A solution of HClO is mixed and found to
have a pH of 4.53. Find what the initial
concentration of HClO was for this solution.
Ka = 3.00 × 10−8
for HClO.
Answer in units of mol/L
please help this is due in 1 hour
pH = 4.53 = -log(H^+). Solve for (H^+) and that's approx 1E-5 but that's just close so you need to do it more accurately.
.........HClO ==> H^+ + ClO^-
I........x........0......0
C.......-y........y.......y
E........x-y......y........y
The problem tells you (actually the (H^+) from above tells you y = 10^-5
Ka = (H^+)(ClO^-)/(HClO)
(H^+) = 1E-5
(ClO^=) = 1E-5
(HClO) = x-1E-5
Solve for x.
To find the initial concentration of HClO, we can start by analyzing the given information.
The pH of a solution is a measure of its acidity and is equal to the negative logarithm base 10 of the concentration of H+ ions in the solution. Mathematically, it is represented as:
pH = -log[H+]
Since we have the pH, we can calculate the concentration of H+ ions using the formula:
[H+] = 10^(-pH)
Now, HClO is a weak acid, and we are given its Ka value, which is the acid dissociation constant. The Ka expression for HClO is:
Ka = [H+][ClO-] / [HClO]
Since HClO is a monoprotic acid, the concentration of H+ ions is equal to the concentration of HClO that has dissociated. Therefore, we can rewrite the equation as:
Ka = [H+][HClO] / [HClO]
= [H+]
Now that we have the concentration of H+ ions, which is equal to the initial concentration of HClO, we can substitute the given pH value into the equation and find the initial concentration of HClO.
[H+] = 10^(-pH)
= 10^(-4.53)
Using a calculator, we can evaluate this:
[H+] ≈ 2.97 x 10^(-5) mol/L
Thus, the initial concentration of HClO for the given solution is approximately 2.97 x 10^(-5) mol/L.