Calculate pH of Buffer CH_3COOH & CH_3COONa

AI Thread Summary
To calculate the pH of the buffer solution made from 55 mL of CH3COOH and 45 mL of CH3COONa, the Henderson-Hasselbalch equation is essential. The acetic acid concentration is 0.10 M for both components, and the dilution of 10 mL of this buffer to 100 mL with deionized water must be considered. The pKa can be calculated using the formula pKa = -log(1.76 x 10^-5). After determining the concentrations post-dilution, the pH can be easily calculated. Understanding these steps will simplify the process of finding the expected pH of the buffer solution.
n77ler
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Homework Statement


Buffer solution : CH_3COOH (55 mL) + CH_3COONa (45 mL)
Ka (CH_3COOH) = 1.76 x 10^-5

Calculate the expected pH of your buffer solution after it has been dilted by making 10 mL of buffer up to 100 mL with deionized water.

Homework Equations


Ice charts
n= c x v
c= n/v

The Attempt at a Solution


I've calculated the pH for other additions of strong acids and strong bases but I'm not sure how to start this one. Little bit confused... Just the first step would be a great help :)
 
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You haven't given enough information to start. You need to know something about the concentrations of acetic acid (55 mL?) and acetate (45 mL?).
 
Sorry, they are both 0.10 M
 
Review the Henderson-Hasselbach equation. Hint: pKa = -log[Ka]
 
n77ler said:

Homework Statement


Buffer solution : CH_3COOH (55 mL) + CH_3COONa (45 mL)
Ka (CH_3COOH) = 1.76 x 10^-5

Calculate the expected pH of your buffer solution after it has been dilted by making 10 mL of buffer up to 100 mL with deionized water.

Homework Equations


Ice charts
n= c x v
c= n/v

The Attempt at a Solution


I've calculated the pH for other additions of strong acids and strong bases but I'm not sure how to start this one. Little bit confused... Just the first step would be a great help :)
the question is a little confusing due to the given volumes.i need help how to attempt the question.
 
As chemisttree already wrote: use Henderson-Hasselbalch equation. Start calculating concentrations of both acid and conjugate base. At this stage it will be simple dilution problem.

(in fact you can solve the problem without calculating these concentrations, but I guess explaining the shorter route will only confuse you more)
 
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