Question

The pH of a solution obtained by mixing $50mL$ of $HCl$ and $430mL$ of $INNaOH$ is $[\log 2.5=0.3979]$

• A. 3.979
• B. 0.6021
• C. 12.042
• D. 1.2042

Answer 1

Answer: (B)

In $HCl=1MHCl.1NNaOH=1MNaOH$
Moles of $HCl=M\times V=50\times 10^{-3}\times 1$
$=5\times 10^{-2}$
Moles of $NaOH=M\times V=30\times 10^{-3}\times 1$
$=3\times 10^{-2}$
As number of moles of $HCl$ are more than $NaOH$ the resulting solution $(30+50=80mL)$ will contain excess of $HCl$
$=5\times 10^{-2}-3\times 10^{-2}=2\times 10^{-2}mol$
$\therefore H^{+}$ (after mixing) $=\frac{2\times 10^{-2}mol}{80\times 10^{-3}L}=\frac{1}{4}M$
$PH=-log{H}^{+}=-log\frac{1}{4}=log4$
$=0.6021$