The pH of a .10M HOBr solution is 4.8

1. What is [H+] in that solution?
2. What is [OBr-]? What is [HOBr]? (where do the H+ and OBr ions come from?)
3. what is the value of Ka for HOBr? what is the value for pKa?

1. 1.58x10^-5 M

1. To determine the [H+] concentration in the solution, we can use the formula for pH:

pH = -log[H+]

Therefore:
4.8 = -log[H+]

Rearranging the equation, we get:
[H+] = 10^(-pH)

Substituting the given pH value into the equation, we have:
[H+] = 10^(-4.8)
[H+] ≈ 1.58 x 10^(-5) M

So, the concentration of [H+] in the solution is approximately 1.58 x 10^(-5) M.

2. In the HOBr solution, H+ and OBr- ions are formed through the dissociation of the HOBr molecule. HOBr can act as an acid, donating a proton (H+ ion) and leaving behind the OBr- ion. Hence:

HOBr ⇌ H+ + OBr-

The [OBr-] concentration will be equal to the [H+] concentration, as one OBr- ion is produced for every H+ ion generated. Therefore, [OBr-] ≈ 1.58 x 10^(-5) M.

The [HOBr] concentration can be calculated using the initial concentration and the [H+] concentration. Since HOBr dissociates to form an H+ ion, the [HOBr] concentration will be equal to the initial concentration of HOBr minus the [H+] concentration:

[HOBr] = Initial concentration of HOBr - [H+]
[HOBr] = 0.10 M - 1.58 x 10^(-5) M
[HOBr] ≈ 0.0999842 M (approximately)

3. To find the Ka (acid dissociation constant) for HOBr, we can use the concentrations of the dissociated products.

Ka = ([H+][OBr-]) / [HOBr]

Substituting the given concentrations into the equation:
Ka = (1.58 x 10^(-5) M) * (1.58 x 10^(-5) M) / (0.0999842 M)

Calculating this expression, we find:
Ka ≈ 2.50 x 10^(-9) M

The pKa is simply the negative logarithm of Ka:
pKa = -log(Ka)

Substituting the calculated value of Ka into the equation, we get:
pKa ≈ -log(2.50 x 10^(-9))
pKa ≈ 8.60

To answer these questions, we need to understand the dissociation of HOBr in water. HOBr is a weak acid that partially dissociates in water, producing H+ and OBr- ions.

1. The pH of a solution is a measure of the concentration of H+ ions. Given that the pH of the solution is 4.8, we know that the concentration of H+ ions can be calculated using the formula:

[H+] = 10^(-pH)

[H+] = 10^(-4.8) = 1.58 x 10^(-5) M

2. In the dissociation of HOBr, H+ ions and OBr- ions are formed. Since the concentration of H+ ions is 1.58 x 10^(-5) M, it means the concentration of OBr- ions is also 1.58 x 10^(-5) M since they are produced in a 1:1 ratio during the dissociation.

So, [OBr-] = 1.58 x 10^(-5) M
And [HOBr] remains at 0.10 M, as it does not dissociate entirely.

3. The Ka value represents the acid dissociation constant of HOBr. As you mentioned that the pH of the solution is 4.8, we can use this information to find the Ka and pKa values.

Ka is calculated using the formula:

Ka = [H+] [OBr-] / [HOBr]

Plugging in the values we found:
Ka = (1.58 x 10^(-5))^2 / 0.10
Ka = 2.50 x 10^(-10)

The pKa value is obtained by taking the negative logarithm (base 10) of the Ka value:
pKa = -log10(2.50 x 10^(-10))
pKa = 9.60

So, the value of Ka for HOBr is 2.50 x 10^(-10) and the pKa value is 9.60.