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

explanation:

i think option no d is the correct answer

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%