Answer:
pH=14−pOH=8.11
Explanation:
(i) using the Henderson-Hasselbalch equation and use HNO2 as the acid and NaOH as the base, (ii) using the Ka and thus assume HNO2 is dissociating in water, or (iii) forget about dilution and use 0.150 M=[NO−2]i. That doesn't make sense.
The pH is 8.11.
By neutralizing to the equivalence point, only NO−2 remains. That comes from
0.150 mol HNO2L×0.025 L=0.00375 mols NO−2 formed
upon exact neutralization of HNO2 with NaOH. This is now in a volume of 25.0 mL+25.0 mL=50.0 mL=0.050 L.
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 Kb:
KaKb=Kw
⇒KwKa=Kb=10−144.50×10−4=2.22×10−11
It is apparent that this Kb is very small, so the small x approximation works here very well. This is true usually when K is on the order of 10−5 or less.
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
