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प्रश्न
The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
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उत्तर
Since pH = 3.19,
`["H"_3"O"^+] = 6.46 xx 10^(-4) "M"`
\[\ce{C_6H_5COOH + H_2O <-> C_6H_5COO- + H_3O}\]
`"K"_"a" = (["C"_6"H"_5"COO"^(-)]["H"_3"O"^(+)])/["C"_6"H"_5"COOH"]`
`["C"_6"H"_5"COOH"]/["C"_6"H"_5"COO"^(-)] = ["H"_3"O"^+]/["K"_"a"]`
`= (6.46 xx 10^(-4))/(6.46 xx 10^(-5)) = 10`
Let the solubility of C6H5COOAg be x mol/L.
Then,
`["Ag"^(+)] = x`
`["C"_6"H"_5"COOH"] + ["C"_6"H"_5"COO"^(-)] = x`
`10["C"_6"H"_5"COO"^(-)] + ["C"_6"H"_5"COO"^(-)] = x`
`["C"_6"H"_5"COO"^(-)] = x/11`
`"K"_("sp")["Ag"^(+)]["C"_6"H"_5 "COO"^(-)]`
`2.5 xx 10^(-13) = x(x/11)`
`x = 1.66 xx 10^(-6) "mol"/"L"`
Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10–6 mol/L.
Now, let the solubility of C6H5COOAg be x’ mol/L
Then `["Ag"^(+)] = x'"M"` and `["CH"_3"COO"^(-)] = x'"M"`
`"K"_("sp") = ["Ag"^+] ["CH"_3"COO"^-]`
`"K"_("sp") = (x')^2`
`x' = sqrt("K"_("sp")) = sqrt(2.5 xx 10^(-13)) = 5xx 10^(-7) "mol"/"L"`
`therefore x/(x') = (1.66 xx 10^(-6))/5xx10^(-7) = 3.32`
Hence, C6H5COOAg is approximately 3.317 times more soluble in a low pH solution.
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