For this question, I know how to answer the first three questions, but not the rest.
12ml 0.025 M of benzoic acid (Ka=6.3x10-5)is titrated with 0.050 M NaOH.
How many moles of benzoic acid are there? I got 0.0003
What volume of NaOH is required?(L) I got 0.006
What is the total volume of the solution at equivalence?(L) I got 0.018
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What is the concentration of the salt solution formed at equivalence?
What is the value of the correct K.
Are we looking for OH- or H3O+?
What is the concentration of the species above at equilibrium?
What is the pH of the solution?
I have tried posting the answer but the site won't let me. I'll try posting them as separate problems.
The first three are right.
4. millimols HB = 0.3
volume NaOH = 6 mL
total volume = 18 mL at equivalence point.
If we call benzoic acid HB, then
.........HB + NaOH ==> NaB + H2O
I.......0.3.............0.....0
add...........0.3................
C.......-0.3..-0.3......0.3.....0.3
E........0.....0........0.3.....0.3
So you have 0.3 millimols salt formed. Therefore, M = mmols/mL = ?
5. I don't know which K you are talking about. K for HB is given in the problem as 6.3E-5. I expect you are talking about the Kb for B^- (benzoate ion) and that is Kb benzoate = (Kw/Ka for HB) = ?
6. When the B^- reacts with H2O it is
............B^- + HOH ==> HB + OH^- so you are looking for OH^- which you let be x.
7. ........B^- + HOH ==> HB + OH^-
I.......0.0167.............0....0
C..........-x..............x....x
E.......0.0167-x...........x....x
Kbase for B^- is (Kw/Ka for HB) = (x)(x)/(0.0167-x)
Solve for x = (OH^-).
The last one is to convert OH^- to pH.
I don't know why I couldn't post all of the answers with one post.
8. Convert OH^- to pH.