for the titration of 10.0 ml of 0.200 m nitrous ... Question For the titration o
ID: 989759 • Letter: F
Question
for the titration of 10.0 ml of 0.200 m nitrous ... Question For the titration of 10.0 mL of 0.200 M nitrous acid HNO2 with 0.100 M NaOH, answer the following questions and construct the titration curve (please use pKa = 3.34 for nitrous acid). a. What volume (in mL) of 0.100 M NaOH is required to neutralize 10.0 mL of 0.200 M nitrous acid? b. What is the pH of the solution before any NaOH is added? c. What is the pH of the half-stoichiometric acid or half-equivalence point? d. What is the pH when 95% of the NaOH required to reach the stoichiometric point has been added? e. What is the pH of the stoichiometric point? f. What is the pH when a 5% excess of NaOH required to reach the stoichiometric point has been added?g attach your graph to this test
Explanation / Answer
a)
M1V1 = M2v2
0.1 x V1 = 0.2 x 10
V1 = 20 ml
hence 20 ml of NaOH is required.
b) before any NaOH added
pH = pKa = 3.34
c) In acid-base titration the ratio between the acid and corresponding base is exactly 1:1 at the half-equivalence point.
Since acid and base react in a 1:1 ratio or in equal volumes or equimolar proportion. total volume = 30
MV = M1V2 +M2V2
M *30 =0.1*20 +0.2*10 = 4
M =4/30 = 0.1333
The reaction is
HNO3+ NaOH-> NaNO3 +H2O
hence half concentration of salt NaNO3 will be 0.066
Hence
KaKb = Kw
pKb =14-pKa
= 14- 3.34 = 10.66
half equivalence point will be 5.33
c)
pH = pKa + log ([base] / [acid]) .
When 95% of the base required to reach the equivalence point has been added, the ratio of propionate ion to acid is 95/5. Thus
pH = pKa + log 95/5 = 3.34+ 1.28 = 4.62
e)
The pH after the equivalence point is solely a function of the excess base added. 105% of the stoichiometric amount = 1.05 x 20.0 mL = 21.0 mL. So 1.0 mL has been added in excess.
mmoles excess base added = M OH- x mL OH- = (0.10)(1.0) = 0.10 mmoles OH-
The solution volume at that point = 31.0 mL.
[OH-] = mmoles OH- / mL of solution = 0.10 / 31.0 = 0.0032 M
pOH = -log [OH-] = -log 0.0032 = 2.49
pH = 14.00 - pOH = 14.00 - 2.49 = 11.51
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