Question

Moles of $CuS{O}_4$ dissolved in water to make the above solution is :

• A. $1\times {10}^{-5}$
• B. $2\times {10}^{-5}$
• C. $2\times {10}^{-4}$
• D. $4\times {10}^{-4}$

Answer 1

The correct option is A $1\times {10}^{-5}$

The equilibrium reaction is as follows:

$\left[Cu\right(H_{2}O{)}_6{]}^{2+}+H_{2}O\rightleftharpoons \left[Cu\right(H_{2}O{)}_5\left(OH\right){]}^{+}+H_3{O}^{+};K={10}^{-5}$
Let $a$ M, $b$ M and $c$ M be the equilibrium concentrations of $\left[Cu\right(H_{2}O{)}_6{]}^{2+},\left[Cu\right(H_{2}O{)}_5\left(OH\right){]}^{+}$ and $H_3{O}^{+}$ respectively.
The expression for the equilibrium constant is $K=\frac{\left[Cu\right(H_{2}O{)}_5\left(OH\right){]}^{+}\left[H_3{O}^{+}\right]}{\left[Cu\right(H_{2}O{)}_6{]}^{2+}}=\frac{b\times c}{a}={10}^{-5}$
$pH=-log\left[H_3{O}^{+}\right]=5$
Hence, $\left[H_3{O}^{+}\right]={10}^{-5}$
Substituting values in the expression for the equilibrium constant, we get
$\frac{b\times {10}^{-5}}{a}={10}^{-5}$
Thus, $a=b$ but $b=c={10}^{-5}$
Hence, $a=b=c={10}^{-5}$
Thus, total concentration of $CuS{O}_4$ will be $2a=2\times {10}^{-5}$ M.
Thus, $1$ L of solution contains $2\times {10}^{-5}$ moles of $CuS{O}_4$.
To prepare the given solution, the number of moles of $CuS{O}_4$ that should be dissolved in $500$ ml of solution is ${10}^{-5}$.