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PH of buffer solution

  • Thread starter pari786
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  • #1
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Homework Statement



A buffer solution was formed by mixing 30ml of 0.1M sodium acetate and 15ml of 0.1M HCl
a) Calculate the pH of this buffer solution.
b) Calculate the pH of this buffer solution after the addition of 15ml of 0.1M HCl.


Homework Equations



pH = pKa + log[salt]/[acid]

3. The Attempt at a Solution



for part a) I substituted the value in the above equation as pKa= 4.75 for acetic acud. [salt]=0.1 and [acid]=0.1

and got pH as 4.75


Is it right ? and I'm stuck with part B) can anyone please help me .... please soon!
 

Answers and Replies

  • #2
symbolipoint
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EDIT: this here is misleading because the resulting concentrations ARE equal: BIG HINT for part 'a' -------- the starting 'molarities' of each solution were the same, but the volumes used from each were NOT the same. Now which reactant is in excess?

30 ml of 0.1M sodium acetate, how many moles?
15 ml of 0.1M HCl, how many moles?
......In what final volume?
 
Last edited:
  • #3
symbolipoint
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I am trying to say that my response was misleading because the resulting concentrations WILL be equal -- I'm still rechecking my computation.
 
  • #4
symbolipoint
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The formality of salt = 0.1/3
Formality of the weak acid = 0.1/3

Therefore, pH=pKa

You could try more precise arithmetic if you account for hydrolysis of the
salt and of the weak acid. The formality of the formed acid will be 0.0333...
and will also be the formality of the remaining salt. In this case, both
formalities F being equal, you can set up the equilibrium constant expression

K=(H)*(F+H)/(F-H)

You obtain the quadratic equation:
H^2 +(F+K)*H - KF =0

From that quadratic equation, find solution for H,
the hydronium ion concentration molarity; then find pH.

[ If the formalities of the salt, Fs, and the weak acid, Fa, were not
the same, then your equilibrium expression equation would be
K=(H)*(Fs + H)/(Fa - H) ]
 
Last edited:
  • #5
Borek
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First of all - START WITH THE REACTION EQUATION.

What happens when you add strong acid (HCl) to the solution of weak acid salt?
 
  • #6
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Sorry , I misunderstood the part b) Calculate the pH of this buffer solution after the addition of 15ml of 0.1M HCl.

I didn't realize that it was 5 ml of 0.1M HCl (instead).

I realized after doing the calculations for three hours. Sorry about any kind of inconvience.
 

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