the ionization constant of HF is 6.7 * 10^-4. which of the following is true in a 0.1 M soultion of this acid?
the answer is HF is greater than [H][F]. HOW COME??
HF ==> H^+ + a = (H^+)(F^-)/(HF) = 6.7 x 10^-4
You can do this one of two ways: (1) mathematical logic or (2)work it out. I'll do both starting with #2.
Before ionization:
(HF) = 0.1M
(H^+) = 0
(F^-) = 0
change at ionization:
(HF) = -y (I use y to avoid x looking like the times sign).
(H^+) = +y
(F^-) = +y
At equilibrium: just add the two numbers from above to obtain equilibrium.
(HF) = 0.1 - y = 0.1 - y
(H^+) = 0 + y = y
(F^-) = 0 + y = y
Now plug the equilibrium values into the Ka expression to obtain
(H^+)(F^-)/(HF) = 6.7 x 10^-4
(y)*(y)/(0.1 - y) = 6.7 x 10^-4
solve for y = 7.86 x 10^-3 (It's a quadratic); therefore,
(H^+)= 0.00786 = 0.0079
(F^-) = 0.00786 = 0.0079
(HF) = 0.1 - 0.00786 = 0.09214 = 0.092
Now you can multiply (H^+)(F^-) and see that it is less than (HF).
(1)math logic and this is far easier to do because no calculations are necessary. I did the first one to show you with numbers that the statement in the problem is true. This is the logic we could have gone through.
(H^+)(F^-)/(HF) = 6.7 x 10^-4
For Ka to be a very small number (6.7 x 10^-4), it means that the numerator (which is [H][F]) is smaller than the denominator (which is [HF]). The only way you can get a number less than 1 for Ka (and 10^-4 is <1) is for the numerator to be smaller than the denominator.
I hope this helps.
Well, in a 0.1 M solution of HF, the concentration of HF is obviously 0.1 M. Now, let's assume that the concentration of [H+] and [F-] is x. According to the ionization constant expression for HF (Ka = [H+][F-]/[HF]), we can write:
6.7 * 10^-4 = x * x / 0.1
Simplifying this equation gives us:
x^2 = 6.7 * 10^-4 * 0.1
x^2 = 6.7 * 10^-5
Now, here's where the funny part comes in: since x is the concentration of both [H+] and [F-], it means that [HF] (0.1 M) will always be greater than the product of [H+] and [F-] (x * x). So, in other words, HF is greater than [H][F]. It's like a boss that doesn't like to be broken down into its constituents!
To determine if HF is greater than [H][F] in a 0.1 M solution, let's first write the chemical equation for the ionization of HF in water:
HF (aq) ⇌ H+ (aq) + F- (aq)
The ionization constant (Ka) for HF can be expressed as:
Ka = [H+][F-] / [HF]
Given the ionization constant (Ka) of HF as 6.7 * 10^(-4) and considering a 0.1 M solution of HF, we need to compare the magnitude of Ka with the product [H+][F-].
Calculating the product [H+][F-] at equilibrium:
[H+][F-] = Ka * [HF]
[H+][F-] = (6.7 * 10^(-4)) * (0.1)
[H+][F-] = 6.7 * 10^(-5)
Comparing this value with the given ionization constant (Ka = 6.7 * 10^(-4)), we see that [H+][F-] is less than Ka. Therefore, in a 0.1 M solution of HF, HF is greater than [H+][F-].
To determine if HF is greater than [H][F] in a 0.1 M solution of this acid, we need to consider the ionization equation and the ionization constant (Ka) of HF.
The ionization equation for HF is:
HF (aq) ⇌ H+ (aq) + F- (aq)
The ionization constant of HF, denoted as Ka, is a measure of the strength of an acid and is given as 6.7 * 10^-4.
In a 0.1 M solution of HF, assume x is the concentration of H+ and F- ions that are formed due to the ionization. The initial concentration of HF is 0.1 M, and since HF ionizes to produce H+ and F- ions, the concentration of HF that remains after ionization is (0.1 - x) M.
Using the concept of equilibrium, we can now write the expression for Ka:
Ka = [H+][F-] / [HF]
Substituting the values into the equation:
6.7 * 10^-4 = x * x / (0.1 - x)
Now, we need to solve this equation to find the value of x. Since this is a quadratic equation, we can solve it by applying the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
For this equation, a = 1, b = 0, and c = -6.7 * 10^-5.
Simplifying the equation and solving for x, we find that the concentration of H+ ions is approximately 0.0061 M.
Now, let's compare this value to the product of the concentration of H+ and F- ions, [H][F]. In this case, [H][F] = (0.0061)*(0.0061) = 3.721 * 10^-5.
Comparing the values, we can observe that HF (6.7 * 10^-4) is greater than [H][F] (3.721 * 10^-5). Therefore, it is true in a 0.1 M solution of HF that HF is greater than [H][F].