Question

What will be the $H^{+}$ ion concentration of a solution prepared by mixing 50 mL of 0.20 M $NaCl$, 25 mL of 0.10 M $NaOH$ and 25 mL of 0.30 M $HCl$?

• A. 0.5 M
• B. 0.05 M
• C. 0.02 M
• D. 0.10 M

Answer 1

The correct option is B 0.05 M

$\underset{0.2M50mL}{NaCl}\underset{0.1M25mL}{NaOH}\underset{0.3M25mL}{HCl}$

Total Volume = (50+25+25) = 100 mL

Since NaCl is a neutral salt. So, it's concentration will not have affect on mixture.

Calculating milliequivalents (meq.) of NaOH and HCl.
meq= Normality $\times$ Volume
Normality=Molarity $\times$ n-factor
Since,
n-factor of HCl $=$ n-factor of NaOH $=$ 1

Normality= Molarity

meq.= Molarity $\times$ Volume
$meq{.}_{NaOH}=0.1\times 25=2.5meq.$
$meq{.}_{HCl}=0.3\times 25=7.5meq.$
$[H{]}^{+}=\frac{meq{.}_{HCl}-meq{.}_{NaOH}}{\text{Total Volume}}=\frac{(7.5-2.5)meq.}{100mL}[H{]}^{+}=0.05N=0.05M$