Question

${K_{sp}}$ values for silver bromide, silver chloride and silver iodide are $5 \times {10^{ - 12}}ml{o^2}d{m^{ - 6}},2 \times {10^{ - 10}}mo{l^2}d{m^{ - 6}},8 \times {10^{ - 17}}mo{l^2}d{m^{ - 6}}$ respectively. The order of solubility of solubility of these silver salts is:
(A) $AgCl > AgBr > AgI$
(B) $AgI > AgBr > AgCl$
(C) $AgCl > AgI > AgBr$
(D) $AgI > AgCl > AgBr$

Answer 1

Hint: At first we will write what is given in the question. We are to find the solubility of silver slats. The solubility will be decided by looking at the ${K_{sp}}$ values of the silver salts. Then in the end we will choose the correct option.

Complete step-by-step answer:
Step1. We are given with three silver chlorides. The ${K_{sp}}$ value of silver bromide is $5 \times {10^{ - 12}}ml{o^2}d{m^{ - 6}}$. The ${K_{sp}}$ value of silver chloride is $2 \times {10^{ - 10}}mo{l^2}d{m^{ - 6}}$ and the . The ${K_{sp}}$ value of silver iodide is $8 \times {10^{ - 17}}mo{l^2}d{m^{ - 6}}$.
Step2. ${K_{sp}}$is known as the solubility product constant. It is the equilibrium constant when the solid substance is dissolved into the aqueous solution. It represents the levels at which a solute is dissolving in the solution. The more value of the ${K_{sp}}$ is more the substance will be soluble.
Step3. Now the solubility is directly proportional to the ${K_{sp}}$. The more the ${K_{sp}}$ the more it will be soluble. So in above given salts. The ${K_{sp}}$ of silver chloride is most so it will be most soluble in the water. Then the second is silver bromide. It will be second most soluble. The least soluble will be silver Iodide as its ${K_{sp}}$ is least.

Hence the correct solution is Option (A) which is $AgCl > AgBr > AgI$.

Note: Silver chloride is a white chemical compound. When we heat it, it changes into the silver colour. It is naturally found as chlorargyrite. In labs it is made by combining silver nitrate with sodium chloride.