What is [H+] concentration in a 4M aq solution of acetic acid (Ka= 1.8 E-5)?
what formula do you use to get the answer?
Set up an ICE chart, substitute into the Ka expression, and solve for the unknown.
could you remind me what an ICE chart is?
If we call acetic acid, HAc, then
................HAc ==> H^+ + Ac^-
initial(I).......4.......0......0
change(C).......-x......x.......x
equil(E).........4-x.....x.......x
Ka = (H^+)(Ac^-)/(HAc)
1.8E-5 = (x)(x)/(4.0-x)
Solve for x = (H^+).
thank you!
To find the [H+] concentration in a solution of acetic acid, we can use the Ka value and initial concentration of the acid. The formula used is based on the equilibrium expression of the acid dissociation reaction:
Kw = [H+][OH-]
For a weak acid like acetic acid, the concentration of the hydroxide ions (OH-) can be assumed to be negligible compared to the concentration of water (H2O). Therefore, we can simplify the equation to:
Kw = [H+]^2
Rearranging the equation, we can solve for the [H+] concentration:
[H+] = √(Kw)
In this case, we don't have the value for Kw, but we can use the Ka value instead. For a weak acid, Kw = Ka * [HA], where [HA] represents the initial concentration of the acid. Rearranging this equation, we get:
[H+] = √(Ka * [HA])
Now, let's substitute the given values into the equation:
Ka = 1.8 E-5 (Given)
[HA] = 4M (Given)
[H+] = √(1.8 E-5 * 4)
To calculate this, we first multiply the Ka value by the initial concentration [HA], then take the square root of the result.