Ksp Problems need help ASAP! Best answer get 10!

Question

1) what is the Ksp of MgCO3 when .0000019 g dissolve in 250 ml of water at 25 degrees C?

2)The Ksp for PbI2is 7.9 x 10^-9. What are the products and their concentration?

please show all steps and explain!

Answer 1

1) MW of MgCO3 is 84.3 g/mole

(1.9x10^-6 g MgCO3) / (84.3 g/mole) (0.250 L) = 9.01x10^-8 mol/L is dissolved at saturation

MgCO3 ↔ Mg++ (aq) + CO3-- (aq)

Ksp = [Mg++] [CO3--]

For every MgCO3 that dissolves, one Mg++ and one CO3-- dissolves. So,

Ksp = [9.01x10^-8 mol/L] [9.01x10^-8 mol/L]

Ksp = 8.13x10^-15 M².

2) PbI2 ↔ Pb++ (aq) + 2 I- (aq)

Ksp = [Pb++] [I-]²

Let molar solubility = x. For every PbI2 that dissolves, one Pb++ dissolves (x) and two I- dissolve (2x). Substituting all we know into the equation,

7.9x10^-9 = (x) (2x)², or

7.9x10^-9 = 4x³

Solving, x = 0.00125 M.

The answers are

[Pb++] = x = 0.00125 M

[I-] = 2x = 0.00251 M.

If the answers are substituted into the equilibrium equation, Ksp is the result, so the answers check.