Question

what is the molar solubility of SrSO4 (Ksp = 7.6 x 10 -7)

Answer 1

the dissociation of SrSO4 follows a weak equilibrium, so\[SrSO_4(s) \rightleftharpoons Sr^{+2}(aq) + SO_4^{-2}(aq)\] with an equilibrium value of 7.6x10-7. This means that:\[[Sr^{+2}]*[SO_4^{_2}] = 7.6*10^{-7}\] since one formula unit of SrSO4 produces one Sr+2 ion and one SO4-2 ion, the number of ions in a solution must be equal. Whatever they are, the concentrations of the two ions have to be equal, and they have to multiply out to 7.6*10-7\[[x]*[x] = 7.6*10^{-7}\]\[x = [Sr^{+2}] = [SO_4^{-2}] = \sqrt{7.6*10^{-7}} = 8.72*10^{-4}M\] the value of x is in terms of concentration, so it gets units of mol/L, or M.