Question

Calculate the pH of the following:0.02mol of methanoic acid (HCOOH) are given Ka(HCOOH)=1.78×10^-14.

Answer 1

I would like to point out that this is a flawed question.
1. The correct Ka value for HCOOH is 1.78 x 10^-4 and not 10^-14; i.e. it is a much strong acid that indicated in the problem and
2, Note that the answer for your solution is correct BUT that a pH of 7.37 is basic whereas this is in reality an acid.
If worked using the correct value for Ka those problems are overcome.
3. I suspect the student just made a typo.

Answer 2

Another typo in the problem is 0.02 mol. We ASSUME that is the molar concentration but the problem uses the wrong terminology. The problem should have been 0.02 M and not mol.

Answer 3

To calculate the pH of the methanoic acid solution, we need to consider that methanoic acid is a weak acid. Therefore, we need to use the expression for the acid dissociation constant (Ka) to find the concentration of the hydronium ions (H3O+):

Ka = [H3O+][HCOO-] / [HCOOH]

Since we are given the Ka value and the concentration of methanoic acid, we can solve for [H3O+]. Let's denote [H3O+] as x:

1.78 × 10^-14 = x^2 / 0.02

Rearranging the equation and solving for x:

x^2 = 0.02 × 1.78 × 10^-14
x = √(0.02 × 1.78 × 10^-14)

Using a calculator, we find:
x ≈ 4.221 × 10^-8 M

Now that we have the concentration of H3O+, we can use the formula for pH:

pH = -log[H3O+]

Substituting the value we found for [H3O+]:

pH = -log(4.221 × 10^-8)
pH ≈ 7.37

Therefore, the pH of the methanoic acid solution is approximately 7.37.

Answer 4

I apologize for the mistake in the Ka value provided in the question. If the correct Ka value for methanoic acid (HCOOH) is 1.78 x 10^-4 as you mentioned, we can recalculate the pH using the correct Ka value.

Using the same steps as before, we can set up the equation:

Ka = [H3O+][HCOO-] / [HCOOH]

1.78 x 10^-4 = (x)(x) / 0.02

Simplifying the equation:

x^2 = (1.78 x 10^-4)(0.02)
x = √(1.78 x 10^-4)(0.02)

Calculating this value, we find:
x ≈ 4.216 x 10^-3 M

Now, we can calculate the pH using the formula:

pH = -log[H3O+]

Substituting the value we found for [H3O+]:

pH = -log(4.216 x 10^-3)
pH ≈ 2.37

Therefore, with the correct value for Ka, the pH of the methanoic acid solution is approximately 2.37.

Answer 5

Thank you for pointing out the error. If the correct concentration is 0.02 M (instead of 0.02 mol), we can recalculate the pH using the correct concentration.

Using the same steps as before and the correct Ka value of 1.78 x 10^-4:

Ka = [H3O+][HCOO-] / [HCOOH]

1.78 x 10^-4 = (x)(x) / 0.02

Simplifying the equation:

x^2 = (1.78 x 10^-4)(0.02)
x = √(1.78 x 10^-4)(0.02)

Calculating this value, we find:
x ≈ 8.435 x 10^-3 M

Now, we can calculate the pH using the formula:

pH = -log[H3O+]

Substituting the value we found for [H3O+]:

pH = -log(8.435 x 10^-3)
pH ≈ 2.07

Therefore, with the correct concentration of 0.02 M and the correct value for Ka, the pH of the methanoic acid solution is approximately 2.07.