Question

A $1.458g$ of $Mg$ reacts with $80.0$ml of a $HCl$ solution whose $pH$ is $-0.477$. The change is $pH$ when all $Mg$ has reacted will be:
Assume constant volume. $Mg=24.3g/mol(log3=0.47,log2=0.301)$

• A. $-0.176$
• B. $+0.477$
• C. $-0.2385$
• D. $0.3$

Answer 1

The correct option is A $-0.176$

$Mg+2HCl\to MgC{l}_2+H_2$

1 mol of $Mg$ reacts with 2 mols of $HCl$.

Here mol of $Mg=\frac{1.458}{24.3}=0.06$

$\therefore$ moles of $H^{+}$ required$=2\times 0.06=0.12$

$\left[H^{+}\right]={10}^{-pH}\approx 3M$

moles of $H^{+}$ present$=80\times 3\times {10}^{-3}=0.24$

$\therefore$ moles of $H^{+}$ left after the reaction$=0.24-0.12=0.12$

$\left[H^{+}\right]=\frac{0.12}{80\times {10}^{-3}}=1.5M$

$\therefore pH=-\log \left[H^{+}\right]=-\log 1.5=-0.176$

$pH$ when all $Mg$ has reacted$=-0.176$