I need help with this question, I don't know how to set up the formula to work it out?
Calculate the dissociation constant of a weak monoprotic acid if a 0.08 M solution of this acid gives a hydrogen-ion concentration of 0.0001 M?
Weak monoprotic acid is HA
..........HA --> H^+ + A^-
I.......0.08.....0.....0
C........-x......x.....x
E......0.08-x....x.....x
Substitute the E line into the Ka expression. The problem tells you that x = 1E-4. You may need to solve the quadratic; i.e., you may not be able to assume 0.08-x = 0.08
Post your work if you get stuck.
I'm confused??
That doesn't help me. What are you confused about. In detail. Show your work as far as you can get and explain what you don't understand about the next step.
To solve this problem, you can use the formula for the dissociation constant (Ka) of a weak monoprotic acid:
Ka = [H+][A-] / [HA]
Where:
- [H+] represents the concentration of hydrogen ions
- [A-] represents the concentration of the dissociated acid ions
- [HA] represents the concentration of the undissociated acid
In this case, you are given the hydrogen-ion concentration ([H+]) of 0.0001 M. However, since the monoprotic acid has a 1:1 stoichiometric ratio (meaning one molecule of acid produces one hydrogen ion), the concentration of the dissociated acid ions ([A-]) is also 0.0001 M.
The only piece of information missing is the concentration of the undissociated acid ([HA]). To find this concentration, you need to subtract the concentration of the dissociated acid ions from the initial concentration of the acid solution. According to the problem, the initial concentration of the acid solution is 0.08 M.
[HA] = Initial concentration - [A-]
[HA] = 0.08 M - 0.0001 M
[HA] = 0.0799 M
Now, plug the values into the formula for Ka:
Ka = [H+][A-] / [HA]
Ka = (0.0001 M)(0.0001 M) / 0.0799 M
Ka = 0.00000001 M^2 / 0.0799 M
Simplifying the expression, we get:
Ka = 1.25 x 10^-7 M
Therefore, the dissociation constant (Ka) of the weak monoprotic acid is 1.25 x 10^-7 M.