find the concentration of h3o in a .86mol/l solution of the weak acid H2O2 (kA=2.4x10-12)
.......H2O2 ==> H^+ + HO2^-
I.....0.86......0......0
C......-x.......x......x
E...0.86-x......x......x
Substitute the E line into the Ka expression and solve for x = (H3O^+).
To find the concentration of H3O+ in a solution of a weak acid like H2O2, we need to use the equilibrium expression for the acid dissociation reaction and the given acid dissociation constant (Ka).
First, let's write the balanced chemical equation for the dissociation of H2O2:
H2O2 ⇌ H+ + HO2-
The equilibrium expression for this reaction can be written as:
Ka = [H+][HO2-] / [H2O2]
Given that the Ka value is 2.4x10^-12 and assuming that the initial concentration of H2O2 is the same as its final concentration (0.86 mol/L), we can consider the change in concentration of H+ and HO2- as "x" (since the dissociation is very small for a weak acid):
Ka = x^2 / (0.86 - x)
Since the dissociation is very small, we can neglect "x" in comparison to the initial concentration of H2O2 (0.86 mol/L). Therefore, we can simplify the equation to:
Ka = x^2 / 0.86
Now, we can solve for "x" (the concentration of H3O+):
2.4x10^-12 = x^2 / 0.86
Rearranging this equation, we have:
x^2 = 2.4x10^-12 × 0.86
x^2 = 2.064x10^-12
Taking the square root of both sides:
x = √(2.064x10^-12)
x ≈ 1.435x10^-6
Therefore, the concentration of H3O+ in the 0.86 mol/L solution of the weak acid H2O2 is approximately 1.435x10^-6 mol/L.