Question

A $1.458$ g 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 D $0.3$

The reaction between magnesium and $HCl$ is as given below.
$Mg\left(aq\right)+2HCl\left(aq\right)\to MgC{l}_2\left(aq\right)+H_2$
The number of moles of magnesium is $\frac{1.458g}{24.3g}=0.06$ moles or 60 mmoles.
HCl solution has pH 3. hence, its molarity is 3 M.
The number o millimoles of HCl is $3\times 80=240millimoles$
After completion of reaction, millimoles of HCl left are $240-60\times 2=120$.
Hence, the molarity of the solution will be $=\frac{120}{80}=1.5M$
The pH of the solution will be $pH=-log\left[H^{+}\right]=-log1.5=-0.176$.
The change in pH will be $=-0.176-(-0.477)=0.3$.