Question

The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated: (a) AgBr: [Ag+] = 5.7 × 10^-7 M, [Br–] = 5.7 × 10^-7 M (b) CaCO3: [Ca^2+] = 5.3 × 10^-3 M, [CO3^2−] = 9.0 × 10^-7 M (c) PbF2: [Pb^2+] = 2.1 × 10^-3 M, [F–] = 4.2 × 10^-3 M (d) Ag2CrO4: [Ag+] = 5.3 × 10^-5 M, 3.2 × 10^-3 M (e) InF3: [In3+] = 2.3 × 10^-3 M, [F–] = 7.0 × 10^-3 M

Answer 1

(a) Solubility product constant or Ksp expression for the reaction is

$AgBr(s) \rightarrow Ag^+(aq) + Br^-(aq)$

$Ksp = [Ag^+][Br^-]$

$Ksp = [5.7 \times 10^{-7}] [5.7 \times 10^{-7}] = 32.49 \times 10^{-14}=3.25 \times 10^{-13}$

(b) Ksp expression for the reaction

$CaCO_3(s)\rightarrow Ca^{2+}(aq)+ CO_3^{2-}(aq)$

$Ksp= [Ca^{2+}] [CO_3^{2-}]$
$Ksp= [ 5.3\times 10^{-3 }][9.0 \times 10^{-7} ]= 47.7 \times 10^{-10}= 4.77 \times 10^{-9}$

(c) Ksp expression for the reaction

$PbF_2 \rightarrow Pb^{2+} + 2F^-$

$Ksp= [Pb^{2+}] [F^{-}]^2$

$Ksp= [2.1 \times 10^{-3} ][4.2 \times 10^{-3}]^2 =37.044 \times 10^{-9}= 3.7 \times 10^{-8}$

(d) Ksp expression for the reaction

$Ag_2CrO_4(s)\rightarrow 2Ag^+ (aq)+ CrO_4^-(aq)$

$Ksp= [Ag^{+}]^2 [CrO_4^{-}]$

$Ksp= [5.3 \times 10^{-5}]^2 [3.2 \times 10^{-3}]= 89.88 \times 10^{-13} = 8.9 \times 10^{-12}$​

(e) Ksp expression for the reaction

$InF_3 (s) \rightarrow In^{3+}(aq)+ 3F^-(aq)$

$Ksp= [In^{3+}] [F^{-}]^3$

​$Ksp= [2.3 \times 10^{-3}] [7.0 \times 10^{-3} ]^3= 788.9 \times 10^{-30}= 7.89 \times 10^{-28}$