Question

100 ml of 0.5 N NaOH solution is added to 10 mi of 3 N H2SO4 solution and 20 ml of 1 N HCl solution.The mixture is

Answer 1

Moles of solute = Volume x Molarity

100 ml of 0.5 N NaOH = 0.100 x 0.5 N = 0.05 equivalent moles.

10 ml of 3 N H2SO4 = 0.10 x 0.3 N = 0.03 equivalent moles.

20 ml of 1 N HCL = 0.20 x 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 H2SO4.

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 = 0.01/0.300 L = 0.033 N NaOH.

Answer 2

Moles of solute = Volume x Molarity

100 ml of 0.5 N NaOH = 0.100 x 0.5 N = 0.05 equivalent moles.

10 ml of 3 N H2SO4 = 0.10 x 0.3 N = 0.03 equivalent moles.

20 ml of 1 N HCL = 0.20 x 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 H2SO4.

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 = 0.01/0.300 L = 0.033 N NaOH.