Question

1.458 g of $Mg$ reacts with 80.0 mL of $HCl$ solution whose $pH$ is $-0.477$. The change in $pH$ after all $Mg$ has reacted is:

(Assume constant volume. $Mg=24.3$ g $mo{l}^{-1}$ and $log3=0.477$)

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

Answer 1

Answer: (D)

$+0.3$

$Mg\left(aq\right)+2HCl\left(aq\right)⟶MgC{l}_2\left(aq\right)+H_2$

Number of moles of $Mg$ is $\frac{1.458}{24.3}=0.06$ moles $=60$ M moles

Molarity of $HCl=3M$

$\Rightarrow 3\times 80=240$ M moles

$HCl$ left after reaction= $240-(60\times 2)=120$ M moles

Hence molarity of solution= $\frac{120}{80}=1.5M$

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

$\therefore$ Change in $pH=-0.176-(-0.477)=0.3$