Explain (using equations) how a solution of 1.0 mol dm^-3 in both CH3COOH and CH3COONa is resistant to changes n pH when we add either small amounts of acid or small amount of base (such solution is called a buffer)

Answer 1

continued from above calculate the initial pH of the acetic acid-sodium acetate solution above.The pKa of acetic acid is 4.76

Answer 2

To do the second part from first principles. Start with the acid dissociation equation:

HAc -> H+ + Ac-

Ka is then

[H+][Ac-]/[HAc]

at the start

HAc -> H+ + Ac-

1.0 M ___0__1.0M

at the end

HAc -> H+ + Ac-
1.0-x__x___1.0+x

so Ka is
Ka=(x)(1.0+x)/(1.0-x)

we can treat this in two ways, we can solve for x

or

Assume that x is small so that

Ka=x(1.0)/(1.0)=x

thus pKa = pH =4.76

Answer 3

To give another perspective, you can calculate the pH of a buffer using the Henderson-Hasselbalch equation.