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Exercise · Q34

Q.Equal volumes of 0.002 M AgNO30.002\ \text{M}\ \text{AgNO}_3 solution and 0.002 M NaCl0.002\ \text{M}\ \text{NaCl} solution are mixed. Given Ksp(AgCl)=1.8×10−10K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}, determine by calculation whether a precipitate of AgCl\text{AgCl} will form.

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When equal volumes of the two solutions are mixed, the total volume doubles, so each original concentration is halved: [Ag+]=0.002/2=0.001 M[\text{Ag}^+] = 0.002/2 = 0.001\ \text{M} and [Cl−]=0.002/2=0.001 M[\text{Cl}^-] = 0.002/2 = 0.001\ \text{M} immediately after mixing (before any precipitation has occurred). The ionic product is then Q=[Ag+][Cl−]=(0.001)(0.001)=1×10−6Q = [\text{Ag}^+][\text{Cl}^-] = (0.001)(0.001) = 1\times10^{-6}. Comparing with Ksp(AgCl)=1.8×10−10K_{sp}(\text{AgCl}) = 1.8\times10^{-10}: since Q (1×10−6)≫Ksp (1.8×10−10)Q\ (1\times10^{-6}) \gg K_{sp}\ (1.8\times10^{-10}), the solution is far more concentrated in Ag+\text{Ag}^+ and Cl−\text{Cl}^- than the …

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