Question

$2M$ solution of $N{a}_{2}C{O}_3$ is boiled in a closed container with excess of $Ca{F}_2$. A very small amount of $CaC{O}_3$ and $Na{F}_2$ is $y$, Find the molar concentration of $F^{⊝}$ in the resulting solution after equilibrium is attained.

• A. $(4y^3/x{)}^{1/2}$
• B. $(6y^3/x{)}^{1/2}$
• C. $(8y^3/x{)}^{1/2}$
• D. $(10y^3/x{)}^{1/2}$

Answer 1

The correct option is C $(8y^3/x{)}^{1/2}$

$N{a}_{2}C{O}_3+Ca{F}_2\left(s\right)⟶CaC{O}_3+2NaF$
Mole taken 2 0 0
Mole left (2-a) a 2a
where a is very-very small and thus assume that $CaC{O}_3$ is in soluble form

Now, $K_{sp}$ of $CaC{O}_3=x=\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]$

Also $\left[C{O}_3^{2-}\right]=2-a+a=2\therefore \left[C{a}^{2+}\right]=\frac{x}{2}$

For $Ca{F}_2\rightleftharpoons C{a}^{2+}+2F^{⊝}$

$K_{sp\left(Ca{F}_2\right)}=\left[C{a}^{2+}\right][F^{⊝}{]}^2=(y\left)\right(2y{)}^2=4y^3.$

Further for $\left[F^{⊝}\right]$, we can have

$\left[F^{⊝}\right]=\left[F^{⊝}\right]$ from, $Ca{F}_2+\left[F^{⊝}\right]$ from NaF

$\left[F^{⊝}\right]=\sqrt{\frac{K_{sp\left(Ca{F}_2\right)}}{\left[C{a}^{2+}\right]}}+$Negligible value

$\left[F^{⊝}\right]=\sqrt{\frac{4y^3}{x/2}}=\sqrt{\frac{8y^3}{x}}$