Question

Part 1. We’re going to titrate formic acid with the strong base, NaOH. There is initially 100. mL of 0.50 M formic acid and the concentration of NaOH is 1.0 M. All work must be shown to receive credit

7) What is the pOH at the equivalence point? ___________

8) If, instead of NaOH being added, 0.05 moles of HCl is added by   

bubbling the gas through the solution. Assume that the volume has

not changed. What is the percent dissociation of formic acid?

Answer 1

I-7) Millimoles of formic acid = (100.0 mL)*(0.50 M) = 50 mmole.

At the equivalence point, formic acid is completely neutralized to sodium formate. The volume of NaOH required to reach the equivalence point = (50 mmole)/(1.0 M) = 50.0 mL.

Millimoles of sodium formate formed = millimoles of formic acid neutralized = 50 mmole.

Sodium formate, Na(HCOO) is the conjugate base of the weak acid, formic acid, HCOOH as establishes equilibrium as

HCOO^- (aq) + H₂O (l) ----------> HCOOH (aq) + OH^- (aq)

Since OH^- is a base, we must work with K_b (base dissociation constant of formate – available in tables or can be calculated from K_a of formic acid). The K_b value is 4.8*10^-11.

[HCOO^-] = (mmoles of HCOONa)/(total volume of solution) = (50 mmole)/[(100.0 + 50.0) mL] = 0.3333 M.

K_b = [HCOOH][OH^-]/[HCOO^-]

====> 4.8*10^-11 = (x).(x)/(0.3333 – x)

Since K_b is small (of the order of 10^-11), we can assume the extent of dissociation is small, i.e, x << 0.3333 M; therefore,

4.8*10^-11 = x²/(0.3333)

====> x² = 1.59984*10^-11

====> x = 3.9998*10^-6 ≈ 4.0*10^-6

The [OH^-] = 4.0*10^-6 and pOH = -log [OH^-] = -log (4.0*10^-6) = 5.3979 ≈ 5.40 (ans).