A 25.0 ml sample of 0.150 M nitrous acid is titrated with a0.150 M NaOH solution. What is the pH at the equivalence point? theKa of Nitrous Acid is 4.50 x 10-4.
3 Answers
See the answer by @Andrea. https://socratic.org/users/andrea-b-3
pH= 8.11
Explanation:
When the titration occurrs you have used 25,0 mL of NaOH so you in total have 50.0 mL of
pH of equivalent point is the pH of this solution of
It is a salt with a basic hydrolysis
pOH = 5.89
pH= 8.11
You should treat this problem with care... most students would
The
By neutralizing to the equivalence point, only
0.150 mol HNO2L×0.025 L=0.00375 mols NO−2 formed
upon exact neutralization of
Therefore, we get a diluted concentration of
[NO−2]i=0.00375 mols0.050 L=0.075 M NO−2 ,
as expected since the volume was doubled (so the concentration should halve).
Now, this nitrite will associate in an equilibrium within water:
NO−2(aq)+H2O(l)⇌HNO2(aq)+OH−(aq)
I 0.075 − 0 0
C −x − +x +x
E 0.075−x − x x
Here we have a BASE, and so we must use the
KaKb=Kw
⇒KwKa=Kb=10−144.50×10−4=2.22×10−11
It is apparent that this
Write the mass action expression:
Kb=2.22×10−11=x20.075−x≈x20.075
Hence,
x≡[OH−]=√0.075Kb
=√0.075⋅2.22×10−11
=1.29×10−6M OH−
Therefore,
pOH=−log[OH−]=5.89
pH=14−pOH=8.11
