Question

The molar solubility of AgBr in pure water is 7.3x10^-7 M. Calculate the Ksp.

Answer 1

Determine the K_sp of silver bromide, given that its molar solubility is 7.3 x 10¯⁷ moles per liter.

When AgBr dissolves, it dissociates like this:

AgBr (s) ⇌ Ag⁺ (aq) + Br¯ (aq)

The K_sp expression is:

K_sp = [Ag⁺] [Br¯]

There is a 1:1 molar ratio between the AgBr that dissolves and Ag⁺ that is in solution. In like manner, there is a 1:1 molar ratio between dissolved AgBr and Br¯ in solution. This means that, when 7.3 x 10¯⁷mole per liter of AgBr dissolves, it produces 7.3 x 10¯⁷ mole per liter of Ag⁺ and 7.3 x 10¯⁷ mole per liter of Br¯ in solution.

Putting the values into the K_sp expression, we obtain:

K_sp = (7.3 x 10¯⁷) (7.3 x 10¯⁷) = 5.329 x 10¯¹³