What will be the resultant pH when $200 ext{ mL}$ of an aqueous solution of $ ext{HCl}$ ($ ext{pH

Question

What will be the resultant pH when $200	ext{ mL}$ of an aqueous solution of $	ext{HCl}$ ($	ext{pH} = 2.0$) is mixed with $300	ext{ mL}$ of an aqueous solution of $	ext{NaOH}$ ($	ext{pH} = 12.0$)?

Professor PC Thomas Classes · Answer

For $	ext{HCl}$ solution: $	ext{pH} = 2.0 \implies [	ext{H}^+] = 10^{-2}	ext{ M}$
Moles of $	ext{H}^+$ = $M 	imes V = 10^{-2} 	imes 200	ext{ mL} = 2	ext{ mmol}$
For $	ext{NaOH}$ solution: $	ext{pH} = 12.0 \implies 	ext{pOH} = 14 - 12 = 2.0 \implies [	ext{OH}^-] = 10^{-2}	ext{ M}$
Moles of $	ext{OH}^-$ = $M 	imes V = 10^{-2} 	imes 300	ext{ mL} = 3	ext{ mmol}$
When mixed, $	ext{H}^+$ and $	ext{OH}^-$ react: $	ext{H}^+ + 	ext{OH}^- ightarrow 	ext{H}_2	ext{O}$
Remaining $	ext{OH}^-$ = $3	ext{ mmol} - 2	ext{ mmol} = 1	ext{ mmol}$
Total volume = $200	ext{ mL} + 300	ext{ mL} = 500	ext{ mL}$
Concentration of resulting $	ext{OH}^-$ = $\frac{1	ext{ mmol}}{500	ext{ mL}} = 2 	imes 10^{-3}	ext{ M}$
$	ext{pOH} = -\log(2 	imes 10^{-3}) = 3 - \log 2 = 3 - 0.3 = 2.7$
$	ext{pH} = 14 - 	ext{pOH} = 14 - 2.7 = 11.3$