The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5 COOH and C 2 H 5 COO − in a 0.441 M propanoic acid solution at equilibrium.

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. (**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.