Question

In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 mL of HCl to completely neutralize the acid?

Answer 1

Answer: The volume of 0.35 M NaOH required to neutralize 45 ml of 0.35 M HCl is 45 ml.

Solution :

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of NaOH solution = 0.35 M

$V_1$ = volume of NaOH solution = ?

$M_2$ = molarity of HCl solution = 0.35 M

$V_2$ = volume of HCl solution = 45 ml

Now put all the given values in the above law, we get the volume of NaOH solution.

$(0.35M)\times V_1=(0.35M)\times (45ml)$

$V_1=45ml$

Therefore, the volume of 0.35 M NaOH required to neutralize 45 ml of 0.35 M HCl is 45 ml.