Question

A 25.5 mL aliquot of 0.0695 MH2SO4(aq) is to be titrated with 0.0461 MNaOH(aq) . What volume ( mL ) of base will it take to reach the equivalence point?

A 25.5 aliquot of 0.0695 is to be titrated with 0.0461 . What volume (mL) of base will it take to reach the equivalence point?
A)76.9
B)38.4
C)19.2
D)0.163
E)3980

Honestly, this question is killing me, and I have no idea where to even begin. I know its an acid-base reaction, and produces salt and water?? Mannnn.

Answer 1

Answer: option A

Step-by-step explanation:

Titration maths are mostly solved through n1/n2 = v1s1/v2s2

H2SO4 + 2NaOH ----> Na2(SO4) + 2H2O

so n1 = 1 & n2 = 2

v1 = 25.5 ml s1 = 0.0695 M s2 = 0.0461 M

Now,

⇒1/2 = (25.5 × 0.0695)/ (v2 × 0.0461)

⇒ v2 = 76.88 ml

≈ 76.9 ml