calculate the solubility of cuco₃ in water at 25 °c. youll find kₛₚ data in the aleks data tab. round your…

Explanation:

Step1: Write dissociation equation

$CuCO_3(s)\rightleftharpoons Cu^{2 + }(aq)+CO_3^{2 - }(aq)$ Let the solubility of $CuCO_3$ be $s$ mol/L. Then $[Cu^{2 + }]=s$ mol/L and $[CO_3^{2 - }]=s$ mol/L.

Step2: Write $K_{sp}$ expression

The $K_{sp}$ expression for $CuCO_3$ is $K_{sp}=[Cu^{2 + }][CO_3^{2 - }]$. Substituting $[Cu^{2 + }]=s$ and $[CO_3^{2 - }]=s$ into the $K_{sp}$ expression, we get $K_{sp}=s\times s=s^{2}$. From the ALEKS Data - tab, $K_{sp}(CuCO_3)=1.4\times10^{-10}$ at $25^{\circ}C$. So, $s^{2}=1.4\times 10^{-10}$.

Step3: Solve for molar solubility $s$

Taking the square - root of both sides of the equation $s^{2}=1.4\times 10^{-10}$, we have $s=\sqrt{1.4\times 10^{-10}}\approx1.2\times 10^{-5}$ mol/L.

Step4: Convert molar solubility to g/L

The molar mass of $CuCO_3$ is $M = 63.55+12.01 + 3\times16.00=123.56$ g/mol. The solubility in g/L is $s_{g/L}=s\times M$. Substituting $s = 1.2\times 10^{-5}$ mol/L and $M = 123.56$ g/mol, we get $s_{g/L}=1.2\times 10^{-5}\text{ mol/L}\times123.56\text{ g/mol}\approx1.5\times 10^{-3}$ g/L.

Answer:

$1.5\times 10^{-3}$ g/L