Question

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?

Answer 1

Answer:

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:

$\frac{8,1x10^{-3} mol}{L}$×$\frac{367,0g}{mol}$×$\frac{0,1L}{100mL}$ = 0,297 g/100mL

c) The concentration in mol/L is:

$\frac{6,6g}{mL}$×$\frac{1mol}{367g}$×$\frac{1000mL}{1L}$ = 18,0 mol/L

I hope it helps!