Question

$100ml$ of $0.2MHCl$ are mixed with $100ml$ of $0.2MC{H}_{3}COONa$, the $pH$ of the resulting solution would be nearly:

• A. $1$
• B. $0.7$
• C. $2.875$
• D. $1.6$

Answer 1

The correct option is A $1$

$C{H}_{3}COONa\to C{H}_{3}CO{O}^{-}+N{a}^{+}$

$100ml$ of $0.2MC{H}_{3}COONa$

Number of mole of $N{a}^{+}$ generated = $100\times 0.2$

$=20$ milli mol

$HCl\to H^{+}+C{l}^{-}$

$100ml$ of $0.2MHCl$

Number of mole of $C{l}^{-}$ generated = $100\times 0.2$

$=20$ milli mol

$N{a}^{+}+C{l}^{-}\to NaCl$

Amount of $\left[H^{+}\right]=20$ milli mol

$\left[H^{+}\right]=\frac{20}{100+100}=0.1M$

$pH=-\log \left[H^{+}\right]=-\log 0.1=1$