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The solubility product of AgClin water is 1.5xx10^(-10). Calculate its solubility in 0.01 M NaCl aqueous solution . |
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Answer» SOLUTION :As NaCl dissociates completely, therefore, in 0.01 M NaCl solution, `[Cl^(-)] = 0.01M` If SOLUBILITY of AgCl in 0.01 M NaCl solution is s MOL `L^(-1)`, then from AgCl that dissolves, `[Ag^(+)]=[Cl^(-)]=s ` mol `L^(-1) :. ` Total `[Cl^(-)]=0.01 + s ~= 0.01M` `K_(SP) ` for AgCl `=[Ag^(+)] [ Cl^(-)] = s xx 0.01 = 0.001 s :. 0.01 s = 1.5 xx 10^(-5) xx 10^(-10) or s= 1.5 xx 10^(-8) M` |
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