Question

What volume of 0.2M H2SO4 solution should be mixed with the 40 mL of 0.1M NaOH solution such that the resulting solution has the concentration of H2SO4 as 6/55 M

Answer 1

hope this helps you....

Answer 2

Answer: 26 ml

Explanation:

The expression used will be :

$n_1M_1V_1-n_2M_2V_2=nMV$

where,

$n_1$ = basicity of $H_2SO_4$  = 2

$n_2$ = acidity of NaOH = 1

$M_1$ = concentration of $H_2SO_4$ = 0.2 M

$M_2$ = concentration of NaOH = 0.1 M

$V_1$ = volume of $H_2SO_4$ = v ml

$V_2$ = volume of NaOH = 40 ml

n = basicity of $H_2SO_4$= 2

M = concentration of resulting $H_2SO_4$ =$\frac{6}{55}=0.109M$

V = volume of $H_2SO_4$ = (v + 40) ml

Now put all the given values in the above law, we get the volume of $H_2SO_4$ added.

$(2\times 0.2M\times v)-(1\times 0.1M\times 40ml)=(2\times 0.109\times (v+40)ml)$

By solving the terms, we get :

$v=26ml$

Thus the volume of 0.2 M $H_2SO_4$ is 26 ml.