Question 66 of 69advanced🔧 ApplyNumerical5 marks

The ionization constant of benzoic acid is 6.46×10−56.46 \times 10^{-5} and KspK_{sp} for silver benzoate is 2.5×10−132.5 \times 10^{-13}. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Correct Answer

Silver benzoate is approximately 3.32 times more soluble in the buffer of pH 3.19 compared to pure water.

Exercise: EXERCISES | Q: 6.70 | (Page 46)
For More Understanding

Explanation

The solubility of a salt of a weak acid increases in an acidic medium because the common ion effect is suppressed by the reaction of the anion with H+ ions. The formula relating solubility S, Ksp, Ka, and [H+] is S = {Ksp * ([H+] + Ka) / Ka}^{1/2}. In pure water, solubility S0 is simply the square root of Ksp. By calculating both and taking their ratio, we find the factor by which solubility increases.

Solution Steps

  1. Step 1: Calculate solubility in pure water (S0S_0) In pure water, Ksp=[Ag+][B−]=S02K_{sp} = [Ag^+][B^-] = S_0^2 S0=Ksp=2.5×10−13=5×10−7S_0 = \sqrt{K_{sp}} = \sqrt{2.5 \times 10^{-13}} = 5 \times 10^{-7} M

  2. Step 2: Calculate [H+][H^+] in the buffer Given pH = 3.19 [H+]=10−3.19=6.46×10−4[H^+] = 10^{-3.19} = 6.46 \times 10^{-4} M

  3. Step 3: Calculate solubility in the buffer (SS) Using the formula S={Ksp([H+]+Ka)Ka}1/2S = \left\{K_{sp} \frac{([H^+] + K_a)}{K_a}\right\}^{1/2} Given Ka=6.46×10−5K_a = 6.46 \times 10^{-5} Notice that [H+]=6.46×10−4=10×Ka[H^+] = 6.46 \times 10^{-4} = 10 \times K_a

    S={2.5×10−13×10Ka+KaKa}1/2S = \left\{2.5 \times 10^{-13} \times \frac{10 K_a + K_a}{K_a}\right\}^{1/2} S={2.5×10−13×11}1/2S = \left\{2.5 \times 10^{-13} \times 11\right\}^{1/2} S=11×2.5×10−13=11×S0S = \sqrt{11} \times \sqrt{2.5 \times 10^{-13}} = \sqrt{11} \times S_0

  4. Step 4: Find the ratio of solubilities

    SS0=11≈3.317\frac{S}{S_0} = \sqrt{11} \approx 3.317

    Thus, silver benzoate is approximately 3.32 times more soluble in the buffer.

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