Question

calculate the concentration of formic acid if at the titration endpoint B=49.5 mL of .180 M NaOH completely reacts with A=12.6 mL of formic acid.

Answer 1

Solution:

Since 49.5 mL of 0.180 M NaOH is required for complete neutralization (since formic acid is a strong acid so complete neutralization occurs) of 12.6 mL of Formic acid of S₁ molar so

49.5 × 0.180 = 12.6 × S₁ (from the formula V₁S₁=V₂S₂)

S₁ = (49.5 × 0.180)/12.6 M

S₁ = 8.91/12.6 M

S₁ = 0.707M This is the strength of 12.6 ml of formic acid.

( This can be thought in another way,

1 equivalent of x volume of y M NaOH is required to complete neutralization of 1 equivalent of x volume of y M formic acid

and given 49.5 ml of 0.180 M NaOH is required to neutralize 12.6 ml formic acid

So now think what will be the strength of 0.180 M NaOH solution if its volume become 12.6ml this will be the strength of formic acid.

So,

49.5 × 0.180 = 12.6 × S₁ (from the formula V₁S₁=V₂S₂)

S₁ = (49.5 × 0.180)/12.6 M

S₁ = 8.91/12.6 M

S₁ = 0.707M This is the strength of 12.6 ml of formic acid.