Question

what is the PH of a solution containing .03 M of acetic acid and .3 mole of sodium acetate (CH3COONa) ?
Ka for CH3COOH = 1.8*10^-5

Answer 1

CH3COONa ----> CH3COO- + Na+
CH3COONa dissociates completely producing 0.3 moles of CH3COO- ion in 1 litre of the solution.So initially the concentration of CH3COO- ion becomes 0.3M

CH3COOH ---> CH3COO- + H+
0.3M 0.3M
CH3COOH ---> CH3COO- + H+ .... Ka=1.8*10^-5
0.3M-x 0.3M+x x

\[([0.3M+x] [x])\div([0.3-x]])\]=Ka = 1.8*10^-5

x= concentration of H+ ions.Since Ka is very small, of the order of 10^-5,the acid dissociates very poorly and x is negligible compared to 0.3.So 0.3-x and 0.3 +x is approx. 0.3 .So those two terms cancel

therefore x=1.8*10^-5
pH= -log(1.8*10^-5)

Answer 2

for buffer systems you can also use the Henderson-Hasselbalch equation:

\(pH=pKa+log\dfrac{[A^-]}{[HA]}\)