Question

a. What is the pKa of a 0.010 M solution with a pH of 5.24?

b. CaCO₃ is dissolved in water. What is the complete charge balance equation for its dissolution?

Answer 1

If we take weak acids of the form HA they dissociate in the following manner -

HA <-> H+ + A-

Making ICE table for this equilibrium we have

HA <-> H+ + A-

I 0.01 0 0

C -0.01x +0.01x +0.01x

E 0.01-0.01x 0.01x 0.01x

So, Ka = [H+][A-]/[HA] where Ka is the equlibrium constant and [H+] ,[A-], [HA] are the equilibrium concentrations of each of them.

So, Ka = (0.01x)*0.01x/(0.01-0.01-x) = 0.01x²/(1-x)

Assuming x is negligible compared to 1 that is x<<1 we can approximate the denominator to be 1

So, Ka = 0.01x² or x = sqrt(Ka/0.01)

[H+] = 0.01x = 0.01*sqrt(Ka/0.01) = sqrt(Ka*0.01)

So, pH = -log(H+) = -log(Ka*0.01) = 5.24 so,

Ka/0.01 = 10^-5.24 , Ka = 10^-7.24 . So, pKa = -logKa = 7.24

b)

To maintain the electrical neutrality of the solution we use charge balance equation. Charge balanced equation has nothing to do with the concentration but it uses the net charge on each species formed during the reactions to form an equation where sum of positive charged species = sum of all negative charged species

When CaCO₃ is dissolve in water CaCO₃ being very less soluble in water dissolves to very less extent to form Ca⁺² and CO₃^2-

CaCO₃ -> Ca⁺² + CO₃^2-

CO₃^2- can react with water to form -

CO₃^2- + H₂O -> HCO₃^- + OH^-

also, HCO₃^- + H₂O -> H₂CO₃ + OH^-

So, the charge balance equation can be written as sum of all positive charged species

2*[Ca⁺²] (as each Ca⁺² is contributing 2 positive charge)

and sum of all negative charged species = 2[CO₃^2-] + [OH^-] + [HCO₃^-]

So,

2*[Ca⁺²] = 2[CO₃^2-] + [OH^-] + [HCO₃^-]