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

a) s = 8,1x10⁻³ mol/L

b) 29,7 g/100mL

c) 18,0 M

Step-by-step 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:

`(8,1x10^(-3) mol)/(L)`×
`(367,0g)/(mol)`×
`(0,1L)/(100mL)` = 0,297 g/100mL

c) The concentration in mol/L is:

`(6,6g)/(mL)`×
`(1mol)/(367g)`×
`(1000mL)/(1L)` = 18,0 mol/L

I hope it helps!