Question

The value of ksp for silver carbonate, ag2co3, is 8.10×10−12. calculate the solubility of ag2co3 in grams per liter.

Answer 1

0.0349 g/L

Step-by-step explanation:

Let s = the molar solubility.

Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq);
`K_(sp)` = 8.10 × 10⁻¹²

E/mol·L⁻¹: 2s s

`K_(sp)` =[Ag⁺]²[CO₃²⁻] = (2s)²×s = 4s^3 = 8.10 × 10⁻¹²

`s^(3)= (8.10 * 10^(-12))/(4) = 2.025 * 10^(-12)`

`s = \sqrt[3]{2.025* 10^(-12)} \text{ mol/L}= 1.27 * 10^(-4) \text{ mol/L}`

`s = 1.27 * 10^(-4) \text{ mol/L} * 275.75 \text{ g/mol} = \text{0.0349 g/L}`