Question

What is the pH solution containing 0.3 M HCOOH and 0.52 M HCOOK?

Answer 1

First, we need the reaction to start the exercise.

HCOOH(aq) + H2O(l) = H3O+(aq) + HCOO-(aq)

HCOOK => HCOO- + H+

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pH)

We have here an ionic equilibrium, HCOOH is a weak acid.

pKa = 3.77 (from tables)

The concentration of HCOOH 0.3 M

The concentration of HCOO- = 0.52 M HCOOK.

If we propose Ka for HCOOH(aq)

`Ka\text{ = }\frac{\lbrack H_3O^+\rbrack x\lbrack HCOO-\rbrack^{_{}}}{\lbrack HCOOH\rbrack}`

We clear (H3O+) and we apply -log and we get this:

`pH\text{ = }pKa+log(([HCOO^-])/([HCOOH]))`
`pH=3.77+log(([0.52])/([0.30]))\text{ }`

Answer: pH = 4.01

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pOH)

pH + pOH = 14

(This formula appears from the ionic equilibrium of water)

pOH = 14 - 4.01 = 9.99

Answer: pOH= 9.99

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