Question

Determine the acid dissociation constant for a 0.020 m formic acid solution that has a ph of 2.74. formic acid is a weak monoprotic acid and the equilibrium equation of interest is hcooh(aq) + h2o(l) ⇌ h3o+(aq) + hco2-(aq). enter your answer in exponential (e) format (sample 1.23e-4) with two decimal places and without units.

Answer 1

Answer:

Ka = 1.82x10⁻⁴

Explanation:

To solve this problem, we need a basis. In this case, we need the overall reaction and a ICE chart. So, let's write the overall reaction first:

HCOOH + H₂O <--------> H₃O⁺ + HCOO⁻    Ka = ?

Now that we have the overall reaction, we need to write the ICE chart. In this way we can determine what data do we have, and what do we have left to determine:

      HCOOH + H₂O <--------> H₃O⁺ + HCOO⁻    Ka = ?

i)        0.02                                0             0

e)       0.02-x                             x              x

The Ka expression is:

Ka = [H₃O⁺] [HCOO⁻] / [HCOOH]

Replacing the given data we have:

Ka = x² / 0.02 - x

Now, the value of x can be calculated because we already have the pH of the formic acid, and with it, we can calculate the [H₃O⁺] with the following expression:

[H₃O⁺] = 10^(-pH)

Replacing we have:

[H₃O⁺] = 10^(-2.74) = 1.82x10⁻³ M

This is the value of x, so replacing in the Ka expression, we can calculate then, the value of Ka:

Ka = (1.82x10⁻³)² / (0.02 - 1.82x10⁻³)

Ka = 1.82x10⁻⁴