What is the pH of 64.1mL of a solution which is 0.81M in KlO and 0.33M in HlO? For HlO use ka=2.3x10^-11.
To determine the pH of the solution, we need to find the concentration of H+ ions present. Since HlO is an acid, it will dissociate and release H+ ions in water according to the following balanced equation:
HlO ⇌ H+ + lO-
The concentration of H+ ions can be calculated using the equilibrium constant expression (Ka) and the concentration of the acid (HlO).
The Ka expression is given as:
Ka = [H+][lO-] / [HlO]
First, let's calculate the concentration of H+ ions using the given Ka and concentration of HlO.
Ka = 2.3 × 10^-11
[HlO] = 0.33 M
Let x be the concentration of H+ ions in M.
[H+] = x
[lO-] = x
Using the given Ka expression:
2.3 × 10^-11 = (x)(x) / 0.33
Simplifying the equation:
2.3 × 10^-11 = x^2 / 0.33
Rearranging the equation:
x^2 = (2.3 × 10^-11)(0.33)
x^2 = 7.59 × 10^-12
x = √(7.59 × 10^-12)
x ≈ 2.76 × 10^-6
Therefore, the concentration of H+ ions in the solution is approximately 2.76 × 10^-6 M.
Now we can calculate the pH using the concentration of H+ ions.
pH = -log[H+]
pH = -log(2.76 × 10^-6)
pH ≈ 5.56
Therefore, the pH of the solution is approximately 5.56.
To determine the pH of the solution, we need to first identify the species that contribute to the acidity. In this case, we have both KlO and HlO.
The equation for the dissociation of HlO is: HlO ⇌ H+ + lO-
Given that the Ka (acid dissociation constant) for HlO is 2.3x10^-11, we can determine the concentration of H+ (hydrogen ion concentration) in the solution using the equation:
Ka = [H+][lO-] / [HlO]
Since the concentration of H+ is the same as the concentration of KlO, we'll label it as 'x'.
Therefore, the equation becomes:
2.3x10^-11 = (x)(0.33) / 0.81
Now, let's solve for 'x' with this equation:
x = (2.3x10^-11)(0.81) / 0.33
Calculating this expression, we find:
x ≈ 5.634x10^-11
Now that we have the concentration of H+, we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(5.634x10^-11)
Calculating this expression, we get:
pH ≈ 10.25
Therefore, the pH of the solution is approximately 10.25.