Calculate the hydronium ion concentration of a 0.250M solution of ethaonic acid (Ka= 1.82 x 10^-5)
My explanation looks simple to me.
Ka = 1,82E5 = (H^+)(Ac^-)/(HAc)
1.82E-5 = (x)(x)/(0.250-x)
Solve for x. What's confusing.
HAc = CH3COOH = ethanoic acid.
.................HAc ==> H^+ + Ac^-
I................0.250.......0........9
C..............-x..............x........x
E...........0.25-x..........x........x
Write the Ka expression for HAc and plug in the E line. Solve for x = (H^+).
I am a little bit confused with it please...thanks for the help...
I need the , quite confused
To calculate the hydronium ion concentration of a solution of ethanoic acid, we can use the expression of the acid dissociation constant (Ka) and assume that the acid dissociates completely.
The equation for the dissociation of ethanoic acid is:
CH3COOH + H2O ⇌ CH3COO- + H3O+
The Ka expression for this equation is:
Ka = [CH3COO-][H3O+] / [CH3COOH]
Since we are assuming complete dissociation, the concentration of the ethanoic acid after dissociation is negligible compared to its initial concentration, so we can approximate [CH3COOH] as the initial concentration of ethanoic acid (0.250M).
Rearranging the equation to solve for [H3O+], we get:
[H3O+] = (Ka * [CH3COOH])^0.5
Plugging in the values, we have:
[H3O+] = (1.82 x 10^-5 * 0.250)^0.5
[H3O+] ≈ 1.35 x 10^-3 M
Therefore, the hydronium ion concentration of the 0.250M ethanoic acid solution is approximately 1.35 x 10^-3 M.