Question
What will be the pOH of $\frac{ M }{100} \quad H _2 SO _4$ solution ?
✓
Answer
$\frac{ M }{100} H _2 SO _4=0.01 M H _2 SO _4$
By ionisation of $H _2 SO _4, 2 H ^{+}$ions are obtained.
Hence $0.01 M =0.01 \times 2=0.02 N$
Therefore
$
\begin{aligned}
{\left[H^{+}\right] } & =0.02=10^{-2} \times 2 \\
pH & =-\log 10^{-2}-\log 2 \\
pH & =2-0.3010 \quad(\log 2=0.3010) \\
pH & =1.6990
\end{aligned}
$
As $pH + pOH =14$
Hence
$
\begin{array}{l}
pOH=14-pH \\
pOH=14-1.6990 \\
pOH=12.3010
\end{array}
$
By ionisation of $H _2 SO _4, 2 H ^{+}$ions are obtained.
Hence $0.01 M =0.01 \times 2=0.02 N$
Therefore
$
\begin{aligned}
{\left[H^{+}\right] } & =0.02=10^{-2} \times 2 \\
pH & =-\log 10^{-2}-\log 2 \\
pH & =2-0.3010 \quad(\log 2=0.3010) \\
pH & =1.6990
\end{aligned}
$
As $pH + pOH =14$
Hence
$
\begin{array}{l}
pOH=14-pH \\
pOH=14-1.6990 \\
pOH=12.3010
\end{array}
$
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