Acid, Base, and Buffer Titration Problem

  • Thread starter MitsuShai
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  • #1
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I missed this question on an exam and I was wondering if anyone can explain to me how to do it.


Homework Statement



The pKa of acetic acid, HC2H3O2, is 4.76. A buffer solution was made using an unspecified amount of acetic acid and 0.300 moles of NaC2H3O2 in enough water to make 2 Liters of solution. Its pH was measured as 4.40 on a meter. How many moles of HC2H3O2 were used?


Homework Equations


K= [products]/[reactions]
pH= pKa + log (A/HA)


The Attempt at a Solution



Ka= 10^(-4.76) = 1.74 x 10 ^(-5)
Kb = 5.75 x 10^(-10)
pH =4.4 [H]=3.98 x 10^(-5)
pOH= 9.6 [OH]= 2.51 x 10^(-10)

moles of OH= 2.51 x 10^(-10) x 2= 5.02 x 10^(-10) moles

Base_________+____H2O -->___Acetic Acid_+_____OH +______Na
.3 moles______________________x_________________0
-5.024 x 10^(-10)__________5.02 x 10^(-10)_______5.02 x 10^(-10)
.2999 moles_______________x+ 5.02 x 10^(-10)______5.02 x 10^(-10)


pH=pKa + log(A/HA)
4.4=4.76+ log (.3/(x+5.02 x 10^(-10)))
=10^-(-.36)= 10^[log (.3/(x+5.02 x 10^(-10)))]
2.29= .3/(x+5.02 x 10^(-10)))

(2.29)(x+5.02 x 10^(-10))=.3
x=.131 moles

.3 + .131= 0.43 moles
 

Answers and Replies

  • #2
Borek
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Most of what you did is irrelevant to the question, and to be honest - I don't plan to try to understand what you did and why. For some reason you decided to take into account base hydrolysis, but you ignored acetic acid dissociation (these both processes take place in the solution). But - and this is the most important hint at solving the problem - when doing buffer questions we usually ignore both dissociation and hydrolysis, assuming they are neglectable.

Start with HH equation, solve it for the concentration of acetic acid - and as you know volume of the solution, you are done.

--
 

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