Question

Which of the following will decrease the $pH$ of a $50mL$ solution of $0.01MHCl$?

• A. Addition of $5mL$ of $1MHCl$
• B. Addition of $50mL$ of $0.01MHCl$
• C. Addition of $50mL$ of $0.002MHCl$
• D. Addition of $Mg$

Answer 1

Answer: (D)

Addition of $Mg$

Millimoles of $50mL$ of $0.01MHCl$ $=50\times 0.01M=0.5$
pH of the solution $=-\log \left(0.5\right)=0.3010$
(a) On adding $5mL$ of $1MHCl$,
Concentration of the resultant solution $=\frac{50\times 0.01+5\times 1}{50+5}=0.1$
$\therefore$ pH of the solution $=-\log \left(0.1\right)=1$
(b) On adding $50mL$ of $0.01MHCl$,
Concentration of the resultant solution $=\frac{50\times 0.01+50\times 0.01}{50+50}=0.01$
$\therefore$ pH of the solution $=-\log \left(0.01\right)=2$
(c) On adding $50mL$ of $0.002MHCl$,
Concentration of the resultant solution $=\frac{50\times 0.01+50\times 0.002}{50+50}=6\times {10}^{-3}$
$\therefore$ pH of the solution $=-\log (6\times {10}^{-3})=2.22$
(d) On adding $Mg$,
$Mg+2HCl⟶Mg{Cl}_2+2H^{+}$
Since, $pH\propto \frac{1}{\left[H^{+}\right]}$
$\therefore$ In this case pH decreases.