You wish to prepare a buffer solution containing NaN_3 and HN_3 (hydrazoic acid)

Question

You wish to prepare a buffer solution containing NaN_3 and HN_3 (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 HN_3(aq) to water to a final volume of 250.0 mL? c. What volume of the HN_3(aq) made in part b is needed to add to 100.0 mL of the NaN_3(aq) solution of made in part a to make a buffer at pH = 5.00?

Explanation / Answer

Solution:- (a) mass of NaN3 = 187 g

molar mass of NaN3 = 65.0 g/mol

187 g x (1mol/65g) = 2.88 mol

volume of solution = 1.0 L

concentration of solution = 2.88 mol/1.0 L = 2.88 M

NaN3 dissociates to give Na+ and N3- that on reacting with water gives OH- as...

N3-(aq) + H2O(l) <------> HN3(aq) + OH-(aq)

I 2.88 0 0

C -X +X +X

E (2.88 - X) X X

Kb = [HN3][OH-]/[N3-]

5.26 x 10-10 = (X)2/(2.88 - X)

since the value of Kb is very low so 2.88 - X could be taken as 2.88

5.26 x 10-10 = (X)2/(2.88 )

X2 = 5.26 x 10-10 x 2.88 = 1.51 x 10-9

taking square root to both sides..

X = 3.89 x 10-5

[OH-] = 3.89 x 10-5 M

pOH = - log(3.89 x 10-5) = 4.41

pH = 14 - pOH

pH = 14 - 4.41

pH = 9.59

(b) Dilution is done here so we will use the dilution equation, M1V1 = M2V2

where M1 and M2 are initial and final molarities and V1 and V2 are initial and final volumes.

6.0 M x 120.0 ml = M2 x 250.0 ml

M2 = (6.0 M x 120.0 ml)/250.0 ml

M2 = 2.88 M

so, the molarity of solution of HN3 is 2.88 M.

(c) 100.0 ml of NaN3 solution from part A means 2.88 M solution are taken to make the buffer solution, moles of NaN3 used = 100.0 ml x (1L/1000ml) x (2.88 mol/L) = 0.288 mol

let's say X moles of HN3 from part B are used.

Handerson equation equation is used for finding out pH of buffer solutions, pH is given here as 5.00

pH = pKa + log(base/acid)

Ka for HN3 is 1.9 x 10-5.

pKa = - log Ka

pKa = - log 1.9 x 10-5 = 4.72

Let's plug in the values in the Handerson equation..

5.00 = 4.72 + log(0.288/X)

5.00 - 4.72 = log(0.288/X)

0.28 = log(0.288/X)

taking antilog....

100.28 = 0.288/X

1.91 = 0.288/X

X = 0.288/1.91 = 0.151

so, 0.151 moles of HN3 are required from part B solution. molarity from part B is 2.88 M that is 2.88 mol/L

so, 0.151 mol x (1L/2.88 mol) x (1000 ml/1L) = 52.4 ml

So, 52.4 ml of HN3 from part B are required.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

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