Calculate the pH of a solution that is 0.15 M CH3COOH and 0.75 M CH3COONa?

Question

Calculate the pH of a solution that is 0.15 M CH3COOH and 0.75 M CH3COONa (For CH3COOH, Ka= 1.8 x 10^-5)

Answer 1

O.K . First, establish the equilibrium


CH3COOH <---> CH3COO- + H+ (This is a weak acid, so there will be very little dissociation: the Ka value attests to that). Also, note that CH3COONa will dissociate fully (since it is a salt).

First, find the amount of CH3COO- and H+ dissociated

CH3COOH--> CH3COO- + H+
Initial 0.15 0.75 0
Final 0.15-x 0.75 + x x

Ka = [CH3COO-][H+]/[CH3CHOOH] (We approximate 0.75 + x to 0.75
since x is a very small number)
1.8 x 10^-5 = [0.75 ] * [x]/[0.15 -x]

2.7 x 10^-6 - 1.8 x 10^-5x = 0.75x (Here, also, approximate 1.8 x 10^-5x to 0)
2.7 x 10^-6 = 0.75 x

x= 3.6 x 10^-6

pH = -log(3.6 x 10^-6) = 5.44.


Or, to simplify life, you could use

pH = pka + log [CH3COO-]/[CH3COOH]


BUT, only if the % dissociation is less than 5%.

% dissociaion = [conjugate base]/ acid * 100% = [CH3C00-]/[CH3COOH] * 100%

Which is in this case, (NOT counting the availability of the 0.75 from the salt),
3.6 x 10^-6 /0.15 = 0.0024%, so the above equation is feasible.

Answer 2

pH is the -log of the concentration of [H+] ions. I'm assuming the Ka is the conc. of H+? Anyway, plug that into your calculator and see what ya get.

Answer 3

pKa = - log Ka = - log 1.8 x 10^-5 = 4.74 pH = pKa + log [CH3COO-]/ [CH3COOH] ( Handerson- Hasselbalch equation) pH = 4.74 + log 0.50/0.50 = 4.74 + log 1 = 4.74 + 0 = 4.74

Answer 4

What Is Ch3cooh