Question

25 mL of 3.0 M $HN{O}_3$ is mixed with 75 mL of 4.0 M $HN{O}_3$. If the volumes are additive, and 10 mL of this mixture is titrated against 2.5 M $NaOH$, find the volume of $NaOH$ needed for complete neutralisation?

• A. 15 mL
• B. 20 mL
• C. 25 mL
• D. 30 mL

Answer 1

The correct option is A 15 mL

Dilution:
$M_1{V}_1+M_2{V}_2=M_f{V}_f$
$3.0\times 25+4.0\times 75=M_f\times 100$
$M_f\left(HN{O}_3\right)=3.75M$
For neutralization:
$HN{O}_3+NaOH\to NaN{O}_3+H_{2}O$
$(M\times V\times n_f{)}_{HN{O}_3}=(M\times V\times n_f{)}_{NaOH}$
$n_f$ is 1 for $HN{O}_3$
$n_f$ is 1 for $NaOH$
$3.75\times 10=2.5\times V_{NaOH}$
$V_{NaOH}=15mL$