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 (C6H5COOAg) 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 C6H5COOAg in buffer of pH
= 3.19 is \( 1.66\times 10^{-6}mol/L\).
For, water: Assuming the solubility of silver benzoate (C6H5COOAg) is x mol/L.
Now, [Ag+] = y’ M and [CH3COO–] = y’M
Ksp =\( [Ag^{+}][C6H5COO^{-}]\)
Ksp = (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: FeSO4, MnCl2, ZnCl2 and CdCl2 . in which of these solutions precipitation will take place?
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