Question

Which of the following solutions will have pH close to $1.0$ ?

• A. $100\, mL $ of $(M/10) HCl + 100\,mL$ of $(M/10) NaOH$
• B. $55 \,mL$ of $(M/10) HCl + 45 \,mL$ of $(M/10) NaOH$
• C. $10 \,mL$ of $(M/10) HCl + 90 \,mL$ of $(M/10) NaOH$
• D. $75 \,mL$ of $(M/5) HCl + 25 \,mL$ of $(M/5) NaOH$

Answer 1

Answer: (D)

$75 \,mL \frac{ M}{ 5 } HCl = 15\, mmol\, HCl$
$25 \,mL \frac{ M}{ 5 } NaOH = 5\, mmol\, NaOH$
After neutralisation, $10 \,mmol\, HCl$ will be remaining in $100 \,mL$ of solution.
Molarity of $HCl$ in the final solution
$ = \frac{ 10 }{ 100 } = 0.10 $
$pH = - log [H^+ ]$
$ = - \log (0.10) = 1.$