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The solubility of `AgCl(s)` with solubility product `1.6 xx 10^(-10)` is in 0.1 M NaCl solution would beA. ZeroB. `1.26 xx 10^(-5)M`C. `1.6 xx 10^(-9) M`D. `1.6 xx 10^(-11) M` |
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Answer» Correct Answer - C `AgCl hArr Ag^(+) + Cl^(-)` `S hArr S + 0.1` `K_(sp) = [Ag^(+)][Cl^(-)]` `1.6 xx 10^(-10) = S xx (S + 0.1) = S xx 0.1` `1.6 xx 10^(-9) = S` |
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