Question

Write a balanced net ionic equation to show why the solubility of Ca(OH)2(s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. Use the pull-down boxes to specify states such as (aq) or (s). + + K =

Answer 1

Solubility equilibrium for Ca(OH)2 is :

Ca(OH)₂(s)<------>Ca²⁺(aq) + 2OH^-(aq), K₁= Ksp = 7.9×10^-6

And in strong acid H₃O⁺ are present which would react with OH^- as follows H₃​​​​​​O⁺(aq) + OH^{​​​​​​-}(aq) <-----> 2H2O(l), K₂=10¹⁴

Combining two times of second equation with Ca(OH)2 equation and cancelling common terms, we get net ionic equation as follows:

Ca(OH)₂(s) + 2H₃​​​​​​O⁺(aq) <-----> Ca²⁺(aq) +4H₂​​​​​​O(l),

Note if you write H+ in place of H3O+ then write 2H+ on left and 2H2O on right side instead of 4H2O.

Keq= K₁×K₂² (as second equation is multiplied by two, so K2 raised to power 2)= 7.9×10^-6×(10^{​​​​​14})²= 7.9×10²², which is very high.

Comment in case of any doubt.