The weak acid HQ has pKa of 4.89. Calculate the (OH-) of 0.500 M HQ
This is what I have:
pKa = 4.89 so Ka = 10^-4.89 = 1.29x10^-5
Ka = 1.29x10^-5 = [H+][Q-] / [HQ]
1.29x10^-5 = x^2 / 0.046 - x
x^2 + 1.29x10^-5x - 5.93x10^-7 = 0
x = [H+] = 7.64x10^-4M
pH = -log[H+] = 3.12
1.29x10^-5 = x^2 / 0.5 - x
x^2 + 1.29x10^-5x - 6.71x10^-6 = 0
x = [H+] = 3.146x10^-3M
[OH-] = 1x10^-14 / 3.146x10^-4 = 3.179x10^-12M = [OH-] answer......????
pKa = 4.89 so Ka = 10^-4.89 = 1.29x10^-5
Ka = 1.29x10^-5 = [H+][Q-] / [HQ]
1.29x10^-5 = x^2 / 0.046 - x
Shouldn't this be
1.29E-5 = x^2/0.5 and you don't need the quadratic. I get 2.54E-3 for x. pH = 2.59 but check me out on that.
x^2 + 1.29x10^-5x - 5.93x10^-7 = 0
x = [H+] = 7.64x10^-4M
pH = -log[H+] = 3.12
Your calculation is partially correct, but there are some errors in your calculation of the concentration of hydroxide ions ([OH-]).
The first step is to determine the concentration of the dissociated form of the weak acid, which is the conjugate base Q-. Since the concentration of the weak acid HQ is given as 0.500 M, and the acid is weak, we can assume that most of it (approximately 100%) remains undissociated. Therefore, the concentration of the undissociated weak acid [HQ] is approximately 0.500 M.
Using the equation for the dissociation of the weak acid, we have:
Ka = [H+][Q-] / [HQ]
Substituting the values, we get:
1.29x10^-5 = [H+] * [Q-] / 0.500
Since the weak acid is a monoprotic acid, the concentration of Q- will be equal to the concentration of [H+], so we can replace [Q-] with [H+] in the equation:
1.29x10^-5 = [H+]^2 / 0.500
Simplifying:
[H+]^2 = 1.29x10^-5 * 0.500
[H+]^2 = 6.45x10^-6
Taking the square root of both sides:
[H+] = sqrt(6.45x10^-6)
[H+] = 2.54x10^-3 M
Now, to calculate the concentration of [OH-], we can use the equation:
Kw = [H+][OH-]
Substituting the values:
1.0x10^-14 = (2.54x10^-3)([OH-])
Simplifying:
[OH-] = 1.0x10^-14 / 2.54x10^-3
[OH-] = 3.94x10^-12 M
Therefore, the concentration of [OH-] in a 0.500 M solution of the weak acid HQ is approximately 3.94x10^-12 M.
To calculate the concentration of hydroxide ions ([OH-]) of a 0.500 M HQ solution, you need to use the equilibrium expression for the ionization reaction of the weak acid.
First, let's write the ionization reaction of HQ as follows:
HQ + H2O ⇌ H3O+ + Q-
The equilibrium constant expression for this reaction is:
Ka = [H3O+][Q-] / [HQ]
Given that the pKa of HQ is 4.89, we can calculate Ka as follows:
Ka = 10^(-pKa) = 10^(-4.89) = 1.29 x 10^(-5)
Now, let's assume that the concentration of [H3O+] is x. Since the initial concentration of HQ is 0.500 M, the concentration of [Q-] is (0.500 - x) M.
Using the equilibrium constant expression, we can substitute the values to obtain:
1.29 x 10^(-5) = x * (0.500 - x)
Rearrange the equation to get:
x^2 + 1.29 x 10^(-5)x - 6.45 x 10^(-6) = 0
You can solve this quadratic equation using the quadratic formula or by factoring. Solving the equation, you will find that the concentration of [H3O+] is approximately 3.146 x 10^(-3) M.
To calculate the concentration of [OH-], you can use the equation:
[H3O+][OH-] = 1 x 10^(-14) (from the autoionization of water)
Rearrange the equation to solve for [OH-]:
[OH-] = 1 x 10^(-14) / [H3O+]
Plug in the value of [H3O+] and calculate [OH-]:
[OH-] = 1 x 10^(-14) / 3.146 x 10^(-3) = 3.179 x 10^(-12) M
Therefore, the concentration of [OH-] in a 0.500 M HQ solution is approximately 3.179 x 10^(-12) M.