Question

How many milliliters of 4.00 M NaOH are required to exactly neutralize 50.0 milliliters of a 2.00 M solution of HNO3 ?

Answer 1

Answer: The volume of $NaOH$ required is 25.0 ml

Explanation:

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity $HNO_3$ = 1

$M_1$ = molarity of $HNO_3$ solution = 2.00 M

$V_1$ = volume of  $HNO_3$ solution = 50.0 ml

$n_2$ = acidity of $NaOH$ = 1

$M_1$ = molarity of $NaOH$ solution = 4.00 M

$V_1$ = volume of  $NaOH$ solution =  ?

Putting in the values we get:

$1\times 2.00\times 50.0=1\times 4.00\times V_2$

$V_2=25.0ml$

Therefore, volume of $NaOH$ required is 25.0 ml