What is the pH of a 0.200M solution for HBrO?

Ka = 2.8 x 10^-9

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Well, if the pH of a solution is like a balance between acids and bases, then let's take a hilarious journey to find the answer!

HBrO is a weak acid, and its acidity is determined by the equilibrium reaction HBrO ⇌ H+ + BrO-. The Ka value given is 2.8 x 10^-9.

Now, we need to determine the concentration of H+ ions in the solution. Since the initial concentration of HBrO is 0.200M and we don't know how much of it reacts, let's call that x. So, the equilibrium concentration of H+ ions would be x M, and that of BrO- would also be x M.

Using the Ka expression for HBrO, we have:

Ka = [H+][BrO-]/[HBrO]

Substituting the equilibrium concentrations:

2.8 x 10^-9 = x * x / (0.200 - x)

Now, because x is small compared to 0.200, we can approximate (0.200 - x) as 0.200.

2.8 x 10^-9 = x^2 / 0.200

Solving for x:

x^2 = 0.200 * (2.8 x 10^-9)

x ≈ 2.35 x 10^-5 M

Since we now know the concentration of H+, to find the pH, we can use the formula pH = -log[H+].

pH = -log(2.35 x 10^-5)

pH ≈ 4.63

So, the pH of a 0.200 M solution of HBrO is approximately 4.63. But hey, don't be salty about it! Just sit back, relax, and enjoy the reaction!

To determine the pH of a solution of HBrO, you need to use the given information about the dissociation constant (Ka). The Ka is a measure of the strength of the acid.

First, you need to write the balanced equation for the dissociation of HBrO:

HBrO ⇌ H+ + BrO-

Next, you can set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of the different species in the solution:

HBrO ⇌ H+ + BrO-
I | 0.200M | 0 | 0
C | -x | +x | +x
E | 0.200M-x | x | x

Since the initial concentration of HBrO is given as 0.200M, the initial concentration is 0.200M. The concentration of H+ and BrO- at equilibrium is represented by x.

The dissociation constant (Ka) expression for this reaction is:

Ka = [H+][BrO-]/[HBrO]

Given that the Ka value is 2.8 x 10^-9, you can substitute the concentrations into the expression:

2.8 x 10^-9 = x * x / (0.200 - x)

The concentration of HBrO (0.200M - x) is only slightly smaller than 0.200M since x is usually small compared to 0.200. Therefore, we can assume that the value of x is small enough to be negligible. The equation can be simplified as:

2.8 x 10^-9 ≈ x * x/ 0.200

Simplifying further:

2.8 x 10^-9 ≈ x^2/0.200

Rearrange the equation to solve for x:

x^2 = 2.8 x 10^-9 * 0.200

x^2 = 5.6 x 10^-10

Taking the square root of both sides:

x ≈ 7.5 x 10^-6

Since x represents the concentration of H+ and BrO- at equilibrium, which is equal due to the 1:1 stoichiometry, the concentration of H+ is approximately 7.5 x 10^-6M.

To calculate the pH, you can use the equation pH = -log[H+]:

pH = -log(7.5 x 10^-6)
pH ≈ -(-5.12)
pH ≈ 5.12

Therefore, the pH of the 0.200M solution of HBrO is approximately 5.12.

HBrO ==> H^+ + BrO^-

Ka = (H^+)(BrO^-)/(HBrO)
Set up and ICE chart, substitute into Ka expression and solve. Post your work if you get stuck.