3. You wish to prepare a buffer solution containing NaN 3 and HN 3 (hydrazoic ac

ID: 872588 • Letter: 3

Question

3. You wish to prepare a buffer solution containing NaN3 and HN3 (hydrazoic acid). You have to prepare the solutions first.

a. 187 g of sodium azide is dissolved in water to make a solution with a final volume of 1.0L. What is the concentration and pH of the solution?

b. What is the molarity a solution made by adding 120.0 mL of a 6.0 M HN3(aq) to water to a final volume of 250.0 mL?

c. What volume of the HN3(aq) made in part b is needed to add to 100.0 mL of the NaN3(aq) solution of made in part a to make a buffer at pH= 5.00?

a) Molecular mass of NaN3: 23 + 3(14) = 65 g/mol

NaN3 in water produces Na+ and N3 -
N3 - hydrolyzes to form HN3 + OH- : N3- + H2O = HN3 + OH-

moles = 187 / 65 = 2.88 moles.

[NaN3] = 2.88 / 1 = 2.88 mol/L

For the pH we need the Kb. The Ka for hydrazoic acid is 1.9x10-5 so the Kb = 1x10-14 / 1.9x10-5 = 5.26x10-10

N3- + H2O = HN3 + OH-

2.88 0 0

2.88 - x x x

Kb = [HN3]*[OH] / [N3] -----> 5.26x10-10 = x2 / 2.88 - x Here, we can ignore X because Kb value is too low so:

5.26x10-10 = x2 / 2.88 ----> (5.26x10-10 x 2.88)1/2 = x x = 3.89x10-5 mol/L = [OH]

pOH = -log (3.89x10-5) = 4.41

pH = 14 - 4.41 = 9.59

b) M = 6x0.120 / 0.25 = 2.88 M

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.