Question

1.Calculate the pH of a 0.075 M soultion of hydrazoic acid, which is a weak acid. Hydrazoic acid, HN3, has an acid dissociation constant Ka, equal to 1.9 x 10-5. Also, calculate the concentrations of the azide ion (N3-) and the hydroxide ion.

2. Calculate the pH that would result if 15.0 mL of 0.075 M NaOH, strong base, were added to 15.0mL of the solution in the problem above.

I have already completed 1, just need help with 2. If you can also show your work I would greatly appreciate it. Please and thank you.

Answer 1

1)For a weak HN3 acid we can write,

                                                              HN₃ < ----------- > H⁺       +            N₃^- .

Initial concentrations                    0.075M                   0                         0

Conc. At eq^m.                                 (0.075 – X)                X                         X

Dissociation constant Ka is given as,

Ka =[H⁺][N₃^-]/[HN₃]………..(1)

Ka = 1.9 x 10^-5 and [H⁺] = X moles/L, [N₃^-]= X moles/L, [HN₃] = (0.075 – X).

Hence, eq.1 takes form

1.9 x 10-5 = X x X / (0.075-X)……………..(2)

As, HN3(hydrazoic acid) is weak acid it ionizes little hence, X <<< 0.075 so (0.075 – X) = 0.0759(apprx.)

Eq.2 takes form,

1.9 x 10-5 = X₂/(0.075)

X² =1.9 x 10-5 x 0.075

X² = 1.425 x 10^-6

X = 1.19 x 10^-3 moles/L

i.e. [H⁺] = X = 1.19 x 10^-3 moles/L and also [N₃^-] = X = 1.19 x 10^-3 moles/L

pH = -log[H+]

pH = -log(1.19 x 10^-3)

pH = 2.92

Hence pH of 0.075 M HN₃ (hydrazoic acid) is 2.92.

We know that

pH + pOH = 14

pOH = 14 – pH

pOH = 14 – 2.92

pOH = 11.08

-log[OH^-] = 11.08

log[OH^-] = -11.08

Taking antilog of both sides,

[OH^-] = Antilog(-11.08)

[OH^-] = 8.32 x 10^-12 moles/L.

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2)

NaOH + HN3 < -------- > NaN3 + H2O.

a)Determine milimoles of acid and bases before reaction,

NaOH = Molarity x volume = (0.075M) x (15mL) = 1.125 milimoles

HN3 = molarity x volume = (0.075M) x (15mL) = 1.125 milimoles.

Addition of equal Milimoles of weak acid and strong base will give the solution whose solution is numerically equal to pKa of acid

Hence,

pH = pKa of HN3 acid

pH = pKa = -log(Ka)

pH = -log(Ka)

pH = -log(1.9 x 10^-5)

pH = 4.72

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