Part 1. We’re going to titrate formic acid with the strong base, NaOH. There is

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?


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?





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?




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?

Explanation / Answer

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 Kb (base dissociation constant of formate – available in tables or can be calculated from Ka of formic acid). The Kb value is 4.8*10-11.

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

Kb = [HCOOH][OH-]/[HCOO-]

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

Since Kb 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).

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---