Compute for the molar solubility of cadmium hydroxide which is slightly soluble in water if the Ksp value is 6.0 x 10-12.
From the balanced net ionic equation;
Cd(aq)2++2OH(aq)−⇋Cd(OH)2(aq)Cd^{2+}_{(aq)}+2OH^{-}_{(aq)}\leftrightharpoons Cd(OH)_{2_{(aq)}}Cd(aq)2++2OH(aq)−⇋Cd(OH)2(aq)
Ksp=[Cd2+][OH−]K_{sp}=[Cd^{2+}][OH^-]Ksp=[Cd2+][OH−]
6.0×10−12=(x).(2x2)6.0\times 10^{-12}=(x).(2x^2)6.0×10−12=(x).(2x2)
4x3=6.0×10−124x^3=6.0\times 10^{-12}4x3=6.0×10−12
x3=1.5×10−12x^3=1.5\times 10^{-12}x3=1.5×10−12
x=3.67×10−6x=3.67\times 10^{-6}x=3.67×10−6 MMM
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