CH3COOH(aq) + H2O(l) ⇌ H3O^+(aq) + CH3COO^–(aq)
[H3O^+] = 4.0 × 10^–3 M
[CH3COOH] = 0.90 M
What is the Ka value for the reaction?
Ka= (4e-3)^2/.9=1.777e-5 M
assuming both the concentrations you gave were at equilibrium
To solve for the Ka value, you need to use the equation for the equilibrium constant, Ka:
Ka = [H3O+][CH3COO-] / [CH3COOH]
Now, substitute the given values into the equation:
Ka = (4.0 × 10^-3 M)(CH3COO-) / 0.90 M
Since the concentration of H3O+ (or [H3O+]) is the same as the concentration of H3O+, we can write:
Ka = (4.0 × 10^-3)([H3O+] / [CH3COOH])
Now, substitute the given values for [H3O+] and [CH3COOH]:
Ka = (4.0 × 10^-3) / 0.90
Simplifying this equation, we get the value for Ka:
Ka = 4.4 × 10^-3 (approximately)
To find the Ka value for the reaction, we can use the equation:
Ka = ([H3O+][CH3COO-]) / [CH3COOH]
Given:
[H3O+] = 4.0 × 10^–3 M
[CH3COOH] = 0.90 M
Now, substitute these values into the equation for Ka:
Ka = (4.0 × 10^–3 M × [CH3COO-]) / (0.90 M)
To find the value of [CH3COO-], we need to use the fact that the concentration of CH3COO- is equal to the concentration of H3O+:
[CH3COO-] = [H3O+] = 4.0 × 10^–3 M
Substituting this value into the equation for Ka:
Ka = (4.0 × 10^–3 M × 4.0 × 10^–3 M) / (0.90 M)
Now, we can calculate the value of Ka:
Ka = 1.78 × 10^–5
Therefore, the Ka value for the reaction is 1.78 × 10^–5.