Question

100 c.c. of $0.5N$ $NaOH$ Solution is added to $10$ c.c. of $3N$ $H_{2}S{O}_4$ solution and $20$ c.c. of $1N$ $HCl$ solution. The solution will be -

• A. Strongly acidic
• B. Alkaline
• C. Neutral
• D. Fairly acidic

Answer 1

The correct option is B Alkaline

Moles of solute = Volume $\times$ Molarity

100 ml of 0.5 N NaOH = 0.100 $\times$ 0.5 N = 0.05 equivalent moles.

10 ml of 3 N $H_{2}S{O}_4$ = 0.10 $\times$ 0.3 N = 0.03 equivalent moles.

20 ml of 1 N HCL = 0.20 $\times$ 0.1 N = 0.01 equivalent moles.

0.05 equivalent moles of NaOH will neutralise 0.02 equivalent moles of HCL and 0.05 equivalent moles of NaOH will neutralise 0.03 equivalent moles of $H_{2}S{O}_4$ .

The unreacted equivalent moles of NaOH = 0.05 – (0.03 + 0.01) = 0.01

The solution would be basic containing 0.01 equivalent moles of NaOH = $\frac{0.01}{0.300}$ L = 0.033 N NaOH.

Hence , option B is correct .