What is the pH at the equivalence point when a 25.0 mL sample of 0.200 M aqueous formic acid (HCO₂H)is titrated with 0.100 M aqueous potassium hydroxide? (K_a of HCO₂H = 1.8 × 10^-4) A) 5.71 B) 7.00 C) 8.28 D) 8.52 E) 10.26

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Formic acid (HCO₂H) is a weak acid that reacts with potassium hydroxide (KOH) to form a salt (potassium formate) and water. The balanced chemical equation for the reaction is: HCO₂H + KOH → KOC(H)=O₋ + H2O At the equivalence point, the number of moles of KOH added is equal to the number of moles of HCO₂H initially present in the solution. This means that all of the HCO₂H has been neutralized and converted to its conjugate base (formate ion) along with water. The resulting solution contains only the salt and the water. The pH at the equivalence point can be calculated using the following formula: pH = pKa + log([base]/[acid]) where pKa is the acid dissociation constant of HCO₂H (1.8 × 10^-4), [base] is the concentration of the formate ion, and [acid] is the concentration of formic acid initially present. At the equivalence point, [base] = [acid], so the equation simplifies to: pH = pKa + log(1) pH = pKa Therefore, the pH at the equivalence point is equal to the pKa of HCO₂H, which is: pH = -log(1.8 × 10^-4) = 3.74 However, this is the pH of the solution at the midpoint of the titration, not the equivalence point. At the midpoint, half the amount of the HCO₂H has been neutralized and converted to its conjugate base. At the equivalence point, all the HCO₂H has been neutralized. To calculate the pH at the equivalence point, we need to consider the reaction stoichiometry. For every one mole of HCO₂H that is neutralized by KOH, one mole of formate ion is formed. This means that the final volume of the solution is the sum of the initial volumes of the HCO₂H and KOH solutions. Before the start of the titration, the HCO₂H solution has a volume of 25.0 mL and a concentration of 0.200 M. This means that it contains: 25.0 mL × 0.200 mol/L = 5.00 mmol HCO₂H During the titration, we are adding 0.100 M KOH solution. At the equivalence point, the number of moles of KOH added is equal to the number of moles of HCO₂H initially present in the solution. This means that we need: 5.00 mmol HCO₂H ÷ (0.100 mol/L) = 50.0 mL KOH The total volume of the solution at the equivalence point is: 25.0 mL + 50.0 mL = 75.0 mL Since we are dealing with a strong base (KOH), the pH of the solution at the equivalence point will be greater than 7.00 (neutral pH). The correct pH can be calculated using the equation: moles of formate ion = moles of HCO₂H neutralized [formate ion] × 75.0 mL = 5.00 mmol [formate ion] = 5.00 mmol ÷ 75.0 mL = 0.067 M pH = pKa + log([base]/[acid]) pH = -log(1.8 × 10^-4) + log(0.067/0.200) pH = 8.28 Therefore, the best choice is C.

What Is the PH at the Equivalence Point When a 25.0