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Question: x *10^-6M is the solubility of silver benzoate in a buffer solution having [H]=6.3×10^-4M. given tha...

x *10^-6M is the solubility of silver benzoate in a buffer solution having [H]=6.3×10^-4M. given that the Ka of benzoic acid is 6.3×10^-5 while the solubility product (Ksp) of silver benzoate is equal to 2.5×10^-13 Find value of x. (Given: root11=3.316)

Answer

1.66

Explanation

Solution

Solution:

  1. In pure water, the solubility SS (mol/L) of Ag-benzoate is given by:

    S=Ksp=2.5×10−13=5.0×10−7 M.S = \sqrt{K_{sp}} = \sqrt{2.5\times10^{-13}} = 5.0\times10^{-7}\,M.

  2. In the buffer solution, benzoate ion reacts with H⁺:

    C6H5COO−+H+⇌C6H5COOH.\text{C}_6\text{H}_5\text{COO}^- + H^+ \rightleftharpoons \text{C}_6\text{H}_5\text{COOH}.

    Thus, the effective concentration of free benzoate is reduced. The relation becomes:

    Ksp=S′2(KaKa+[H+]),K_{sp} = S'^2\left(\frac{K_a}{K_a + [H^+]}\right),

    where S′S' is the solubility in the buffer.

  3. Rearranging,

    S′=Ksp(Ka+[H+]Ka).S' = \sqrt{K_{sp}\left(\frac{K_a + [H^+]}{K_a}\right)}.

  4. Substitute the given values:

    Ka=6.3×10−5,[H+]=6.3×10−4,Ksp=2.5×10−13.K_a = 6.3\times10^{-5},\quad [H^+] = 6.3\times10^{-4},\quad K_{sp}=2.5\times10^{-13}.

    Compute the ratio:

    Ka+[H+]Ka=6.3×10−5+6.3×10−46.3×10−5=6.93×10−46.3×10−5=11.\frac{K_a + [H^+]}{K_a} = \frac{6.3\times10^{-5} + 6.3\times10^{-4}}{6.3\times10^{-5}} = \frac{6.93\times10^{-4}}{6.3\times10^{-5}} = 11.

  5. Therefore,

    S′=2.5×10−13×11=2.75×10−12≈1.66×10−6 M.S' = \sqrt{2.5\times10^{-13} \times 11} = \sqrt{2.75\times10^{-12}} \approx 1.66\times10^{-6}\,M.

  6. Since the solubility is given as x×10−6 Mx \times 10^{-6}\,M, we have:

    x≈1.66.x \approx 1.66.

Summary:

  • Explanation: Use the relation Ksp=S′2(KaKa+[H+])K_{sp}=S'^2\left(\dfrac{K_a}{K_a+[H^+]}\right) to solve for S′S'. Substitute the values and simplify to obtain S′≈1.66×10−6 MS' \approx 1.66\times10^{-6}\,M. Thus, x≈1.66x \approx 1.66.