What is the pH of a 0.350M HBrO Solution? HBrO, Ka=2.3 x 10^-9
........HBrO ==> H^+ + BrO^-
I.......0.350M....0......0
C..........-x.....x......x
E......0.350-x.....x.....x
Substitute the E line into the Ka expression and solve for H^+. Then pH = log(HT^+)
To find the pH of a solution, we need to know the concentration of H+ ions in the solution.
HBrO is a weak acid, and we can use its Ka value to calculate the concentration of H+ ions.
The Ka expression for HBrO is:
Ka = [H+][BrO-] / [HBrO]
Since we are given the value of Ka (2.3 x 10^-9), we can set up the equation:
2.3 x 10^-9 = [H+][BrO-] / (0.350M)
To simplify the equation, we can assume that the concentration of BrO- is equal to the concentration of H+. This is because HBrO dissociates into one H+ ion and one BrO- ion.
Therefore, we can write:
2.3 x 10^-9 = [H+]^2 / (0.350M)
Rearranging the equation to solve for [H+], we get:
[H+]^2 = (2.3 x 10^-9) * (0.350M)
[H+]^2 = 8.05 x 10^-10
Taking the square root of both sides, we find:
[H+] ≈ 2.84 x 10^-5 M
Since the pH is defined as the negative logarithm (base 10) of the concentration of H+ ions, we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(2.84 x 10^-5)
Using a calculator, we can find that the pH of the 0.350M HBrO solution is approximately 4.55.
To determine the pH of the HBrO solution, we need to first calculate the concentration of H+ ions in the solution using the dissociation constant (Ka) value provided.
The dissociation of HBrO can be represented by the equation:
HBrO ⇌ H+ + BrO-
The Ka expression for this reaction can be written as:
Ka = [H+][BrO-] / [HBrO]
Given that the concentration of HBrO is 0.350 M, let's assume the change in concentration of H+ and BrO- is "x" (as the solution dissociates partially). So, at equilibrium:
[H+] ≈ x M
[BrO-] ≈ x M
[HBrO] ≈ (0.350 - x) M
Now, substitute these values into the Ka expression:
Ka = (x)(x) / (0.350 - x)
Since the value of Ka is known (2.3 x 10^-9), we can use this equation to solve for "x".
2.3 x 10^-9 = (x)(x) / (0.350 - x)
To simplify the equation, we can assume that x is much smaller than 0.350. Therefore, we can ignore "x" in the denominator.
2.3 x 10^-9 ≈ (x)(x) / 0.350
Rearranging the equation gives:
(0.350)(2.3 x 10^-9) ≈ x^2
0.805 x 10^-9 ≈ x^2
Now, take the square root of both sides:
√(0.805 x 10^-9) ≈ √(x^2)
x ≈ √(0.805 x 10^-9)
x ≈ 8.97 x 10^-5 M
Since we assumed x to be much smaller than 0.350 M, this approximation is reasonable.
Now, the concentration of H+ ions is approximately 8.97 x 10^-5 M. To calculate the pH, we need to take the negative logarithm (base 10) of the H+ concentration:
pH = -log[H+]
pH ≈ -log(8.97 x 10^-5)
Using a calculator, the pH can be estimated to be approximately 4.048.
Therefore, the pH of the 0.350 M HBrO solution is approximately 4.048.