A 0.00100 M solution of a weak acid HX is 6% ionized. Calculate Ka for the acid.
Oh, calculating Ka, huh? That's like trying to calculate the exact number of jellybeans in a jar by shaking it vigorously. It's gonna give you a headache. But fear not, my friend, the Clown Bot is here to help!
First, let's break down the problem. You've got a 0.00100 M solution of a weak acid HX that is 6% ionized. So, if we assume x is the amount of HX that ionizes, we can say that 0.06x is the concentration of H+ ions and 0.06x is the concentration of X- ions (because of a 1:1 ratio).
Now, since the acid is weak, we can make the approximation that x is very small compared to 0.00100 M. So, we can say that the concentration of HX remaining (0.00100 M - x) is approximately equal to 0.00100 M.
Now, let's set up an expression for the equilibrium constant (Ka):
Ka = [H+][X-] / [HX]
Substituting the values we found earlier, we get:
Ka = (0.06x)(0.06x) / (0.00100)
Simplifying this equation, we have:
Ka = (0.0036x^2) / 0.00100
And since x is small compared to 0.001, we can approximate it as:
Ka ≈ (0.0036x^2) / 0.00100
Now, here comes the tricky part. We need to find the value of x. However, it's not given in the problem. But wait! We know that the acid is 6% ionized. So, we can say that 0.06x is equal to 6% of 0.00100. That is:
0.06x = 0.06(0.00100)
Solving for x, we get:
x ≈ 0.001/0.06
x ≈ 0.0167 M
Now, we can plug this value of x into our earlier expression for Ka:
Ka ≈ (0.0036(0.0167)^2) / 0.00100
And after shaking the equation like a jar of jellybeans, we find:
Ka ≈ 0.00226
So, the approximate value of Ka for the acid is 0.00226.
To calculate Ka for the acid (HX), we first need to calculate the concentration of the ionized acid (H+) in the solution.
Given:
Initial concentration of HX (C0) = 0.00100 M
% ionization = 6% = 0.06
Step 1: Calculate the concentration of H+ ions (C(H+))
C(H+) = C0 × % ionization
C(H+) = 0.00100 M × 0.06
C(H+) = 0.00006 M
Step 2: Calculate the concentration of the unionized acid (HX) (C(unionized HX))
C(unionized HX) = C0 - C(H+)
C(unionized HX) = 0.00100 M - 0.00006 M
C(unionized HX) = 0.00094 M
Step 3: Write the equilibrium equation for the ionization of the weak acid HX:
HX ⇌ H+ + X-
Step 4: Write the expression for the equilibrium constant (Ka):
Ka = [H+][X-] / [HX]
Step 5: Substitute the concentrations into the expression for Ka:
Ka = (C(H+) × C(X-)) / C(unionized HX)
Ka = (0.00006 M × 0.00006 M) / 0.00094 M
Ka = 0.0000036 M2 / 0.00094 M
Ka ≈ 0.00383 M
Therefore, the Ka value for the weak acid HX is approximately 0.00383 M.
To calculate the Ka for the weak acid HX, we first need to find the concentration of the ionized form of the acid, which in this case is X-.
Given:
- HX is a weak acid with a concentration of 0.00100 M
- The acid is 6% ionized
Step 1: Find the concentration of the ionized form (X-)
Since the weak acid is 6% ionized, this means that only 6% of HX has dissociated into X-. Therefore, we can calculate the concentration of X- by multiplying the concentration of HX (0.00100 M) by the percent ionization (6% = 0.06):
Concentration of X- = 0.00100 M x 0.06 = 0.00006 M
Step 2: Find the concentrations of both the undissociated form (HX) and the ionized form (X-)
The concentration of undissociated HX can be found by subtracting the concentration of the ionized form (X-) from the initial concentration of the acid (0.00100 M):
Concentration of HX = 0.00100 M - 0.00006 M ≈ 0.00094 M
Step 3: Set up the Ka expression
The Ka expression for a weak acid is given by: Ka = [H+][X-] / [HX]
However, since the concentration of [H+] is very small compared to the concentrations of [HX] and [X-], we can assume that [H+] is negligible.
Thus, the expression simplifies to: Ka = [X-] * [H+]/[HX] ≈ [X-]² / [HX]
Step 4: Calculate Ka
Now that we have the concentrations of both [X-] and [HX], we can substitute these values into the Ka expression to calculate Ka:
Ka = [X-]² / [HX] = (0.00006 M)² / (0.00094 M) ≈ 4.04 x 10^-6
Therefore, the Ka for the weak acid HX is approximately 4.04 x 10^-6.
.........HX | H2O ==> H3O^+ X^-
I.......0.001..........0.....0
C........-x...........x......x
E......0.001-x........x......x
Ka = (H3O^+)(X^-)/(HX)
(H3O^+) = 0.001*0.06
(X^-) = 0.001*0.06
(HX) = (0.001-(0.001*0.06)