# chem- acids and bases

sorry to take up your time again, acids and bases just really confuse me, i'd really appreciate your help again.

"What is the pH at the end point for the titration of 0.20M HNO2 by 0.20M NaOH? Ka nitrous acid= 4.5 x 10^-4."

this is what i do know:
- end point, so it's the salt solution of a weak acid/strong base
- i THINK the formula for the ICE chart is NO2^- + H2O ==> HNO2 + OH^-
- i think i need to use the Ka to solve for x now. is this thinking wrong?

- how do i find the concentration of HNO2, is it 0.20M?
- is the OH- added 0.20M also?

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1. correction for typo: "- how do i find the concentration of HNO2, is it 0.20M? "

i meant NO2^- instead of HNO2

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2. You're on the right track.
............NO2^- + HOH ==> HNO2 + OH^-
initial.....0.1.............0........0
change......-x...............x.......x
equil......0.1-x.............x........x

Kb for NO2^- = (Kw/Ka for HNO2) =
(HNO2)(OH^-)/(NO2^-)
(HNO2) = (OH^-) = x
(NO2^-) = 0.1-x.
Solve for x which is OH, convert that to pH.
Most students ask, "Where did the 0.1M come from for NO2^-?" If we have 0.2M acid titrated with 0.2M base, then the salt formed is 0.2M EXCEPT that it has been diluted. If you do it in moles it is easier to see.
We start with an easy number like 100 mL x 0.2M acid = 20 millimoles acid. That reacts exactly with 20 mmoles base to form 20 mmoles salt. What is the concn of the salt? It is M = mmoles/mL = 20 mmoles/200 mL = 0.1M. You are looking for shortcuts. When you titrate acids with bases and the concns are the same, as they are in this problem, then the concn of the salt is just 1/2 of the starting molarity of the acid (or base). So 0.1M HCl titrated with 0.1M NaOH produces 0.05M NaCl.
0.3M HNO2 + 0.3M NaOH produces 0.3/2 = 0.15M NaNO2.

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3. you're the best, thank you so much!

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