Calculate the solubility in mol/l of pbbr2(s) (ksp = 2.1 • 10-6, mw = 367.0 g/mol)

(b) convert your answer to part a to g/100 ml

(c) what is the concentration (in mol/l) of pbbr2(s) given that its density is 6.66 g/ml?

a) s = 8,1x10⁻³ mol/L

b) 29,7 g/100mL

c) 18,0 M

Explanation:

For the equilibrium:

PbBr₂ ⇄ Pb²⁺ + 2 Br⁻

You can take the molar solubility of PbBr₂ (s) as:

[Pb²⁺] = s

[Br⁻] = 2s

As ksp is defined as:

ksp = [Pb²⁺] [Br⁻]²

Replacing:

2,1x10⁻⁶ = s * (2s)²

2,1x10⁻⁶ = 4s³

s = ∛2,1x10⁻⁶/4

s = 8,1x10⁻³ mol/L

b) The answer is g/100mL is:

×× = 0,297 g/100mL

c) The concentration in mol/L is:

×× = 18,0 mol/L

I hope it helps!

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