The ionization constant for benzoic acid is 6.5x10^-5. find the hydrogen ion concentration at equilibrium when the initial concentration is .29M
Look at the others I've done. This follows the same guidelines.
To find the hydrogen ion concentration at equilibrium for the given conditions, we can use the ionization constant (Ka) of benzoic acid.
The ionization of benzoic acid can be represented by the following equation:
C6H5COOH (aq) ⇌ C6H5COO- (aq) + H+ (aq)
The expression for the ionization constant (Ka) is given by:
Ka = [C6H5COO-] [H+] / [C6H5COOH]
We are given that the ionization constant (Ka) is 6.5x10^-5, and the initial concentration of benzoic acid ([C6H5COOH]) is 0.29 M.
Let's assume the change in concentration at equilibrium is x M. At equilibrium, the concentration of benzoic acid will be ([C6H5COOH] - x) M, and the concentrations of benzoate ([C6H5COO-]) and hydrogen ions ([H+]) will be x M.
Substituting the given values into the equilibrium expression, we have:
6.5x10^-5 = (x)(x) / (0.29 - x)
Solving this quadratic equation will give us the value of x, which represents the concentration of H+ ions at equilibrium.
To solve the equation, rearrange it to the form:
x^2 + (6.5x10^-5)x - (6.5x10^-5)(0.29) = 0
We can solve this quadratic equation using the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
Here, a = 1, b = 6.5x10^-5, and c = -(6.5x10^-5)(0.29).
Let's calculate the values:
a = 1
b = 6.5x10^-5
c = -(6.5x10^-5)(0.29)
Discriminant, D = b^2 - 4ac
D = (6.5x10^-5)^2 - 4(1)(-(6.5x10^-5)(0.29))
Now, substitute the values of a, b, c, and D into the quadratic formula:
x = [-b ± √(D)] / (2a)
Calculate x using the values:
x = [-6.5x10^-5 ± √(D)] / (2)
Once we calculate x, we can substitute this value back into the expression for the hydrogen ion concentration [H+] to find its value at equilibrium.
To find the hydrogen ion concentration at equilibrium, we can use the ionization constant (Ka) expression of benzoic acid:
Ka = [H+][C6H5COO-] / [C6H5COOH]
Where:
[H+] is the hydrogen ion concentration.
[C6H5COO-] is the concentration of the benzoate ion.
[C6H5COOH] is the concentration of benzoic acid.
Since benzoic acid is a weak acid, we can assume that the concentration of the benzoate ion is equal to the concentration of the hydrogen ions formed when benzoic acid ionizes.
Let's assume that x is the concentration of the hydrogen ions at equilibrium. Therefore, [H+] = x.
Given that the initial concentration of benzoic acid is 0.29 M, and since it's a weak acid, we can assume that the concentration of [C6H5COOH] at equilibrium will be (0.29 - x) M.
Plugging these values into the Ka expression:
6.5x10^-5 = [H+][C6H5COO-] / [C6H5COOH]
6.5x10^-5 = x * x / (0.29 - x)
Next, we can solve this equation for x to find the hydrogen ion concentration at equilibrium. Since this is a quadratic equation, we can solve it using the quadratic formula.
6.5x10^-5 = x^2 / (0.29 - x)
Rearranging the equation, we get:
x^2 = 6.5x10^-5 * (0.29 - x)
x^2 = 1.885x10^-5 - 6.5x10^-5 * x
x^2 + 6.5x10^-5 * x - 1.885x10^-5 = 0
Now, we can solve this quadratic equation for x. Plugging the coefficients into the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Where a = 1, b = 6.5x10^-5, and c = -1.885x10^-5.
Plugging in the values and calculating, we get:
x ≈ 1.139x10^-3 M (approximately)
Therefore, the hydrogen ion concentration at equilibrium is approximately 1.139x10^-3 M.