · Problem 3.3

Q.A weak monobasic acid is 12% dissociated in 0.05 M solution. What is percent dissociation in 0.15 M solution.

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From the 0.05 M data, Ka=7.2×10−4K_a = 7.2\times10^{-4}; at 0.15 M this gives α2=0.0693\alpha_2 = 0.0693, i.e. 6.93%6.93\% dissociation -- more concentrated, less dissociated.

Step 1. At the first concentration, α1=12%=0.12\alpha_1 = 12\% = 0.12 and c1=0.05c_1 = 0.05 M. By the small-α\alpha Ostwald relation (Eq. 3.7), Ka=α12c1=(0.12)2×0.05=0.0144×0.05=7.2×10−4K_a = \alpha_1^2 c_1 = (0.12)^2 \times 0.05 = 0.0144\times0.05 = 7.2\times10^{-4}.

Step 2. KaK_a is a true equilibrium constant -- the same at 0.15 M. So α22=Kac2=7.2×10−40.15=0.0048\alpha_2^2 = \dfrac{K_a}{c_2} = \dfrac{7.2\times10^{-4}}{0.15} = 0.0048.

Step 3. α2=0.0048=0.0693\alpha_2 = \sqrt{0.0048} = 0.0693 (a fraction, dimensionless). …

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