Question

What volume (in mL) of 0.51 M H2SO4(aq) solution is necessary to completely react with 103 mL of 0.36 M KOH(aq)

Answer 1

Answer: 36.3 ml

Explanation:

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity $H_2SO_4$ = 2

$M_1$ = molarity of $H_2SO_4$ solution = 0.51 M

$V_1$ = volume of  $H_2SO_4$ solution = ?

$n_2$ = acidity of $KOH$ = 1

$M_1$ = molarity of $KOH$ solution = 0.36 M

$V_1$ = volume of  $KOH$ solution = 103 ml

Putting in the values we get:

$2\times 0.51\times V_1=1\times 0.36\times 103$

$V_1=36.3ml$

Therefore, volume of $H_2SO_4$ required is 36.3 ml