The pH of a 0.060 M weak monoprotic acid, HA is 3.44. Calculate the Ka of the acid.
My calculator is broken. You calculate H^+ from pH = -log(H^+). I will call that value, whatever it is, x
.....................HA ==> H^+ + A^-
Initial..........0.06..........0........0
change..........-x............x..........x
equilibriium..0.06-x.......x..........x
Ka = (H^+)(A^-)/(HA)
Substitute the E line into Ka expression and solve for Ka.
Post your work if you get stuck.
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To calculate the Ka (acid dissociation constant) of the weak monoprotic acid, HA, we can use the equation for the acid dissociation reaction:
HA ⇌ H+ + A-
The pH of a solution can be related to the concentration of hydronium ions [H3O+] using the equation:
pH = -log[H3O+]
We can use this equation to find the concentration of hydronium ions, [H3O+]. From the given pH of 3.44, we can calculate [H3O+] as follows:
[H3O+] = 10^(-pH)
[H3O+] = 10^(-3.44)
[H3O+] = 3.70 x 10^(-4) M
Since the acid, HA, is a weak acid, we can assume that the concentration of [H3O+] comes mainly from the dissociation of HA. We assume that the initial concentration of HA is equal to the initial concentration of [H3O+].
To calculate the concentration of HA that dissociates, let x be the concentration of [H3O+] that comes from the dissociation of HA. Then, the equilibrium concentrations of [H3O+], [A-], and [HA] can be expressed as follows:
[H3O+] = x
[A-] = x
[HA] = 0.060 - x
Using the equilibrium expression for Ka:
Ka = [H+][A-] / [HA]
Substituting the equilibrium concentrations into the Ka expression:
Ka = (x)(x) / (0.060 - x)
Since a weak acid typically undergoes limited dissociation, we can assume that the change in concentration of HA, x, is small compared to the initial concentration of HA. Therefore, we can approximate 0.060 - x as approximately 0.060.
Ka = (x)(x) / 0.060
To solve for x, we can use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
In this case, a = 1, b = 0, and c = -Ka(0.060). Plugging these values into the quadratic formula:
x = [-0 ± sqrt(0 - 4(1)(-Ka*0.060))] / (2*1)
Simplifying,
x = [sqrt(0.24Ka)] / 2
Since we assumed that the change in concentration of HA, x, is small, we can neglect the negative square root term:
x = sqrt(0.24Ka) / 2
Now, we substitute this value of x back into the expression for Ka:
Ka = (sqrt(0.24Ka) / 2) * (sqrt(0.24Ka) / 2) / 0.060
Simplifying,
Ka = (0.24Ka / 4) / 0.060
Ka = (0.24Ka) / (0.24)
Ka = 0.060
Therefore, the value of Ka for the given weak monoprotic acid, HA, is 0.060.
To calculate the Ka of the acid, we need to use the equation for acid dissociation:
HA ⇌ H+ + A-
The Ka is the equilibrium constant for this reaction, which can be calculated using the formula:
Ka = [H+][A-] / [HA]
We know the concentration of the weak acid, HA, which is 0.060 M. We also know the pH of the solution, which is 3.44, and we can use this value to find the concentration of H+ ions in the solution.
To convert the pH to [H+], we must use the following formula:
[H+] = 10^(-pH)
Plugging in the value, we get:
[H+] = 10^(-3.44) = 3.52 x 10^(-4) M
Since the weak acid, HA, completely dissociates into H+ and A-, we can assume that [HA] = initial concentration of HA.
Using the given concentration of HA, which is 0.060 M, and the concentration of [H+], which is 3.52 x 10^(-4) M, we can now calculate the concentration of [A-]:
[A-] = [H+] = 3.52 x 10^(-4) M
Plugging these values into the Ka formula, we get:
Ka = ([H+][A-]) / [HA]
= (3.52 x 10^(-4) M)(3.52 x 10^(-4) M) / 0.060 M
= 1.98 x 10^(-8)
Therefore, the Ka of the weak acid HA is 1.98 x 10^(-8).