1)A solution has a [OH-] of 5.2 x 10-4. What is the [H3O+] in the solution?




2) A 0.25 M solution of a monoprotic acid, HA, has a pH of 2.54. What is Ka for this acid?
The dissociation of HA is: HA + H2O <--> H3O+ + A-
(Hint: write the expression for pH)



3 A solution has a pH of 11.95. What is the [OH-] in the solution?




4) Calculate the pH of a 0.0054 M KOH solution.

1) To find the [H3O+] in the solution, we can use the relationship between [H3O+] and [OH-]. In water, the product of [H3O+] and [OH-] is constant at 1.0 x 10^-14. Therefore, we can calculate [H3O+] using the given [OH-] value:

[H3O+] x [OH-] = 1.0 x 10^-14

[H3O+] = 1.0 x 10^-14 / [OH-]

Plugging in the given [OH-] value of 5.2 x 10^-4:

[H3O+] = 1.0 x 10^-14 / 5.2 x 10^-4

[H3O+] ≈ 1.9 x 10^-11

So, the [H3O+] in the solution is approximately 1.9 x 10^-11.

2) The pH of a solution is defined as the negative log of the [H3O+] concentration. In this case, the pH is given as 2.54, so we can use this information to calculate the [H3O+] concentration:

pH = -log[H3O+]

2.54 = -log([H3O+])

[H3O+] = 10^(-pH)

[H3O+] = 10^(-2.54)

Now that we have the [H3O+] concentration, we can use it to calculate the Ka for the acid.

The Ka expression is Ka = [H3O+][A-] / [HA]

Since the acid HA is monoprotic, the concentration of the conjugate base A- is equal to the concentration of the acid HA (0.25 M).

Ka = [H3O+][A-] / [HA]

Ka = [H3O+]^2 / [HA]

Plugging in the values:

Ka = (10^(-2.54))^2 / 0.25

Ka ≈ 5.7 x 10^(-5)

So, the Ka for the acid HA is approximately 5.7 x 10^(-5).

3) The pH of a solution is related to the [H3O+] concentration by the equation:

pH = -log[H3O+]

Given that the pH of the solution is 11.95, we can calculate the [H3O+] concentration using the inverse of the logarithmic function:

[H3O+] = 10^(-pH)

[H3O+] = 10^(-11.95)

Now, to find the [OH-] concentration, we can use the relationship between [H3O+] and [OH-]:

[H3O+] x [OH-] = 1.0 x 10^-14

Since we have the value of [H3O+], we can rearrange the equation to find [OH-]:

[OH-] = 1.0 x 10^-14 / [H3O+]

Plugging in the [H3O+] value we calculated:

[OH-] = 1.0 x 10^-14 / (10^(-11.95))

[OH-] ≈ 5.01 x 10^(-4)

So, the [OH-] concentration in the solution is approximately 5.01 x 10^(-4).

4) To calculate the pH of a solution, we need to determine the [H3O+] concentration. In this case, we are given the concentration of KOH, which is a strong base and completely dissociates in water to produce hydroxide ions (OH-).

KOH → K+ + OH-

The concentration of hydroxide ions ([OH-]) is the same as the concentration of the KOH solution, which is 0.0054 M.

To find the [H3O+] concentration, we can use the relationship:

[H3O+] x [OH-] = 1.0 x 10^-14

[H3O+] = 1.0 x 10^-14 / [OH-]

Plugging in the [OH-] concentration of 0.0054 M:

[H3O+] = 1.0 x 10^-14 / 0.0054

[H3O+] ≈ 1.85 x 10^-12

Finally, to find the pH, we can use the equation:

pH = -log[H3O+]

pH = -log(1.85 x 10^-12)

pH ≈ 11.73

Therefore, the pH of the 0.0054 M KOH solution is approximately 11.73.