Question

$V_1$ L of a 2.5 M NaOH solution and $V_2$ L of a 0.4 M NaOH solution were mixed to make 3 L of a 1 N NaOH, without adding any water. Find the ratio of $V_1:V_2.$

• A. 5:2
• B. 2:5
• C. 3:1
• D. 5:3

Answer 1

The correct option is B 2:5

Molarity of NaOH solution = Normality of NaOH solution (n-factor is 1 for NaOH)
Let $N_1$ and $N_2$ be the normalities corresponding to $V_1$ and $V_2$ (volumes) of the NaOH solutions, respectively. Let $N_T$ and $V_T$ be the normality and volume of the resultant solution.
$N_1{V}_1+N_2{V}_2=N_T{V}_{T}2.5V_1+0.4V_2=1\times 32.5V_1+0.4(3-V_1)=32.5V_1+1.2-0.4V_1=32.1V_1=1.8V_1=\frac{1.8}{2.1}=\frac{6}{7}{V}_2=3-\frac{6}{7}=\frac{15}{7}\frac{V_1}{V_2}=\frac{6}{7}\times \frac{7}{15}=2:5$