Question

The ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water is:
(Given: Solubility product of AgCl = 10⁻¹⁰)

• A. 10⁻⁴
• B. 10⁻⁶
• C. 10⁻⁹
• D. 10⁻⁵

Answer 1

The Correct Option is A

The question involves calculating the ratio of the solubility of AgCl in a 0.1 M KCl solution to its solubility in pure water, using the given solubility product of silver chloride (AgCl), which is \(10^{-10}\).

For the solubility in water, let's define:

• \(S_w\) = solubility of AgCl in pure water

Since AgCl dissociates as:

• \(AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}\)

The solubility product expression is:

• \(K_{sp} = [Ag^+][Cl^-] = S_w \cdot S_w = S_w^2\)\)

Hence,

• \(S_w^2 = 10^{-10}\)
• \(S_w = \sqrt{10^{-10}} = 10^{-5}\)

Now consider AgCl's solubility in 0.1 M KCl solution:

• Initial concentration of Cl\(^{-}\) from KCl = 0.1 M
• So let the solubility of AgCl in this solution be \(S_{KCl}\)

The total Cl\(^-\) concentration becomes:

• \([Cl^-] = 0.1 + S_{KCl}\)

Assume \(S_{KCl}\) is very small compared to 0.1 M, such that:

• \([Cl^-] \approx 0.1\)

Solubility product assumption gives:

• \(K_{sp} = [Ag^+][Cl^-] = S_{KCl} \cdot 0.1 = 10^{-10}\)

Solving for \(S_{KCl}\):

• \(S_{KCl} = \frac{10^{-10}}{0.1} = 10^{-9}\)

The ratio of the solubility of AgCl in 0.1 M KCl solution to its solubility in water is then:

• \(\frac{S_{KCl}}{S_w} = \frac{10^{-9}}{10^{-5}} = 10^{-4}\)

Hence, the correct answer is: 10⁻⁴.