Following four solutions are prepared by mixing different volumes of NaOH and HCl of different conce

Question

Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?
I. $75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$
II. $100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$
III. $55 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+45 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$
IV. $60 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+40 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$

Professor PC Thomas Classes · Answer

(a): $\mathrm{NaOH}+\mathrm{HCl} ightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}$ 25 mL of NaOH will react with 25 mL of HCl . Total volume of solution $=75+25$ $ =100 \mathrm{~mL} $ Millimoles of HCl left $=\frac{75}{5}-\frac{25}{5}=10$ Concentration of HCl $ \begin{aligned} & =\frac{	ext { Millimoles }}{	ext { Volume }}=\frac{10}{100}=0.1 \mathrm{M} \ \mathrm{pH} & =-\log \left[\mathrm{H}^{+}ight]=-\log [0.1]=1 \end{aligned} $