If 0.0067g of calcium carbonate is dissolved in 1.0L of water, calculate Ksp.

Question

If 0.0067g of calcium carbonate is dissolved in 1.0L of water, calculate Ksp.

CaCO3(s) ⇌ Ca^2+(aq)+CO3^2−(aq)

a)1.34*10^-4

b)4.5*10^-5

c)none of these

d)6.7*10^-5

Answer 1

The first step would be to write out equation for the K_sp of this substance, which would be...

K_sp = [Ca²⁺][CO₃^2-]

(CaCO₃ is not included in the K_sp since it is a solid)

We know from the chemical equation that however many moles of CaCO₃ dissolve, we will get the same number of moles of each of the ions. Therefore...

K_sp = x * x = x²

Here, x represents the molar solubility of the compound, which is related to the solubility of the compound. Once you find x, you can determine the K_sp. You are given that 0.0067g of CaCO₃ dissolve in 1L of water, so the solubility is 0.0067g/L, but to get molar solubility, we must divide by the molar mass of CaCO₃. After finding that the molar mass is around 100.0868g/mol, we divide 0.0067g/L by this to get...

(0.0067g/L)/(100.0868g/mol) = 6.69E-5mol/L = x

Now we just square this to get k_sp...

K_sp = x² = (6.69E-5)² = 4.48E-9

When we round for sig figs, we get 4.5E-9