part a.
0.24g/molar mass benzoic acid = ?
M benzoic acid initially = mols/0.116L = ? which I will call z.
............HBz ==> H^+ + Bz^-
initial......z.......0.....0
change.......-x......x.....x.
aquil.......z-x.......x....x
Ka = (H^+)(Bz^-)/(HBz)
Substitute and solve for x = (H^+) and convert that to pH.
part b.
HBz + NaOH ==> NaBz + H2O
Find the equivalence point.
mLHBz x MHBz = mLNaOH x M NaOH.
Solve for mL NaOH.
Each of the ions = mols/L soln
part C.
The pH at the equivalence point is determined by the hydrolysis of the salt, NaBz. The (NaBz) = mols NaBz/total volume in L. I will call that w.
...........Bz^- + HOH ==> HBz + OH^-
initial.....w...............0....0
change.....-x...............x.....x
equil.....w-x...............x.....x
Kb for Bz^- = (Kw/Ka for benzoic acid) = (HBz)(OH^-)/(Bz^-).
Substitute from the ICE chart above and solve for x = (OH^-) and convert that to pH.
Post your work if you have questions.
Assume you dissolve 0.240 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.16 102 mL of solution and then titrate the solution with 0.153 M NaOH.
C6H5CO2H(aq) + OH -(aq) C6H5CO2-(aq) + H2O(l)
(a) What was the pH of the original benzoic acid solution?
(b) What are the concentrations of the following ions at the equivalence point?
1.Na+
2.H3O+
3.OH-
4.C6H5CO2-
(c) What is the pH of the solution at the equivalence point?
3 answers
I'm sorry thank you so much for all your help but I seem to be getting the wrong answer for part c and I don't understand how to get the individual ion's moles for part b. I have the correct volume of 0.00197L.
For part c I am doing as you say and I get a pOH of 5.28 (x=5.15e-06) which gives me a pH of 8.72 but that is apparently wrong.
For part c I am doing as you say and I get a pOH of 5.28 (x=5.15e-06) which gives me a pH of 8.72 but that is apparently wrong.
Since you didn't show your work I have no idea what you've done; however, my best quick guess (and my guess is all I have to go on) is that you didn't use the right volume at the equivalence point. The volume at the eq pt is 128.84 mL (I'm guessing you didn't add in the 1.16E2) and the (benzoate ion) = 0.01525M. That gives me a pOH of 5.8 and a pH of 8.2. I used 6.14E-5 as Ka for benzoic acid but you should use the one in your text or notes. My book is about 15 years old and these things don't always stay the same.