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Q.
$20 cm^3$ of xM solution of Hcl is exactly neutralised by 40 $cm^3$ of 0.05 M NaOH solution, the pH of HCl solution is
Equilibrium
Solution:
Moles of $NaOH =0.05\times40\times10^{-3}$
Moles o f HCl $=20\times10^{-3}\times x$
$0.05\times 40\times10^{-3}=20\times x\times10^{-3}$
or $x=\frac{0.05\times40}{20}=0.1$ or $1\times10^{-1}$
$\therefore [HCl]=[H_{3}O^{+}]=1\times10^{-1}$
$\therefore PH=-log(1\times10^{-1})=1$