the solubility of cuco₃ in water at 25 °c is measured to be 0.0019 $\frac{g}{l}$. use this information to…

Explanation:

Step1: Calculate molar solubility

First, find the molar mass of $CuCO_3$. $M_{Cu}=63.55\ g/mol$, $M_{C}=12.01\ g/mol$, $M_{O}=16.00\ g/mol$. So $M_{CuCO_3}=63.55 + 12.01+3\times16.00=123.56\ g/mol$. Molar solubility $s=\frac{0.0019\ g/L}{123.56\ g/mol}=1.54\times 10^{-5}\ mol/L$.

Step2: Write dissociation equation

The dissociation of $CuCO_3$ in water is $CuCO_3(s)\rightleftharpoons Cu^{2 +}(aq)+CO_3^{2 -}(aq)$.

Step3: Determine ion - concentrations

From the dissociation equation, $[Cu^{2+}]=s$ and $[CO_3^{2 -}]=s$.

Step4: Calculate $K_{sp}$

$K_{sp}=[Cu^{2+}][CO_3^{2 -}]$. Substituting $[Cu^{2+}]=s$ and $[CO_3^{2 -}]=s$, we get $K_{sp}=s\times s=(1.54\times 10^{-5})^2 = 2.37\times10^{-10}$.

Answer:

$2.4\times 10^{-10}$