Calculating pH of Buffer Solution with HCl Addition

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Discussion Overview

The discussion revolves around calculating the pH of a buffer solution formed by mixing sodium acetate and hydrochloric acid (HCl), specifically addressing the initial pH and the pH after additional HCl is added. The scope includes theoretical calculations and homework-related queries.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the initial pH using the Henderson-Hasselbalch equation, assuming equal concentrations of salt and acid, resulting in a pH of 4.75.
  • Another participant points out that the concentrations of the reactants are not equal due to differing volumes, suggesting that the moles of each component should be calculated to determine which is in excess.
  • A later reply discusses the formalities of the salt and weak acid, proposing that both will be equal and suggesting a quadratic equation approach to find the hydronium ion concentration and subsequently the pH.
  • One participant emphasizes the importance of starting with the reaction equation when adding a strong acid to a buffer solution.
  • Another participant admits to misunderstanding the problem regarding the amount of HCl added in part b), initially calculating with incorrect values and later correcting themselves.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions regarding the concentrations and moles of the reactants. There is no consensus on the correct approach or final pH values, and the discussion remains unresolved.

Contextual Notes

Participants mention the need for precise arithmetic and consideration of hydrolysis in their calculations. There are also references to the importance of correctly identifying the amounts of reactants involved in the buffer solution.

pari786
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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!
 
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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:
I am trying to say that my response was misleading because the resulting concentrations WILL be equal -- I'm still rechecking my computation.
 
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:
First of all - START WITH THE REACTION EQUATION.

What happens when you add strong acid (HCl) to the solution of weak acid salt?
 
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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