A 0.500M solution of a weak acid, HX, is only partially ionized. The [H+] was found to be 4.02 x 10^-3 M. Find the dissociation constant for this acid.
HX ==> H^+ + X^-
Ka = (H^+)(X^-)/(HX)
Set up an ICE chart, substitute for each component in the Ka expression and solve for Ka. I get 3.26 x 10^-3. Don't forget to subtract 4.02 x10^-3 from 0.500 M to obtain unionized (HX); i.e., it isn't 0.500, it is 0.500 - 0.00402.
Well, well, well, we have a "dissociation constant" question here. Looks like we're diving into the chemistry clownery!
To find the dissociation constant (Ka) of the weak acid HX, we need to know the concentration of the acid that has actually dissociated. Thankfully, we have the concentration of the hydronium ions [H+], which is 4.02 x 10^-3 M.
Now, let's assume that the concentration of the undissociated HX is represented by 'x'. Since the weak acid is only partially ionized, we can assume that the concentration of HX that has dissociated is also 'x'. Therefore, the concentration of hydronium ions [H+] can be written as x.
Now, an important equation to remember is:
Ka = [H+] * [A-] / [HA]
Since the weak acid HX is partially ionized, we can assume that [A-] is equal to [H+], which is x. So, we can simplify the formula to:
Ka = x^2 / (0.500 - x)
Remember, the concentration of the acid HX that actually dissociates is 'x', so we plug in the given value of [H+] as x.
Now, let's do some math and calculate the dissociation constant Ka for this acid!
To find the dissociation constant (Ka) for the weak acid HX, we can use the equilibrium expression:
Ka = [H+][X-] / [HX]
Given that the concentration of [H+] is 4.02 x 10^-3 M, we can assume that the concentration of [X-] is also 4.02 x 10^-3 M since HX ionizes partially and produces equal amounts of H+ and X- ions.
The concentration of HX can be calculated by subtracting the concentration of [H+] from the initial concentration of the weak acid. In this case, the initial concentration of the weak acid is 0.500 M.
[HX] = [initial weak acid concentration] - [H+]
[HX] = 0.500 M - 4.02 x 10^-3 M
[HX] = 0.496 M
Now we can substitute the values into the equilibrium expression to find Ka:
Ka = (4.02 x 10^-3 M) * (4.02 x 10^-3 M) / (0.496 M)
Ka ≈ 3.25 x 10^-5
Therefore, the dissociation constant (Ka) for the weak acid HX is approximately 3.25 x 10^-5.
To find the dissociation constant, we can use the equation for the ionization of a weak acid:
HX ⇌ H+ + X-
The dissociation constant, Ka, is defined as the ratio of the concentrations of the products (H+ and X-) to the concentration of the undissociated acid (HX):
Ka = [H+][X-] / [HX]
In order to find the value of Ka, we need the concentrations of H+ and X-, as well as the concentration of HX.
Given that the concentration of H+ ([H+]) is 4.02 x 10^-3 M, we can use this value for [H+] in the equation for Ka.
Now, we need to determine the concentration of X-. Since HX is a weak acid, it only partially ionizes, and based on the information given, we know that the concentration of X- is the same as the concentration of H+:
[X-] = [H+] = 4.02 x 10^-3 M
Finally, we need to determine the concentration of HX. It is given that the solution is 0.500 M in HX. Since HX only partially ionizes, we can assume that the concentration of HX is equal to its initial concentration:
[HX] = 0.500 M
Now, we can substitute the values into the equation for Ka:
Ka = (4.02 x 10^-3 M)(4.02 x 10^-3 M) / (0.500 M)
Simplifying the calculation:
Ka = 1.6164 x 10^-5 / 0.500
Ka ≈ 3.2328 x 10^-5
Therefore, the dissociation constant for the weak acid HX is approximately 3.2328 x 10^-5.