i) pH at equivalence point Reaction between HN3 and NaOH is HN3 + NaOH ------> N
ID: 700213 • Letter: I
Question
i) pH at equivalence point
Reaction between HN3 and NaOH is
HN3 + NaOH ------> NaN3 + H2O
this is 1:1 molar reaction
So, 25ml of 0.150M HN3 react with 25ml of 0.150M NaOH
Equivalence point is 25ml
at equivalence point all the HN3 is converted into NaN3
Total Volume = 25ml + 25ml = 50ml
Dilution of NaN3 concentration = 2 time
Concentration of NaN3 = 0.150M/2 = 0.075M
N3- is partly hydrolysed by water
N3- + H2O <-------> HN3 + OH-
Kb = [HN3][OH-]/[N3-] = 1.9×10^-5
Kb = Kw/Ka
= 1.00 ×10^-14/1.9 ×10^-5
= 5.26×10^-10
at equillibrium
[HN3] = X
[OH-] = X
[ N3-] = 0.075 - X
X^2/(0.075 - X) = 5.26 ×10^-10
we can assume 0.075 - X = 0.075 because x is small value
X^2/0.075 = 5.26 ×10^-10
X^2 = 3.95 ×10^-11
X= 6.28×10^-6
[OH-] = 6.28×10^-6M
pOH = 5.20
pH = 14 - pOH
= 14 - 5.20
= 8.80
ii) pH at 26ml of NaOH solution addition
out of 26 ml , 25ml used for neutralization of HN3
excecess volume of NaOH = 1ml
No of mole of excess OH- added = (0.150mole/1000ml)×1ml = 0.00015
Total volume = 51ml
[OH-] = (0.00015mol/51ml)×1000ml = 0.00294M
pOH = 2.53
pH = 14 - 2.53
= 11.47
Explanation / Answer
A 25.0 mL sample of 0.150 M hydrazoic acid (HN3) is titrated with a 0.150 M NaOH solution. Calculate the pH at equivalence and the pH after 26.0 mL of base is added. The Ka of hydrazoic acid is 1.9x10^-5.
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