You wish to prepare a buffer solution containig NaN3 and HN3. You have to prepar

ID: 1066637 • Letter: Y

Question

You wish to prepare a buffer solution containig NaN3 and HN3. You have to prepare the solution first .

a) 187g 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 of a solution made by adding 120.0 mL of a 6.0M HN3 to water to a final volume of 250.0 mL ?

c) What volume of the NH3 made in part B is needed to add to 100.0 mL of the NaN3 solution of made in part A to make a buffer at pH=5.00 ?

molar mass of sodium azide = 65 g/mol

now, 187g * 1mol/65g = 2.88 moles/L = 2.88 M

N3^- + H2O --> HN3 + OH^-

Ka hydrazoic acid = 2.5*10^-5

Kb azide = 4*10^-10

Kb = 4*10^-10 = [H3N][OH-]/[NaN3] = (x)(x)/2.88

x^2 = 11.52*10^-10

x = [OH-] = 3.39*10^-5

pOH = 4.46

pH = 9.53

b) (120ml)(6M) = (250ml)(x M) and x = 2.88 M HN3

c) pH = pKa + log [NaN3]/[HN3]

5 = pKa + log [NaN3]/[HN3]

log [NaN3]/[HN3] = 5 - 4.6 = 0.4

[NaN3]/[HN3] = 2.5

0.1L * 2.88mol/L = 0.288 mole NaN3

0.288/x = 2.5

x = 0.115

(xL)(2.88mol/L) = 0.115 mol

x = 0.04 L

Need to add 40 ml of 2.88 M HN3

Now we will end with 2.05 M NaN3 and 0.82 M HN3 and ratio will be 2.5 : 1 as needed

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