Answer:
Step-by-step explanation:
Formic acid (HCOOH) is a weak acid and KOH is a strong base. The titration reaction is:
HCOOH + KOH → HCOO^- + H2O
Before adding any KOH, the solution is just formic acid, so the pH is determined by the dissociation of formic acid:
HCOOH + H2O ⇌ HCOO^- + H3O^+
The equilibrium constant expression is:
Ka = [HCOO^-][H3O^+] / [HCOOH]
Since formic acid is a weak acid, we can assume that [HCOO^-] is negligible compared to [HCOOH]. Therefore, the equation simplifies to:
Ka = [H3O^+]^2 / [HCOOH]
Taking the negative logarithm of both sides:
-pKa = log([H3O^+]^2 / [HCOOH])
2pKa - pH = log([HCOOH] / [H3O^+])
At the start of the titration, there is 0.10 mol/L formic acid in the solution. Therefore, the initial concentration of formic acid is:
[formic acid] = 0.10 mol/L
The initial pH can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([formic acid] / [formate ion])
At the equivalence point, all of the formic acid has reacted with the KOH to form sodium formate (HCOONa). The pH of the solution will be determined by the hydrolysis of the sodium formate:
HCOO^- + H2O ⇌ HCOOH + OH^-
The equilibrium constant expression is:
Kb = [HCOOH][OH^-] / [HCOO^-]
Since sodium formate is the salt of a weak acid and a strong base, we can assume that the concentration of OH^- is equal to the concentration of NaOH that has been added. Therefore:
Kb = [HCOOH][NaOH] / [HCOONa]
Taking the negative logarithm of both sides:
-pKb = log([HCOOH][NaOH] / [HCOONa])
At the equivalence point, the moles of NaOH added are equal to the moles of formic acid in the solution:
moles of NaOH = 0.10 mol/L × 0.05000 L = 0.00500 mol
At this point, all of the formic acid has been neutralized and converted to sodium formate, so the final concentration of formate ion is:
[formate ion] = 0.00500 mol / 0.07500 L = 0.0667 mol/L
The concentration of formic acid is zero at the equivalence point, so the final pH is determined by the hydrolysis of the sodium formate:
Kb = [HCOOH][NaOH] / [HCOONa]
-pKb = log([HCOOH][NaOH] / [HCOONa])
-pKb = log([HCOOH][NaOH] / [0.0667])
Since pKb for formate ion is 10.37, we can substitute this value and solve for the pH:
10.37 = log([HCOOH][NaOH] / [0.0667])
[HCOOH][NaOH] / [0.0667] = 10^-10.37
[HCOOH][NaOH] = 6.66 × 10^-11
Substituting the concentration of NaOH at the equivalence point:
[H