What is [H3O+ ] (in M) in a solution of 0.078 M HBrO2 and 0.036 M NaBrO2? (Assume Kw = 1.01 ✕ 10^−14.)
HBrO_2(aq) + H_2O(l) <--> H_3O^+(aq) + BrO_2^−(aq); Ka = 3.7 ✕ 10−4
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid). You have Ka, convert to pKa.
You have the acid (HBrO2) and the base(BrO2^- from NaBrO2).
Post your work if you get stuck.
To determine the concentration of [H3O+] in the given solution, we need to solve the equilibrium expression using the given values of concentrations and the dissociation constant (Ka).
The equilibrium expression for the reaction is:
[H3O+][BrO2-] / [HBrO2]
Since the concentrations of HBrO2 and NaBrO2 are given, we can assume the concentration of NaBrO2 to be negligible due to the small value provided (0.036 M).
Now, let x be the change in the concentration of H3O+, which is also equal to the concentration of BrO2- formed. The concentration of HBrO2 will be equal to the initial concentration minus x.
[H3O+] = [BrO2-] = x
[HBrO2] = 0.078 M - x
The equilibrium expression becomes:
[x][x] / (0.078 - x)
The given Ka value for the reaction is 3.7 x 10^-4. Therefore, we can set up the following equation:
Ka = [H3O+][BrO2-] / [HBrO2]
3.7 x 10^-4 = x * x / (0.078 - x)
To solve this equation, we can make the approximation that x is much smaller than 0.078 M (since it is a small value compared to the initial concentration). Thus, we can ignore "x" in the denominator, and the equation becomes:
3.7 x 10^-4 = x^2 / 0.078
Rearranging the equation:
0.078 * 3.7 x 10^-4 = x^2
x^2 = 2.886 x 10^-6
Taking the square root of both sides:
x ≈ 5.37 x 10^-3 M
Therefore, the concentration of [H3O+] in the solution is approximately 5.37 x 10^-3 M.