Question
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)
Answers
continued from above calculate the initial pH of the acetic acid-sodium acetate solution above.The pKa of acetic acid is 4.76
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
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
To give another perspective, you can calculate the pH of a buffer using the Henderson-Hasselbalch equation.
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