Find the [H+] in a 0.0255 M solution of HF if the Ka = 7.2 X 10^ -4
To find the [H+] in a solution of HF, we need to use the equilibrium expression for the dissociation of HF:
HF ⇌ H+ + F-
The equilibrium constant, Ka, is defined as:
Ka = [H+][F-] / [HF]
Given that the Ka is 7.2 x 10^-4 and the concentration of HF is 0.0255 M, we can set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations at equilibrium.
Let x be the concentration of H+ and F- at equilibrium. Initially, the concentration of HF is 0.0255 M, and there is no H+ or F-. Therefore, the initial concentrations are:
HF: 0.0255 M
H+: 0 M
F-: 0 M
Due to the dissociation of HF, x mol/L of HF will dissociate, resulting in x mol/L of H+ and F-:
HF: 0.0255 - x M
H+: x M
F-: x M
The equilibrium concentration of HF is then given by 0.0255 - x. Plugging these values into the equilibrium expression, we have:
Ka = [H+][F-] / [HF] = x * x / (0.0255 - x)
Simplifying, we get:
7.2 x 10^-4 = x^2 / (0.0255 - x)
Now, we can solve this equation to find the value of x, which represents the concentration of H+. Rearranging the equation, we have:
7.2 x 10^-4 * (0.0255 - x) = x^2
Multiplying out and rearranging terms:
0.01836 - 7.2 x 10^-4x = x^2
Rearranging to a quadratic equation form:
x^2 + 7.2 x 10^-4x - 0.01836 = 0
Now, we can solve this quadratic equation using the quadratic formula. Using the formula, we have:
x = (-b ± √(b^2 - 4ac)) / 2a
where a = 1, b = 7.2 x 10^-4, and c = -0.01836. Plugging in these values and solving for x, we get two possible values. One value will be positive and represent the concentration of H+, while the negative value can be ignored in this context since concentrations cannot be negative.
Using the quadratic formula, we find that x ≈ 0.00497 M.
Therefore, the concentration of [H+] in the 0.0255 M solution of HF is approximately 0.00497 M.