chemistry question (acid/base buffers)

[Pdentalstudent]

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I am not understanding the answer for this question can someone explain?:

A certain buffer solution is 3 M in HF and 2 M in NaF. Calculate the pH of this buffer given that the Ka of HF = 7.0*10^-4.

The solution:
STEP 1: Ka = [H+][F-]/[HF] --> I don't understand this first step, because in the other problems I did, the [H+] is always equal to the [A-] so why is [H+] different from [F-]?

STEP 2:
7.0*10^-4=[H+][2]/[3]

STEP 3: [H+] = 1.05*10^-3 and pH = 2.98

Thanks in advance

 

[buckey]

You can treat a metal associated with a nonmetal as not being there. In this case the Na will dissociate from the F in solution leaving you with 2M F-. HF is a weak acid so it does not dissociate completely. H+ will equal A- only if it is a strong acid/ conjugate base buffer which this is not, this is a weak acid conjugate base so H+ does not necessarily equal A-

 

[Demps]

Use the equation

pH = pKa + log ([base] / [acid])

 

[jefff]

A certain buffer solution is 3 M in HF and 2 M in NaF. Calculate the pH of this buffer given that the Ka of HF = 7.0*10^-4.

The solution:
STEP 1: Ka = [H+][F-]/[HF] --> I don't understand this first step, because in the other problems I did, the [H+] is always equal to the [A-] so why is [H+] different from [F-]?

the way I see it is they are giving you HF (acid) and NaF (conj. base complexed with Na)

automatically you should try and think HF = H+ and NaF = A-

so keep this in mind: its a buffer and only when [acid] = [base] does pH = pKa

thats not the case here.

recall equation: pH = pKa + log ([A-]/[H+])

here we have:

Ka = 7e^-4.

pKa = -log Ka

(think since the Ka is ^-4 , the pKa will have a 3 in front of a decimal (4-1) ...then you have to look @ the 7, so after the decimal youll have a number less than 3 (you should memorize this trend -- if its 5e^-1 then log it will be .3 and since here its 7, which is greater than 5 its less than .3 so i chose .1)

pKa = 3.1

plug in

pH = 3.1 + log ([2]/[3])

pH = 3.1 + log (.66)

pH = 3.1 + log (6.6e^-1)

*same log 'trick'*

pH = 3.1 - .2

pH ~ 2.9

can someone please correct this if i made a mistake in this logic

edit: i was typing away...yes what demps said 🙂

 

[Pdentalstudent]

thanksss! i get it now!