A solution contains $10 \mathrm{~mL} 0.1 \mathrm{~N} \mathrm{NaOH}$ and $10 \mathrm{~mL}$ $0.05 \mathrm{~N}…
A solution contains $10 \mathrm{~mL} 0.1 \mathrm{~N} \mathrm{NaOH}$ and $10 \mathrm{~mL}$ $0.05 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{pH}$ of this solution is :
less than 7
7
zero
greater than 7
Solution
milliequivalent of \(\mathrm{NaOH}=10 \times 0.1=1\)
milliequivalent of \(\mathrm{H}_2 \mathrm{SO}_4=10 \times 0.05=0.5\)
Left milliequivalent of \(\mathrm{NaOH}=1-0.5=0.5 \mathrm{M}\)
\(\therefore\left[\mathrm{OH}^{-}ight]=\frac{0.5}{10+10}=2.5 \times 10^{-2}\)
we know the relation
\(\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}ight]=-\log \left(2.5 \times 10^{-2}ight)=1.6020\)
Also,
\(\begin{aligned}
& \mathrm{pH}=14-\mathrm{pOH} \\
& \mathrm{pH}=14-1.6020=12.398
\end{aligned}\)
/