Question

consider the following two salts, whose $k_{sp}$ values have been reported at $25^{circ}c$:
$agcl (k_{sp}=1.6\times10^{-10})$
$ag_{2}co_{3}(k_{sp}=8.1\times10^{-12})$
which of these salts has the lower solubility in water at $25^{circ}c$?
a. $agcl$
b. $ag_{2}co_{3}$

Explanation:

Step1: Write the solubility - product expressions

For \(AgCl\), the dissolution equation is \(AgCl(s)
ightleftharpoons Ag^{+}(aq)+Cl^{-}(aq)\). Let the solubility of \(AgCl\) be \(s_1\). Then \(K_{sp}(AgCl)=[Ag^{+}][Cl^{-}]\), and since \( [Ag^{+}]=[Cl^{-}]=s_1\), we have \(K_{sp}(AgCl)=s_1^{2}\).
For \(Ag_2CO_3\), the dissolution equation is \(Ag_2CO_3(s)
ightleftharpoons 2Ag^{+}(aq)+CO_3^{2 - }(aq)\). Let the solubility of \(Ag_2CO_3\) be \(s_2\). Then \(K_{sp}(Ag_2CO_3)=[Ag^{+}]^{2}[CO_3^{2 - }]\), and since \( [Ag^{+}]=2s_2\) and \( [CO_3^{2 - }]=s_2\), we have \(K_{sp}(Ag_2CO_3)=(2s_2)^{2}\times s_2 = 4s_2^{3}\).

Step2: Calculate the solubilities

For \(AgCl\):

$$s_1=\sqrt{K_{sp}(AgCl)}=\sqrt{1.6\times 10^{-10}} = 1.26\times 10^{-5}\space mol/L$$

For \(Ag_2CO_3\):

$$s_2=\sqrt[3]{\frac{K_{sp}(Ag_2CO_3)}{4}}=\sqrt[3]{\frac{8.1\times 10^{-12}}{4}}=\sqrt[3]{2.025\times 10^{-12}}\approx1.27\times 10^{-4}\space mol/L$$

Answer:

A. \(AgCl\)