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Calculate the pH for the following weak acid.

A solution of HCOOH has 0.15M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium?

Answer :

joseaaronlara

Answer:

The answer to your question is pH = 2.28

Explanation:

Chemical reaction

                    HCOOH(ac)  +  H₂O  ⇔   H₃O⁺   +  HCOO⁻

I                        0.15                 --              0               0

C                       - x                   --             +x              +x

F                     0.15 - x             --               x                x

Write the equation of equilibrium

                             [tex]ka = \frac{[H3O][HCOO]}{[HCOOH]}[/tex]

Substitution

                             [tex]1.8 x 10^{-4} = \frac{[x][x]}{0.15 - x}[/tex]

But     0.15 - x ≈   0.15

                             [tex]1.8 x 10^{-4} = \frac{x^{2} }{0.15}[/tex]

Solve for x

                             x² = (0.15)(1.8 x 10⁻⁴)

Simplification

                             x² = 0.000027

Result

                             x = 0.0052

                          pH = -log [H₃O⁺]

Substitution

                          pH = - log [0.0052]

Simplification and result

                         pH = 2.28

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