$K_{s p}$ of $AgBr$ is $5.0 \times 1 0^{- 13}$ (molar mass $188 g mo l^{- 1}$ ). What volume of water, in litres, is needed to dissolve $10.0 mg$ of $AgBr$ ? (nearest integer)

Numerical answer type, answer in L, tolerance 0.5.

Answer: 75 L

s = 5.0 \times 1 0^{- 13} = 7.1 \times 1 0^{- 7} M

. Moles of

AgBr

= 0.0100/188 = 5.3 \times 1 0^{- 5}

. Volume

= n / s = 75.2 L

, i.e. about 75 L.