Henderson Haselback quick problem

[discowisco]

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What is the pH after 30 ml of 1.00 M NaOH has been added to 100 mL 0.50 M HOAc? HOAc has a pka = 4.74?

I understand Ph = pka + log base/acid , i just dont understand what to put for base and acid.

I calculated the moles of base to be (.030*1) = .030 and the moles of acid to be .1*.5= .5

Thus shouldnt is be ph = 4.74 + log .03/.5??

The answer is 4.92 instead

 

[cheesier]

When you add 30 ml of 1 M NaOH to 100 ml of 0.5 M HOAc, 60 mL of the HOAc gets neutralized, leaving 40 mL of acid.

4.74 + log(60/40) = 4.92

Remember, when you use the henderson hasselbach equation you plug in the conjugate acid and base, so you wouldn't enter NaOH in the log as you did because we are looking at HOAc. The only role that NaOH plays is in neutralizing some of the HOAc.

 

[wehttamrrek]

Ok so so what you have is .03 mol of a strong base and .05 mol of the weak acid. The base will depronate the HOAc and you will end up with .02 mol of HOAc remaining with .03 mol of its conjugate base present.


pH=4.74+log(.03/.02)=4.916=4.92

 

[discowisco]

Alright I think Im getting it but so what about something like this:

What is the pH after 70 ml of 0.20 M HCL has been added to 50 ml 0.60 M H3CNH2? H3CHN2 has a pkb = 3.42.

I understand how TBR answers this qualitatively but if we were to do this henderson hasselback way how would we do so?

 

[wehttamrrek]

Use the pOH version of the equation: pOH=pKb+log(con.acid/base)

HCl=.014 mol

Remaining base after HCl: .03-.014=.016 mol

pOH=3.42+log(.014/.016)=3.36

pH=14-3.36=10.64

Let me know if that's wrong, a little rusty hah

 

[discowisco]

Thanks alot ! also one more quick question so i know that at the half equivalence point the ph = pka because there is equal amounts of conj acid/ conj acid since ph = pka + log 1/1.

But for the equivalence point isnt there equal amounts of moles of acid and base too? Why isnt the ph = pka at the equivalence point?

I guess what i am trying to ask is how would you solve for PH at the equivalence point then?

 

[wehttamrrek]

Thanks alot ! also one more quick question so i know that at the half equivalence point the ph = pka because there is equal amounts of conj acid/ conj acid since ph = pka + log 1/1.

But for the equivalence point isnt there equal amounts of moles of acid and base too? Why isnt the ph = pka at the equivalence point?

I guess what i am trying to ask is how would you solve for PH at the equivalence point then?

Ok so at 1/2 the eq. Point the ratio is 1 like you said, so by adding another half of the acid/base you are changing that ratio so then your pH will not equal the pKa/pKb.

Make sense? Hope this helps.