HA(aq) <-> H+(aq) + A-(aq) K = 5.0 x 10 -9
The initial concentration of HA is 0.30 M What is the final equilibrium concentrations of HA, H+, A-?
Show work, solve if you can
To solve this problem, we need to set up an ICE table and use the equilibrium constant expression to calculate the final equilibrium concentrations. Let's define the following variables:
HA: initial concentration of HA (0.30 M)
[H+]: final concentration of H+ (unknown)
[A-]: final concentration of A- (unknown)
K: equilibrium constant (5.0 x 10^-9)
Now let's set up the ICE table:
HA(aq) <-> H+(aq) + A-(aq)
Initial (M): 0.30 0 0
Change (x) -x +x +x
Equilibrium: 0.30 - x x x
Using the equilibrium constant expression, we can write:
K = [H+][A-]/[HA]
Substituting the known values:
5.0 x 10^-9 = x*x/(0.30 - x)
Now we'll solve the equation for x. Since the equilibrium constant is very small, we can assume that x is negligible compared to 0.30. This allows us to simplify the expression to:
5.0 x 10^-9 ≈ x*x/0.30
Rearranging the equation:
x*x = 5.0 x 10^-9 * 0.30
x^2 ≈ 1.5 x 10^-9
Taking the square root of both sides:
x ≈ √(1.5 x 10^-9)
x ≈ 1.22 x 10^-5 M
We can now determine the final equilibrium concentrations:
[HA] = 0.30 - x = 0.30 - 1.22 x 10^-5 ≈ 0.2999 M
[H+] = x ≈ 1.22 x 10^-5 M
[A-] = x ≈ 1.22 x 10^-5 M
Therefore, the final equilibrium concentrations of HA, H+, and A- are approximately 0.2999 M, 1.22 x 10^-5 M, and 1.22 x 10^-5 M, respectively.
To solve this question, we'll use the information provided and the concept of equilibrium concentrations.
Let's assume that the final equilibrium concentration of HA is x M. After dissociation, the concentration of H+ and A- will also be x M each.
We can set up an ICE (Initial, Change, Equilibrium) table to track the changes in concentration:
HA <-> H+ + A-
Initial 0.30 0 0
Change -x +x +x
Equilibrium 0.30 - x x x
Now, we can use the equilibrium constant expression (K) to set up an equation:
K = [H+][A-] / [HA]
K = 5.0 x 10^-9
Plugging in the equilibrium concentrations, we get:
5.0 x 10^-9 = x * x / (0.30 - x)
Simplifying the equation, we get:
5.0 x 10^-9 = x^2 / (0.30 - x)
Next, we can cross-multiply and rearrange the equation:
5.0 x 10^-9 * (0.30 - x) = x^2
1.5 x 10^-9 - 5.0 x 10^-9x = x^2
Rearranging and simplifying, we get a quadratic equation:
x^2 + 5.0 x 10^-9x - 1.5 x 10^-9 = 0
At this point, we can solve the quadratic equation using the quadratic formula. However, this equation has coefficients involving very small numbers ("10^-9"), which means that x will be much smaller than 0.30. Therefore, we can assume that "0.30 - x" is approximately equal to 0.30:
x^2 + 5.0 x 10^-9x - 1.5 x 10^-9 ≈ 0
Simplifying further:
x^2 + 5.0 x 10^-9x - 1.5 x 10^-9 ≈ 0
x(x + 5.0 x 10^-9) - 1.5 x 10^-9 ≈ 0
Approximating 5.0 x 10^-9 to 0 gives us:
x^2 - 1.5 x 10^-9 ≈ 0
Solving this simplified equation, we can take the square root of both sides:
x ≈ √(1.5 x 10^-9)
Calculating this value, we find:
x ≈ 3.87 x 10^-5
Therefore, the final equilibrium concentrations are approximately as follows:
HA ≈ 0.30 - x ≈ 0.30 - 3.87 x 10^-5 ≈ 0.2999613 M
H+ ≈ A- ≈ x ≈ 3.87 x 10^-5 M
So, the final equilibrium concentration of HA is approximately 0.29996 M, and the final equilibrium concentrations of H+ and A- are approximately 3.87 x 10^-5 M.
...................HA ==> H^+ + A^-
I..................3.0.........0.........0
C................-x............x.........x
E................3.0-x.......x.........x
Write the K expression, plug in the numbers and solve for x. Use the value of x to determine the values of HA, H^+ and A^-.
Post your work if you get stuck.