Find the concentration of (H3O+) and (OH-) for a solution of weak acid (HA) where, (KA) = 2×10-⁸ and 8 moles of acid are present in 1 litre of solution ❓❓❓
To find the concentration of (H3O+) and (OH-) ions for a solution of a weak acid (HA), you can use the equilibrium expression for the ionization of the acid, which is:
HA ⇌ H+ + A-
First, calculate the initial concentration of HA in the solution. You have 8 moles of HA in 1 liter, so the initial concentration [HA]0 = 8 moles/L.
Now, use the equilibrium constant (Ka) for the ionization of HA, which is given as Ka = [H+][A-]/[HA]. You are given that Ka = 2×10^-8.
Assuming that x represents the concentration of H+ ions produced when the acid ionizes, the equilibrium concentrations are:
[H+] = x
[A-] = x
[HA] = [HA]0 - x = 8 - x
Now, you can set up the equilibrium expression and solve for x:
Ka = [H+][A-]/[HA]
2×10^-8 = x * x / (8 - x)
Now, you can solve this equation for x. We'll use the quadratic formula:
x = [-b ± √(b² - 4ac)] / 2a
Where:
a = 8
b = 0
c = -2×10^-8
x = [0 ± √(0 - 4 * 8 * (-2×10^-8))] / (2 * 8)
x = [± √(6.4×10^-7)] / 16
x ≈ ± 0.0004 M
Since the concentration of H+ ions can't be negative in this context, you take the positive value:
[H+] ≈ 0.0004 M
Now, to find the concentration of (OH-) ions, you can use the fact that in water, [H+][OH-] = 1.0 x 10^-14 (at 25°C).
[OH-] = 1.0 x 10^-14 / [H+]
[OH-] ≈ 1.0 x 10^-14 / 0.0004
[OH-] ≈ 2.5 x 10^-11 M
So, the concentration of (H3O+) ions is approximately 0.0004 M, and the concentration of (OH-) ions is approximately 2.5 x 10^-11 M in the given weak acid solution.
To find the concentration of (H3O+) and (OH-) ions for a solution of a weak acid (HA) with a given Ka (acid dissociation constant) value and 8 moles of acid in 1 liter of solution, you can use the equilibrium expression for the dissociation of the weak acid:
HA ⇌ H+ + A-
Where:
HA represents the weak acid.
H+ represents the hydronium ion (H3O+).
A- represents the conjugate base of the acid.
Given that Ka = 2 × 10^-8, this means that at equilibrium, you can assume that [H+][A-] / [HA] = Ka.
Now, you know that 8 moles of HA are present in 1 liter of solution. So, the initial concentration of HA ([HA]initial) is 8 moles/L.
At equilibrium, some of the HA will dissociate into H+ and A-. Let's assume that x moles of HA dissociate into H+ and A-. So, [HA] at equilibrium will be (8 - x) moles/L, [H+] will be x moles/L, and [A-] will also be x moles/L.
Now, you can use the Ka expression:
Ka = [H+][A-] / [HA]
Plug in the values:
2 × 10^-8 = (x * x) / (8 - x)
Solve for x:
x^2 = 2 × 10^-8 * (8 - x)
x^2 = 1.6 × 10^-7 - 2x × 10^-8
x^2 + 2x × 10^-8 - 1.6 × 10^-7 = 0
You can solve this quadratic equation for x. Once you have the value of x, you can find the concentration of [H3O+] and [OH-].
Keep in mind that in a weak acid solution, [H3O+] and [OH-] will not be equal to each other, but their product will still be related to the value of Ka, following the equation [H3O+][OH-] = Kw (the ion product of water), which is approximately 1.0 x 10^-14 at 25°C.
Answers & Comments
Verified answer
Answer:
To find the concentration of (H3O+) and (OH-) ions for a solution of a weak acid (HA), you can use the equilibrium expression for the ionization of the acid, which is:
HA ⇌ H+ + A-
First, calculate the initial concentration of HA in the solution. You have 8 moles of HA in 1 liter, so the initial concentration [HA]0 = 8 moles/L.
Now, use the equilibrium constant (Ka) for the ionization of HA, which is given as Ka = [H+][A-]/[HA]. You are given that Ka = 2×10^-8.
Assuming that x represents the concentration of H+ ions produced when the acid ionizes, the equilibrium concentrations are:
[H+] = x
[A-] = x
[HA] = [HA]0 - x = 8 - x
Now, you can set up the equilibrium expression and solve for x:
Ka = [H+][A-]/[HA]
2×10^-8 = x * x / (8 - x)
Now, you can solve this equation for x. We'll use the quadratic formula:
x = [-b ± √(b² - 4ac)] / 2a
Where:
a = 8
b = 0
c = -2×10^-8
x = [0 ± √(0 - 4 * 8 * (-2×10^-8))] / (2 * 8)
x = [± √(6.4×10^-7)] / 16
x ≈ ± 0.0004 M
Since the concentration of H+ ions can't be negative in this context, you take the positive value:
[H+] ≈ 0.0004 M
Now, to find the concentration of (OH-) ions, you can use the fact that in water, [H+][OH-] = 1.0 x 10^-14 (at 25°C).
[OH-] = 1.0 x 10^-14 / [H+]
[OH-] ≈ 1.0 x 10^-14 / 0.0004
[OH-] ≈ 2.5 x 10^-11 M
So, the concentration of (H3O+) ions is approximately 0.0004 M, and the concentration of (OH-) ions is approximately 2.5 x 10^-11 M in the given weak acid solution.
Answer:
your answer
Explanation:
To find the concentration of (H3O+) and (OH-) ions for a solution of a weak acid (HA) with a given Ka (acid dissociation constant) value and 8 moles of acid in 1 liter of solution, you can use the equilibrium expression for the dissociation of the weak acid:
HA ⇌ H+ + A-
Where:
HA represents the weak acid.
H+ represents the hydronium ion (H3O+).
A- represents the conjugate base of the acid.
Given that Ka = 2 × 10^-8, this means that at equilibrium, you can assume that [H+][A-] / [HA] = Ka.
Now, you know that 8 moles of HA are present in 1 liter of solution. So, the initial concentration of HA ([HA]initial) is 8 moles/L.
At equilibrium, some of the HA will dissociate into H+ and A-. Let's assume that x moles of HA dissociate into H+ and A-. So, [HA] at equilibrium will be (8 - x) moles/L, [H+] will be x moles/L, and [A-] will also be x moles/L.
Now, you can use the Ka expression:
Ka = [H+][A-] / [HA]
Plug in the values:
2 × 10^-8 = (x * x) / (8 - x)
Solve for x:
x^2 = 2 × 10^-8 * (8 - x)
x^2 = 1.6 × 10^-7 - 2x × 10^-8
x^2 + 2x × 10^-8 - 1.6 × 10^-7 = 0
You can solve this quadratic equation for x. Once you have the value of x, you can find the concentration of [H3O+] and [OH-].
Keep in mind that in a weak acid solution, [H3O+] and [OH-] will not be equal to each other, but their product will still be related to the value of Ka, following the equation [H3O+][OH-] = Kw (the ion product of water), which is approximately 1.0 x 10^-14 at 25°C.
I know this is long method
but hope you understand
Thanks