Final answer:
The pH at the equivalence point in the titration of 50.0 mL of 0.50 M HCO₂H with 1.0 M KOH can be calculated using the hydrolysis of the conjugate base of the weak acid. Taking into account the equilibrium constant expression, the concentration of hydroxide ions can be determined, leading to the calculation of the pOH and pH of the solution. The pH at the equivalence point is 9.04.
Step-by-step explanation:
The pH at the equivalence point in the titration of 50.0 mL of 0.50 M HCO₂H (Ka = 1.810-4) with 1.0 M KOH can be calculated using the concept of the dissociation of the weak acid. At the equivalence point, the moles of acid will be equal to the moles of base added. This means that all of the HCO₂H will be converted to HCO₂⁻, making the solution a salt solution. Since HCO₂⁻ is the conjugate base of a weak acid, it will undergo hydrolysis in water.
The hydrolysis reaction can be represented by the equation: HCO₂⁻ + H₂O ⇌ HCOOH + OH⁻
To calculate the pH, we need to consider the equilibrium constant expression for the hydrolysis reaction. The expression can be written as follows: Ka = [HCOOH][OH⁻] / [HCO₂⁻]
Since we have all the concentrations except for [OH⁻], we rearrange the equation to solve for [OH⁻]: [OH⁻] = (Ka * [HCO₂⁻]) / [HCOOH]
Substituting the given values into the equation, we get [OH⁻] = (1.810-4 * 0.0500) / 0.100 = 9.05 x 10⁻⁵ M
To calculate the pOH, we can use the equation: pOH = -log[OH⁻]
Substituting the calculated value of [OH⁻] into the equation, we get pOH = -log(9.05 x 10⁻⁵) = 4.96
Finally, to calculate the pH, we can use the equation: pH = 14 - pOH
Substituting the calculated value of pOH into the equation, we get pH = 14 - 4.96 = 9.04