Search any question & find its solution
Question:
Answered & Verified by Expert
$1 \mathrm{~L}$ of $0.02 \mathrm{M}$ aqueous $\mathrm{HCl}$ is mixed with $1 \mathrm{~L}$ of $0.01 \mathrm{M}$ aqueous $\mathrm{H}_2 \mathrm{SO}_4$ solution. Assuming complete dissociation and no change in the volume upon mixing, the $\mathrm{pH}$ of resultant solution is $\left(\log _{10} 2=0.3\right)$
Options:
Solution:
1276 Upvotes
Verified Answer
The correct answer is:
1.7
Given,
$\mathrm{HCl} 1 \mathrm{~L} 0.02 \mathrm{M}$
$$
\mathrm{H}_2 \mathrm{SO}_4 1 \mathrm{~L} 0.01 \mathrm{M}
$$
Moles of $\mathrm{H}^{+}$from $\mathrm{HCl}=1 \times 0.02=0.02$
Moles of $\mathrm{H}^{+}$from $\mathrm{H}_2 \mathrm{SO}_4=1 \times 0.01 \times 2=0.02$
Total moles in mixture $=0.02+0.02=0.04$
$$
\begin{aligned}
{\left[\mathrm{H}^{+}\right] } & =\frac{\text { Total moles }}{\text { Total volume }} \\
& =\frac{0.04}{2}=0.02 \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right]=-\log 0.02=1.7
\end{aligned}
$$
$\mathrm{HCl} 1 \mathrm{~L} 0.02 \mathrm{M}$
$$
\mathrm{H}_2 \mathrm{SO}_4 1 \mathrm{~L} 0.01 \mathrm{M}
$$
Moles of $\mathrm{H}^{+}$from $\mathrm{HCl}=1 \times 0.02=0.02$
Moles of $\mathrm{H}^{+}$from $\mathrm{H}_2 \mathrm{SO}_4=1 \times 0.01 \times 2=0.02$
Total moles in mixture $=0.02+0.02=0.04$
$$
\begin{aligned}
{\left[\mathrm{H}^{+}\right] } & =\frac{\text { Total moles }}{\text { Total volume }} \\
& =\frac{0.04}{2}=0.02 \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right]=-\log 0.02=1.7
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.