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The solubility product of AgCl is Ksp = 1.8 x 10^-10 at 298 K.…
Question
The solubility product of AgCl is Ksp = 1.8 x 10^-10 at 298 K. Calculate the molar solubility (in mol L^-1) of AgCl in a 0.10 M NaCl solution due to the common-ion effect. Give the answer in scientific form.
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Step-by-step solution
Let solubility in 0.10 M NaCl be s mol L^-1. Then [Ag+] = s and [Cl-] = 0.10 + s ~ 0.10 (since s is tiny). Ksp = [Ag+][Cl-] = s(0.10) = 1.8e-10, so s = 1.8e-10/0.10 = 1.8e-9 mol L^-1. Final answer: 1.8e-9
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