The ka value for acetic acid, ch3cooh(aq), is 1.8× 10–5. calculate the ph of a 1.60 m acetic acid solution.

To determine the pH of the solution given, we make use of the acid equilibrium constant (Ka) given. It is the ratio of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the CH3COOH acid would be as follows:

CH3COOH = CH3COO- + H+

The acid equilibrum constant would be expressed as follows:

Ka = [H+][CH3COO-] / [CH3COOH] = 1.8× 10^–5

To determine the equilibrium concentrations we use the ICE table,
         CH3COOH             H+             CH3COO-
I            1.60                    0                     0
C             -x                    +x                   +x

E         1.60-x                   x                     x 

1.8× 10^–5 =  [H+][CH3COO-] / [CH3COOH] 
1.8 x 10^-5 = [x][x] / [0.160-x] 

Solving for x,

x = 1.69x10^-3 = [H+] = [F-]

pH = -log [H+] = -log [1.69x10^-3] = 2.8

i think its 5049.

i hope this !