Here, pH = 3.19

\( [H_3O^{+}] = 6.46\times 10^{-4}\)

\( C_6H_5COOH + H_2O \)

\( →C6H5COO^{-} + H_3O\)

\( K_{a}=\dfrac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{C_{6}H_{5}COOH}\)

\( K_{a}=\dfrac{[C_{6}H_{5}COOH]}{C_{6}H_{5}COO^{-}} \)

\( = \dfrac{[H_{3}O^{+}]}{K_{a}} \)

\( = \dfrac{6.46\times 10^{-4}}{6.46\times 10^{-5}} = 10\)

Assuming the solubility of silver benzoate (C_{6}H_{5}COOAg) is y mol/L.

Now, [Ag^{+}] = y

\( [C_6H_5COOH] = [C_6H_5COO^{-}] = y\)

\( 10[C_6H_5COO^{-}] + [C_6H_5COO^{-}]= y\)

\( [C_6H_5COO^{-}] = \dfrac{y}{11}\)

\( Ksp= [Ag^{+}][C_6H_5COO^{-}]2.5\times 10^{-13} \)

\( = y\dfrac{y}{11}\)

y =\( 1.66\times 10^{-6}mol/L\)

Hence, solubility of C_{6}H_{5}COOAg in buffer of pH

= 3.19 is \( 1.66\times 10^{-6}mol/L\).

For, water: Assuming the solubility of silver benzoate (C_{6}H_{5}COOAg) is x mol/L.

Now, [Ag^{+}] = y’ M and [CH_{3}COO^{–}] = y’M

K_{sp} =\( [Ag^{+}][C6H5COO^{-}]\)

K_{sp} = (y’)^{2} \( = \sqrt{K_{sp}}\)

= \( \sqrt{2.5\times 10^{-13}} \)

\( = 5\times 10^{-7} mol/L\)

\( \dfrac{y}{x} = \dfrac{1.66\times 10^{-6}}{5\times 10^{-7}} \)

Thus, the solubility of silver benzoate in water is 3.32 times the solubility of silver benzoate in pH = 3.19.

Answered by Abhisek | 1 year agoThe concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^{–19} M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO_{4}, MnCl_{2}, ZnCl_{2} and CdCl_{2} . in which of these solutions precipitation will take place?

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What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10^{–18}).

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_{sp} = 7.4 × 10^{–8} ).

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