A 25.0-mL sample of 0.35 M HCOOH is titrated with 0.20 M KOH. What is the pH of the solution after 25.0 mL of KOH has been added to the acid? K_a = 1.77 × 10⁻⁴ A)4.00 B)3.88 C)3.63 D)3.51 E)3.47

Question

Quizplus · Accepted Answer

Step 1: Write the balanced chemical equation for the reaction between HCOOH and KOH:
HCOOH + KOH → HCOOK + H2O
Step 2: Determine the number of moles of HCOOH:
moles HCOOH = Molarity × Volume (in liters) = 0.35 mol/L × 0.0250 L = 0.00875 mol
Step 3: Determine the number of moles of KOH:
moles KOH = Molarity × Volume (in liters) = 0.20 mol/L × 0.0250 L = 0.005 mol
Step 4: Determine the limiting reagent:
Since the stoichiometric ratio of HCOOH to KOH is 1:1, KOH is the limiting reagent because there are fewer moles of KOH than HCOOH.
Step 5: Determine the number of moles of HCOOK produced:
moles HCOOK = moles KOH = 0.005 mol
Step 6: Determine the concentration of HCOOH after the addition of KOH:
Volume of HCOOH = 25.0 mL = 0.0250 L
Total volume after the addition of KOH = 25.0 mL + 25.0 mL = 0.0500 L
moles HCOOH = initial moles of HCOOH - moles of KOH used = 0.00875 mol - 0.005 mol = 0.00375 mol
Molarity of HCOOH after the addition of KOH = moles HCOOH / Volume of HCOOH = 0.00375 mol / 0.0250 L = 0.150 M
Step 7: Calculate the pKa of HCOOH using the given Ka value:
Ka = [HCOO-][H+] / [HCOOH]
1.77 × 10^-4 = x^2 / 0.150
x = 0.0112 M
pH = -log[H+] = -log(0.0112) = 1.95
pKa = -log(Ka) = -log(1.77 × 10^-4) = 3.75
Step 8: Calculate the amount of HCOOK formed and the concentration of HCOO-:
moles HCOOK = moles KOH used = 0.005 mol
moles HCOO- = moles HCOOK = 0.005 mol
Volume of solution = 0.0500 L
Concentration of HCOO- = moles HCOO- / Volume of solution = 0.005 mol / 0.0500 L = 0.100 M
Step 9: Use the Henderson-Hasselbalch equation to calculate the pH of the solution:
pH = pKa + log([HCOO-] / [HCOOH])
pH = 3.75 + log(0.100 / 0.150) = 3.88
Therefore, the pH of the solution after 25.0 mL of 0.20 M KOH has been added to 25.0 mL of 0.35 M HCOOH is 3.88. The correct answer is (B).

A 250-ML Sample of 0