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$0.1(\mathrm{M}) \mathrm{HCI}$ and $0.1(\mathrm{M}) \mathrm{H}_2 \mathrm{SO}_4$ each of volume $2 \mathrm{ml}$ are mixed and the volume is made up to $6 \mathrm{ml}$ by adding $2 \mathrm{ml}$ of $0.01(\mathrm{~N}) \mathrm{NaCl}$ solution. The $\mathrm{pH}$ of the resulting mixture is
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Hints: Mili moles of $\mathrm{H}^{+}=0.1 \times 2+0.1 \times 2 \times 2=0.6$
Total volume in $\mathrm{ml}=6$
$$
\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=-\log \left(\frac{0.6}{6}\right)=-\log 0.1=1
$$
Total volume in $\mathrm{ml}=6$
$$
\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=-\log \left(\frac{0.6}{6}\right)=-\log 0.1=1
$$
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