Question

What will be the $H^{+}$ concentration in a solution prepared by mixing $50.0ml$ of $0.20MNaCl,25ml$ of $0.10MNaOH$ and $25.0ml$ of $0.30MHCl$?

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

Answer 1

The correct option is B $0.05M$

First we have to calculate the number of moles of $HCl$ and $NaOH$

Number of moles of $HCl=\frac{25.0mL}{1000mL/L}\times 0.30M=0.0075M$

Number of moles of $NaOH=\frac{25.0mL}{1000mL/L}\times 0.10M=0.0025M$

0.0025 moles of $NaOH$ will neutralize 0.0025 moles of $HCl$.

$0.0075-0.0025=0.0050$ moles of $HCl$ will remain.

0.0050 moles of HCl will ionize to give 0.0050 moles of $H^{+}$ ions.

Total volume $=25.0mL+25.0mL+50.0mL=100.0mL$

$\left[H^{+}\right]=\frac{0.0050mol}{100mol}\times 1000mL/L$

$\left[H^{+}\right]=0.050M$

NaCl is a salt of a strong acid with a strong base. It does not hydrolyze. Hence, it will not affect $\left[H^{+}\right]$

Therefore option B is correct answer.