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.