Question

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. See announcement for due date.

a) What is the initial pH of the formic acid solution?​​​___________

b) What is the percent ionization under initial conditions? ​​​__________%

c) After the addition of 10 mL of NaOH, what is the pH? ​​        ___________

d) After the addition of 25 mL of NaOH, what is the pH? Think ​        ___________

  about where in the titration this brings you.

e) What volume of NaOH is required to reach the equivalence point?   ​       ​ ___________

f) What is the pOH at the equivalence point?     ​        ​  ___________

g) What is the pH at the equivalence point?   ​   ​___________

h) 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

Formic acid is a strong acid which completely dissociates in water.

a)

HCOOH ---> H⁺ + HCOO^-

Molarity of H⁺ = 0.5 M

pH = -log [H+] = -log(0.5)

pH = 0.301

b) % ionization = 100

c)

Milimoles of NaOH added = 10 x 1 = 10

Milimoles of formic acid = 100 x 0.5 = 50

Milimoles of formic acid neutralized = 10

Milimoles of Formic acid left = Milimoles of H+ = 50-10=40

Total volume of solution = 110 mL

[H+] = 40 / 110 = 0.3636 M

pH = -log [H+] = -log(0.3636)

pH = 0.439

d)

Milimoles of NaOH added = 25 x 1 = 25

Milimoles of formic acid = 100 x 0.5 = 50

Milimoles of formic acid neutralized = 25

Milimoles of Formic acid left = Milimoles of H+ = 50-25 = 25

Total volume of solution = 125 mL

[H+] = 25 / 125 = 0.2 M

pH = -log [H+] = -log(0.2)

pH = 0.698

e)

Milimoles of NaOH required to reach equivalence point = Milimoles of formic acid = 50

Molarity of NaOH = 1 M

Volume of NaOH required = 50 / 1 = 50 mL

f) & g)

At equivalence point,

pH = pOH = 7