Question

Exactly 100 mL of .10 M HNO2 are titrated with a .10 M NaOH solution. What is the pH at the equivalence point? Ka (HNO2) = 5.6*10^-4

Answer 1

Could someone provide the reactions as well?

Answer 2

pH = -log([H+])

Answer 3

HNO2 + NaOH -> H2O + NaNO2

Answer 4

notice it is a 1 to 1 reaction

Answer 5

Do you need to react NaNO2 with H2O ?

Answer 6

Ka = [A-][H+]/[HA]

Answer 7

https://en.wikipedia.org/wiki/Acid_dissociation_constant

Answer 8

ok to the actual problem

Answer 9

https://en.wikipedia.org/wiki/Equivalence_point

Answer 10

http://www.chemguide.co.uk/physical/acidbaseeqia/sasb2.gif

Answer 11

that link will explain to you what the equivalence point is, it is the point in which 50%% of the acid is protonated and 50% of the acid is deprotonated

Answer 12

You can solve this problem by simply finding the pKa

Answer 13

https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation

Answer 14

pKa is the point at which a chemical is 50% protonated and 50% deprotonated

Answer 15

Well. The reaction you provided produced a basic salt. So that product should react with water next, in order to produce OH-. NO2- + H2O -- HNO2 + OH-. So I can use the Kb to find [OH-]. And from there I can find pOH and subtract it from 14 to get pH.

Answer 16

I got it now. Thank you!

Answer 17

yeah Im rusty with this stuff I'm not going to lie and it is 4am here, sorry for not providing you with a very sound answer, im much too tired to be coherent

Answer 18

Haha no worries. I'm just cramming for an exam tomorrow.
Have a good night!

Answer 19

\[pKa = -\log(5.6*10^{-4})\]
Pka = 7.48757

pH = pKa + log[A/HA]

A = 0.1M
Ha = 0.1M

Log[0.1/0.1] = 0

therefore,

pH = pKa

like I state earlier

Answer 20

stated*

Answer 21

This is because the reaction is 1 to 1 as you should see above therefore the equivalence point is when the concentration of the conjugate base is equal to the concentration of the acid

Answer 22

I gave you the answer but I failed to prove to you why pKa = pH at which an acid is at its equivalence point

Answer 23

Since,
pH = pKa

Pka = -log(Ka)

therefore

pH = -log(Ka)

at the equivalence point

Answer 24

Therefore the pH at the equivalence point is 7.48757

Answer 25

sorry for ignoring sig figs