Question

The $K_a$ values of $CaC{O}_3$ and $Ca{C}_2{O}_4$ in water are $4.7\times {10}^{-9}$ and $1.3\times {10}^{-9}$, respectively, at ${25}^{o}C$. If a mixture of two is washed with $H_{2}O$ , what is $C{a}^{2+}$ ion concentration in water?

• A. $7.746\times {10}^{-5}$
• B. $5.831\times {10}^{-5}$
• C. $6.856\times {10}^{-5}$
• D. $3.606\times {10}^{-5}$

Answer 1

Answer: (A)

$7.746\times {10}^{-5}$

$\underset{x}{CaC{O}_3}\to \underset{x}{C{a}^{2+}}+\underset{y}{C{O}_3^{2-}}$
$\underset{y}{Ca{C}_2{O}_4}\to \underset{y}{C{a}^{2+}}+\underset{y}{C_2{O}_4^{2-}}$
$\left[C{a}^{2+}\right]=x+y$

$K_{sp,CaC{O}_3}=\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]=(x+y)x=4.7\times {10}^{-9}$......(1)
$K_{sp,Ca{C}_2{O}_4}=\left[C{a}^{2+}\right]\left[C_2{O}_4^{2-}\right]=y(x+y)=1.3\times {10}^{-9}$......(2)

Divide equation (2) by equation (1)
$\frac{y(x+y)}{x(x+y)}=\frac{1.3\times {10}^{-9}}{4.7\times {10}^{-9}}$
$\frac{y}{x}=0.2766$
$y=0.2766x$....(3)

Substitute this in equation (2)

$0.2766x(x+0.2766x)=1.3\times {10}^{-9}$
$0.3531x^2=1.3\times {10}^{-9}$
$x=6.067\times {10}^{-5}$ Substitute this in equation (3)
$y=0.2766x=0.2766\times 6.067\times {10}^{-5}=1.678\times {10}^{-5}$
$\left[C{a}^{2+}\right]=x+y=6.067\times {10}^{-5}+1.678\times {10}^{-5}$

$=7.746\times {10}^{-5}$ M