Question

A student wants to now how many mL of 0.361 M sulfuric acid is to be added to 46.96 mL of 0.436 M sodium hydroxide solution to neutralize it completely. The reaction is:2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) +2H20(I)

Answer 1

2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) +2H20(I)

To solve this problem we use this formula:

M1 x V1 = M2 x V2 (1)

1 is for the acid and 2 is for the base.

M1 = 0.361 M sulfuric acid

V1 = unknown

M2 = 0.436 M sodium hydroxide

V2 = 46.96 mL

We clear V1 from (1) ==> V1 = M2 x V2/M1

===> V1 = 0.436 M x 46.96 mL/0.361 M = 56.72 mL

Answer: V1 = 56.72 mL H2SO4