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?

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

Substitution

But     0.15 - x ≈   0.15

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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

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