Chemistry help pliz
posted by Petty on .
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)

continued from above calculate the initial pH of the acetic acidsodium 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
1.0x__x___1.0+x
so Ka is
Ka=(x)(1.0+x)/(1.0x)
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 HendersonHasselbalch equation.