Question

A solution of formic acid (HCOOH) has a pH of 2.11.

How many

grams of formic acid are there in

100.0 mL

of the solution?

Answer 1

Let, We have given, pH of the formic acid solution is 2.11.

We know; pH = -log[H₃O⁺]

[H₃O⁺] = 10^-pH

= 10^-2.11

  [H₃O⁺]= 0.007762M

Now; Suppose "y" is the initial concentration of HCOOH, So The ICE table would be as;

HCOOH(aq) + H₂O_(l) HCOO^-_(aq) + H₃O⁺_(aq)

I y 0 0

C -x +x +x

E y-x x x

We know;

[H₃O⁺] = x = 0.007762M

Also, [HCOO^-] = x = 0.007762M

[HCOOH] = y-x = y-0.007762 M

From the balanced reaction above;

Ka = [H₃O⁺][HCOO^-] /[HCOOH]

We know; Ka for formic acid =1.8 x10^-4

1.8x 10^-4 = [0.007762][0.007762]/[y-0.007762]

= [6.02486 x10^-5] / y y-0.007762 y

y = [6.02486 x10^-5]/ (1.8 x10^-4)    According to the 5% Rule

y = 0.3347 M

So, The concentration of formic acid was 0.3347M, it means 0.3347 moles in 1L of solution.

Thus, 0.03347 moles in 0.1L(100mL)

Now, Converting these 0.03544 moles of Formic acid to grams of formic acid by using molar mass of Formic acid as;

0.03347 moles x( 46.03g /1mol) =1.54 g of Formic Acid

So, 1.54 g of Formic acid is there in 100mL of solution.

[Note: If we take Ka for formic acid as; 1.77 x10^-4 The final answer will be 1.56 g

and If we take Ka for formic acid as; 1.70 x10^-4, The final answer will be 1.63 g]