Question

What product is formed when dilute $H_{2}S{O}_4$, a non oxidising acid is added to the $Cu(OH{)}_2$ compound?

• A. $Cu(OH{)}_2+2H^{+}\to C{u}^{2+}+2H_{2}O$
• B. $Cu(OH{)}_2+H^{+}\to C{u}^{+}+H_{2}O$
• C. $Cu(OH{)}_2+H^{+}\to C{u}^{2+}+H_{2}O$
• D. None of these

Answer 1

The correct option is A $Cu(OH{)}_2+2H^{+}\to C{u}^{2+}+2H_{2}O$

$Cu(OH{)}_2+H_{2}S{O}_4\to CuS{O}_4+2H_{2}O$.

In $CuS{O}_4$, the $Cu$ is in $+2$ oxidation state as $C{u}^{2+}$.

Hence option A is correct.