Question

The solubility product of AgBr is 5.2 × 10⁻¹³ Calculate its solubility in mol dm⁻³ and g dm⁻³ (Molar mass of AgBr = 187.8g mol⁻¹)

Answer 1

The solubility product of AgBr is:

AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)

x = 1, y = 1

K_sp = [Ag⁺][Br⁻] = S² 

`"S" = sqrt ("K"_(sp)) = sqrt (5.2 × 10^(−13))` 

= 7.2 × 10⁻⁷ Mol dm⁻³ 

The solubility in g dm⁻³  = molar solubility in mol dm⁻³ × molar mass g mol⁻¹ 

S = 7.2 × 10⁻⁷ mol dm⁻³ × 187.8 g mol⁻¹ 

= 1.35 × 10⁻⁴ g dm⁻³