The solubility-product constant for silver chromate in equilibrium with its constituent ions in water at 25°C is 1.1 × 10^−12 M³. Calculate the value of [Ag^+] that results when pure water is saturated with Ag2CrO4(s).

Question

The solubility-product constant for silver chromate in equilibrium with its constituent ions in water at 25°C is [latex]1.1 × 10^{−12}\ M^3[/latex]. Calculate the value of [latex][Ag^+][/latex] that results when pure water is saturated with [latex]Ag_{2}CrO_{4}(s)[/latex].

Accepted Answer

The [latex]Ag_{2}CrO_{4}(s)[/latex] solubility equilibrium can be described by

[latex]Ag_{2}CrO_{4}(s) ⇋ 2\ Ag^+(aq) + CrO_{4}^{2−}(aq)[/latex]

and the corresponding solubility-product expression is

[latex]K_{sp} = [Ag^+]^2[CrO_{4}^{2−}] = 1.1 × 10^{−12}\ M^3[/latex]

Each [latex]Ag_{2}CrO_{4}(s)[/latex] formula unit that dissolves yields two [latex]Ag^+(aq)[/latex] ions and one [latex]CrO_{4}^{2−}(aq)[/latex] ion; thus, [latex][Ag^+] = 2s[/latex] and [latex][CrO_{4}^{2−}] = s[/latex]. Substitution of this result into the [latex]K_{sp}[/latex] expression for [latex]Ag_{2}CrO_{4}(s)[/latex] yields

[latex]K_{sp} = (2s)^2(s) = 1.1 × 10^{−12}\ M^3[/latex]

Solving for s yields

[latex]s= 6.5 × 10^{−5}\ M[/latex]

The value of [latex][Ag^+][/latex] is

[latex][Ag^+] = 2s = 1.3 × 10^{−4}\ M[/latex]

Question 22.2: The solubility-product constant for silver chromate in equil......

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