d
\(\mathrm{CaCO}_{3} \rightarrow \mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-}\)

\(\quad \quad \quad \quad \quad \quad x \quad \quad x\)

\(\mathrm{CaC}_{2} \mathrm{O}_{4} \rightarrow \mathrm{Ca}^{2+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)

\(\quad \quad \quad \quad \quad \quad y \quad \quad y\)

Now, \(\left[\mathrm{Ca}^{2+}\right]=x+y\)

and \(x(x+y)=4.7 \times 10^{-9}\)

\(y(x+y)=1.3 \times 10^{-9}\)

Dividing equation \((i)\) and \((ii)\) we get \(\frac{x}{y}=3.6\)

\(\therefore \quad x=3.6 y\)

Putting this value in equation \((ii)\), we get

\(y(3.6 y+y)=1.3 \times 10^{-9}\)

On solving, we get \(y=1.68 \times 10^{-5}\) and \(x=3.6 \times 1.68 \times 10^{-5}=6.048 \times 10^{-5}\)

\(\therefore \quad\left[\mathrm{Ca}^{2+}\right]=(x+y)=\left(1.68 \times 10^{-5}\right)+\left(6.048 \times 10^{-5}\right)\)

\(\therefore \quad\left[\mathrm{Ca}^{2+}\right]=7.728 \times 10^{-5}\, \mathrm{M}\)