The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?
pH = 2.03 = -log(H^+). That will be about 0.01 but you need a more exact answer.
.................HF --> H^+ + F^-
I................0.01.....0.........0
C................-x........x..........x
E..............0.01-x....x..........x
Ka = (H^+)(F^-)/(HF). You know x, evaluate 0.01-x, plug into Ka expression and solve for Ka.
Post your work if you get stuck.
Can you solve it fully and explain it
To find the value of Ka for HF, we can use the relationship between the concentrations of the dissociated and undissociated forms of the acid, as well as the pH of the solution. The relationship is given by the equation:
Ka = ([H+][F-]) / [HF]
In this case, we know the concentration of HF is 0.25 M, and the pH of the solution is 2.03. To calculate the concentration of H+ ions, we need to convert the given pH to a concentration.
The pH is defined as the negative logarithm (base 10) of the concentration of H+ ions:
[H+] = 10^(-pH)
Plugging in the given pH value:
[H+] = 10^(-2.03)
[H+] = 7.01 x 10^(-3) M
Now, we can plug the values into the equation for Ka:
Ka = ([H+][F-]) / [HF]
= (7.01 x 10^(-3) M)(7.01 x 10^(-3) M) / 0.25 M
Ka = 1.96 x 10^(-6)
Therefore, the value of Ka for HF is 1.96 x 10^(-6).
To find the value of Ka for hydrofluoric acid (HF) in the given aqueous solution, you can use the pH of the solution and the concentration of the acid.
Ka represents the acid dissociation constant, which is a measure of the extent to which an acid dissociates in water. It can be calculated using the equation:
Ka = [H+][F-] / [HF]
Where [H+] represents the concentration of hydrogen ions, [F-] represents the concentration of fluoride ions, and [HF] is the concentration of hydrofluoric acid.
In this case, you are given the pH of the solution, which is 2.03. The pH is a logarithmic scale that measures the concentration of hydrogen ions in a solution. The formula to convert pH to [H+] concentration is:
[H+] = 10^(-pH)
Substituting the given pH value into the equation:
[H+] = 10^(-2.03)
Now, since HF is a weak acid, it will not fully dissociate into H+ and F- ions. Instead, only a fraction of the HF molecules will dissociate, and the concentrations of H+ and F- ions will be equal. Therefore:
[H+] = [F-]
Let's denote the concentration of hydrogen ions as [H+]. Since you have a 0.25 M solution of hydrofluoric acid, the concentration of HF is 0.25 M.
Plugging all of this into the Ka equation:
Ka = [H+][F-] / [HF]
= [H+][H+] / (0.25)
= (0.25)(10^(-2.03))^2
Now you can calculate Ka by substituting the value for [H+]:
Ka = (0.25)(10^(-2.03))^2
Using this equation, you can find the value of Ka for hydrofluoric acid.