What is the pH of a 5.06E-2 M aqueous solution of sodium acetate?
I know that I would set this up the same way I did for the previous question that I posted. I am just confused on how to find the Ka. I know it is kw/kb but for what compound is what I do not know.
Chemistry(Please help) - DrBob222, Saturday, April 7, 2012 at 10:58pm
You want to find Kb for acetate. That is Kw/Ka and Ka is for acetic acid.
Chemistry(Please help) - Hannah, Saturday, April 7, 2012 at 11:03pm
ok thank you!
Chemistry(Please help) - Hannah, Saturday, April 7, 2012 at 11:10pm
so for Kb
Chemistry(Please help) - Hannah, Saturday, April 7, 2012 at 11:12pm
So for Kb I did kw/ka = 1.0e-14/1.8e-5 = 5.55e-10 . Now I would multiply this by 5.06e-6 and take the square root and then covert to pH by taking the -log. Is this correct?
Chemistry(Please help) - DrBob222, Saturday, April 7, 2012 at 11:26pm
Yes, however the number is 5.06E-2 and not -6. Note, too, that a quadratic may be necessary. You will substitute 0.0506-x for (acetate) in the equation and the x may (or may not) be negligible.
Chemistry(Please help) - Hannah, Sunday, April 8, 2012 at 4:30am
I set this up as 5.55e-10 = x^2/0.0506-x
After converting to pH I got 5.27 but it said that this was incorrect so maybe I do need to use a quadratic, but I am not sure how to go about this.
Probably not. I could tell better if you had shown your work. I think you solve for x but failed to realize that x = OH^-. Remember
Ac^- + HOH ==> HAc + OH^-
So your 5.27 probably is pOH. Subtract from 14 to get pH.
To find the pH of a 5.06E-2 M aqueous solution of sodium acetate, you first need to calculate the concentration of hydroxide ions (OH-) in the solution. Sodium acetate is the conjugate base of acetic acid, so it will react with water to produce hydroxide ions.
To find the concentration of hydroxide ions, you need to determine the Kb value for acetate. Kb is the equilibrium constant for the reaction of the acetate ion with water to produce hydroxide ions.
The Kb value is related to the Ka value for acetic acid by the equation Kb = Kw/Ka, where Kw is the ion product constant for water (= 1.0E-14).
In this case, the Ka value for acetic acid is needed to find the Kb value for acetate. The Ka value for acetic acid is 1.8E-5.
So, Kb = 1.0E-14/1.8E-5 = 5.55E-10.
Next, you can use the Kb value to determine the concentration of hydroxide ions from the sodium acetate.
You can assume that the sodium acetate fully dissociates in water, so the concentration of the acetate ion (C2H3O2-) will be equal to the initial concentration of the sodium acetate.
Let's assume x is the concentration of hydroxide ions.
Then, the concentration of acetate ions is (5.06E-2 - x) M.
Using the equation Kb = [OH-][C2H3O2-] / [C2H3O2-], we can substitute the values:
5.55E-10 = x * (5.06E-2 - x) / (5.06E-2)
Now you have a quadratic equation that you can solve to find the concentration of hydroxide ions (x). Once you find the value of x, you can use it to calculate the pOH and then convert it to pH using the equation pH = 14 - pOH.