What is the pH of a solution that contains 0.30 M HC2H3O2 and 0.95 M C2H3O2-? Use Ka = 1.8 × 10-5 for HC2H3O2.
HC2H3O2 = HAc = acetic acid
C2H3O2^- = Ac^- = acetate ion
...........HAc --> H^+ + Ac^-
I........0.30......0....0.95
C..........-x......x......x
E........0.30-x....x....0.95+x
Ka = (H^+)(Ac^-)/(HAc)
1.8E-5 = (x)(0.95+x)/(0.30-x)
Solve for x and convert to pH.
To find the pH of the solution, we need to use the equilibrium expression for the ionization of acetic acid (HC2H3O2) in water:
HC2H3O2 ⇌ H+ + C2H3O2-
The equilibrium constant for this reaction is called the acid dissociation constant (Ka), which is given as 1.8 × 10^(-5).
The expression for Ka is:
Ka = [H+][C2H3O2-] / [HC2H3O2]
Now, we are given the concentrations of HC2H3O2 and C2H3O2- in the solution. Let's assume that the concentration of H+ at equilibrium is x. Therefore, the concentration of C2H3O2- at equilibrium will also be x. The concentration of HC2H3O2 will be 0.30 M - x.
Now, we can substitute these concentrations into the Ka expression:
Ka = x * x / (0.30 - x)
Since the value of Ka is very small (1.8 × 10^(-5)), we can assume that the value of x will be much smaller than 0.30. Therefore, we can approximate 0.30 - x as 0.30.
Now, we can rearrange the equation to solve for x:
(1.8 × 10^(-5)) = (x^2) / 0.30
Multiply both sides by 0.30:
(0.30) * (1.8 × 10^(-5)) = x^2
x^2 = 5.4 × 10^(-6)
Take the square root of both sides:
x = √(5.4 × 10^(-6))
x ≈ 0.002323
This value represents the concentration of H+ ions in the solution at equilibrium. To find the pH, we need to take the negative logarithm (base 10) of the concentration of H+:
pH = -log[H+]
pH ≈ -log(0.002323)
pH ≈ 2.63
Therefore, the pH of the solution is approximately 2.63.