Question

A $1.458g$ of $Mg$ reacts with $80.0ml$ of a $HCl$ solution whose $pH$ is $-0.477$, The change is $pH$ when all $Mg$ has reacted will be$:$

Assume constant volume $Mg=24.3g/mol(\log 3=0.47,\log 2=0.301)$

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

Answer 1

The correct option is D $0.3$

$pH=-0.477$

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

$\therefore$ Moles of $HCl=3\times 0.080$

$=0.24$ moles

Moles of $Mg=\frac{1.458}{24.3}=0.06$ moles

$Mg+2HCl⟶Mg{Cl}_2+H_2$

$0.06$ $2\times 0.06$

$\therefore 0.12$ moles of $HCl$ is required for $0.06$ moles of $Mg$

$\therefore$ Moles of $HCl$ left $=0.24-0.12$

$=0.12$

$\therefore$ New concentration of $\left[H^{+}\right]=\frac{0.12}{0.08}=1.5M$

$\therefore$ New $pH=-\log \left[H^{+}\right]=-\log \left(1.5\right)=-\log \left(\frac{3}{2}\right)$

$=-\log 3+\log 2$

$=-0.47+0.301$

$=-0.169$

$\therefore$ Change in $pH=-0.169-(-0.477)$

$=0.3$