Answer to Question #233652 in General Chemistry for aishwary

Question

Using 1M NaOH as the titrant and varying the molarity of 1L of CH3CO2H from 0.25M to 1.5M, how does the equivalence point (in titrant volume) correspond to the titrand molarity?

1) The equivalence point is numerically double the titrand molarity

2) The equivalence point is numerically equal to the titrand molarity

3) There is no relationship between the two

Answer 1

Moles NaOH = Moles CH₃CO₂H

M₁V₁ = M₂V₂

1M×V₁ = 1M×1L

V_equivalane =V₁ = 1L

The molarity of CH₃CO₂H Varying from

0.25 M to 1.5 M .When molarity reaches 1M CH₃CO₂H under varying condition.

Then the equivalence point in titrant Volume is numerical equal to the titrant molarity