a 25.0 ml sample of 0.50 m hno2 is titrated with 0.25 m koh. what is the ratio of conjugate base to weak acid, at the point where 10.0 ml of koh has been added during the titration? ka for nitrous acid is 5.6 x 10-4.

Respuesta :

The ratio of conjugate base to weak acid, at the point where 10.0 ml of KOH has been added during the titration is 8.11

Using the Henderson-Hasselbalch equation and using HNO2 as the acid and NaOH as the base, (ii) using the Ka and thus assuming HNO2 is dissociating in water, or (iii) forgetting about dilution and using 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 equilibrium with the 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, 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

To learn more about Henderson equation visit:

brainly.com/question/13423434

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