We know, $$\Delta $$G^o = – 2.303 RT log K_sp

$$ \therefore $$ 63300 = – 2.303 × 8.314 × 298 log K_sp

$$ \Rightarrow $$ log K_sp = -11.09

$$ \Rightarrow $$ K_sp = 10^-11.09

= 8.0 × 10^–12

Na₂CO₃ is salt of strong base, NaOH and weak acid, H₂CO₃ hence the pH value of the solution will be high.

Given, K_a = 1 × 10^–4

pK_a = – log (1× 10^–4) = 4

pH = pK_a + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ 5 = 4 + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 1

$$ \Rightarrow $$ $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 10 = 10 : 1

CaCO3	$$ \to $$	Ca2+	+	CO32-
		x		x

CaC2O4	$$ \to $$	Ca2+	+	C2O42-
		y		y

[Ca²⁺] = x + y

Now, K_sp (CaCO₃) = [Ca²⁺] [CO₃ ^2-]

or 4.7 × 10^–9 = (x + y)x .......(1)

similarly, K_sp (CaC₂O₄) = [Ca²⁺] [C₂O₄^2–]

or 1.3 × 10^–9 = (x + y)y .......(2)

Dividing equation (1) and (2), we get

$${x \over y} = 3.6$$

$$ \therefore $$ x = 3.6y

Putting in equation (2) we get

y(3.6y + y) = 1.3 × 10^–9

$$ \Rightarrow $$ y = 1.68 $$ \times $$ 10^-5

and x = 3.6 $$ \times $$ 1.68 $$ \times $$ 10^-5 = 6.048 $$ \times $$ 10^-5

$$ \therefore $$ [Ca²⁺] = (x + y) = (6.048 $$ \times $$ 10^-5) + (1.68 $$ \times $$ 10^-5)

$$ \therefore $$ [Ca²⁺] = 7.746 × 10^–5 M