Question

Calculate the $pH$ of a solution which contains $9.9ml$ of $1MHCl$ and $100ml$ of $0.1MNaOH$.

Answer 1

$NaOH+HCl⟶NaCl+H_{2}O$

$1mol,NaOH$ reacts with $1molHCl$

$1mol,HCl$ in $99ml$ of $1.0M$ solution$=\frac{9.9}{1000}\times 1=0.0099mol,HCl$

$1mol,NaCl$ in $100ml,0.1M$ solution$=\frac{100}{1000}\times {10}^{-1}=0.01mol,NaOH$

$0.0099mol+0.01molHClNaOH\downarrow 0.0099molHClneutralises$

$⟶0.01-0.0099=1\times {10}^{-4}molNaOH$ would be dissolves in $109.9ml$ solution.

Molarity of $NaOH$ solution$=\frac{{10}^{-4}}{0.1099}=9.099\times {10}^{-4}M$

$pOH=-log(9.099\times {10}^{-4})$

$pOH=3.04$

$pH=14-3.04$

$pH=10.69$