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Question: A solution contains \(1.0 \times 1{0^{ - 2}}\)mol of KI and 0.10 mol of KCl per liter. \(AgN{O_3}\)is gradually added to this solution. Which forms first, solid AgI or solid AgCl?

Short Answer

Expert verified

The solid AgI will form first.

Step by step solution

01

Find which will form first:

For AgCl,

\(\begin{array}{*{20}{c}}{{K_{sp}}(AgCl) = \left( {A{g^ + }} \right)\left( {C{l^ - }} \right) = 1.8 \times {{10}^{ - 10}}}\\{\left( {A{g^ + }} \right) = \frac{{1.8 \times {{10}^{ - 10}}}}{{0.10M}} = 1.8 \times {{10}^{ - 9}}M}\end{array}\)

For AgI,

\(\begin{array}{*{20}{c}}{{K_{sp}}(AgI) = \left( {A{g^ + }} \right)\left( {{I^ - }} \right) = 1.5 \times {{10}^{ - 16}}}\\{\left( {A{g^ + }} \right) = \frac{{1.5 \times {{10}^{ - 16}}}}{{1 \times {{10}^{ - 2}}M}} = 1.5 \times {{10}^{ - 9}}M}\end{array}\)

\(\left( {A{g^ + }} \right)(AgCl) > \left( {A{g^ + }} \right)(AgI)\)

Therefore the solution is solid AgI will form first.

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