The Ka of hypochlorous acid (HClO) is 3.00×10-8 at 25.0°C. Calculate the pH of a 0.0385 M hypochlorous acid solution.
a.1.41
b.7.52
c.-1.41
D.4.47
E. 8.94
I got B 7.52 but i feel like im wrong
Why are you posting this again. I thought we took care of this below. Your 4.47 was/is correct.
To calculate the pH of a solution of hypochlorous acid (HClO), you need to consider the dissociation of the acid in water. The Ka value represents the equilibrium constant for the dissociation reaction:
HClO ⇌ H+ + ClO-
The equilibrium expression is:
Ka = [H+][ClO-] / [HClO]
Given that the Ka of HClO is 3.00×10^-8, you can set up the equation using the concentrations of the species, where [H+] is the concentration of hydronium ions and [HClO] is the initial concentration of hypochlorous acid.
Let x represent the concentration of H+ ions formed during the dissociation.
Then, the concentration of ClO- ions formed will also be x, given that the ratio between them is 1:1.
Thus, the equilibrium expression can be written as:
3.00×10^-8 = x * x / (0.0385 - x)
Since the value of x will be small compared to 0.0385, we can approximate 0.0385 - x ≈ 0.0385.
We can now solve for x using the quadratic equation:
3.00×10^-8 = x^2 / 0.0385 ≈ 7.79x10^-7
x^2 ≈ 7.79x10^-7 * 0.0385 ≈ 3.00x10^-8
x ≈ √(3.00x10^-8) ≈ 5.48x10^-5
Now, we can calculate the pH using the equation:
pH = -log[H+]
pH = -log(5.48x10^-5)
pH ≈ 4.26
Therefore, the correct answer is D. 4.47.