Question

The normality of solution obtained by mixing 100 ml of 0.2m h2so4 with 100 ml of 0.2m naoh is

Answer 1

molarity of H2SO4 = 0.2M

so, normality of H2SO4 = n-factor × molarity = 2 × 0.2 = 0.4N

similarly, normality of 0.2M of NaOH = n-factor × 0.2M = 1 × 0.2 = 0.2N

When $V_1$ ml of an acid of normality $N_1$ is mixed with $V_2$ ml of base of normality $N_2$. then, normality of resultant solution is given by,

$N=\frac{N_1V_1-N_2V_2}{V_1+V_2}$ when, $N_1V_1 > N_2V_2$

= (0.4 × 100 - 0.2 × 100)/(100 + 100)

= 0.2 × 100/200

= 0.1N

hence, normality of the solution is 0.1N

Answer 2

explanation given above ...

normality of solution is 0.1 N