Question

A $100mL$ solution of $0.1NHCl$ was titrated with $0.2NNaOH$ solution. The titration was discontinued after adding $30mL$ of $NaOH$ solution. The titration was completed by adding $0.25NKOH$ solution. The volume of $KOH$ required for completing the titration is:

• A. $70mL$
• B. $32mL$
• C. $35mL$
• D. $16mL$

Answer 1

Answer: (D)

$16mL$

Total moles of $H^{+}=\frac{100}{1000}\times 0.1={10}^{-2}moles$

Moles of $H^{+}$ neutralized by $NaOH=\frac{30}{1000}\times 0.2=6\times {10}^{-3}$ moles

Moles of $H^{+}$ remained after $NaOH=4\times {10}^{-3}$ moles

$4\times {10}^{-3}=\left(0.25\right)\times V$ in lit of $KOH$

$V=16\times {10}^{-3}$ lit

$=16ml$ of $KOH$