Q.The solubility product of AgBr is $5.2 \times 10^{-13}$. Calculate its solubility in mol $\mathrm{dm^{-3}}$ and g $\mathrm{dm^{-3}}$(Molar mass of AgBr = 187.8 g $\mathrm{mol^{-1}}$)

For a sparingly soluble salt in equilibrium with its own saturated solution, e.g. $B_xA_y(s)\rightleftharpoons xB^{y+}(aq)+yA^{x-}(aq)$, the solubility product is $K_{sp}=[B^{y+}]^x[A^{x-}]^y$ -- the product of the equilibrium ion concentrations, each raised to the power of its own stoichiometric coefficient. If S is the molar solubility (moles that dissolve per litre of saturated solution), the equilibrium ion concentrations are $[B^{y+}]=xS$ and $[A^{x-}]=yS$, so $K_{sp}=x^xy^yS^{x+y}$: for a 1:1 salt (AgCl, AgBr) this is simply $K_{sp}=S^2$; for a 1:2 salt (PbI2) it is $K_{sp}=4S^3$; for a 1:3 salt (Al(OH)3) it is $K_{sp}=27S^4$. The related ionic product (IP) is defined identically but is evaluated using the ACTUAL ion concentrations present under any given condition, not necessarily equilibrium/saturated concentrations -- comparing IP to Ksp for a specific mixture predicts what will happen: $IP=K_{sp}$ means exactly saturated (equilibrium); $IP>K_{sp}$ means the solution is supersaturated and precipitation will occur; $IP<K_{sp}$ means the solu …