calculate the PH of a mixture containing 30 mL, 0.10M HC2H3O2? (Ka=1.8x10^-8 of HC2H3O2)
If this is a mixture you didn't post all of the problem.
yeah i forgot to put
+ 30 mL, 0.10 M of NaOH
You also listed Ka wrong. That's 1.8E-5.
acetic acid is HAc = HC2H3O2
millimols HAc = mL x M = 30 x 0.1 = 3
millimols NaOH = 3
........HAc + NaOH ==> NaAc + H2O
I.......3.......0........0.....0
add.............3.............
C......-3......-3........3......3
E.......0.......0........3......3
So you have at equilibrium a solution of sodium acetate, NaAc, and the molarity is 3 millimols/60 mL = 0.05M
The acetate ion hydrolyzes in water as follows:
..........Ac^- + HOH ==> HAc + OH^-
I.......0.05..............0.....0
C.........-x..............x.....x
E......0.05-x.............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.05-x)
Solve for x = (OH-) and convert to pH.
To calculate the pH of a mixture containing 30 mL of 0.10 M HC2H3O2, you need to consider the dissociation of HC2H3O2 and the acidic properties of the solution. The dissociation of HC2H3O2 can be represented as follows:
HC2H3O2 ⇌ H+ + C2H3O2-
The equilibrium constant for this dissociation is given by the acid dissociation constant (Ka) of HC2H3O2, which is 1.8x10^-8.
To calculate the concentration of H+ ions in the solution, you need to determine the degree of ionization of HC2H3O2. This can be achieved by using the equilibrium expression for the reaction:
Ka = [H+][C2H3O2-] / [HC2H3O2]
Since the initial concentration of HC2H3O2 is 0.10 M and the concentration of H+ ions is initially zero, we can assume that the concentration of H+ ions at equilibrium is x M (where x is the degree of ionization).
Therefore, the equilibrium expression becomes:
Ka = x * x / (0.10 - x)
Since x is typically much smaller than the initial concentration of HC2H3O2 (0.10 M), we can assume that 0.10 - x ≈ 0.10.
Simplifying the expression:
1.8x10^-8 = x^2 / 0.10
Rearranging and solving for x:
x^2 = 0.10 * 1.8x10^-8
x^2 = 1.8x10^-9
Taking the square root of both sides:
x ≈ 1.34x10^-5 M
The concentration of H+ ions is approximately 1.34x10^-5 M.
Now that you have the concentration of H+ ions, you can calculate the pH using the formula:
pH = -log[H+]
Plugging in the value:
pH = -log(1.34x10^-5)
pH ≈ 4.87
Therefore, the pH of the mixture containing 30 mL of 0.10 M HC2H3O2 is approximately 4.87.