Question

1)_What is the pH of the medium when 10.00 mL of 0.1000M NaOH is added to 18.00 mL of a 0.1000M benzoic acid (C6H5COOH) solution?
(Ka = 6.28×10^-5 for C6H5COOH)



2)_How many grams of NHCl (53.491 g/mol) should be added to 200 mL of 0.180M NH3 solution in order to prepare a buffer solution with pH 8.5?

Answer 1

Let's call benzoic acid HBz. Then HBz + NaOH --> NaBz + H2O and this is a buffer solution with NaBz as the base and HBz as the acid.
Millimols NaOH added = millimols NaBz formed = mL x M = 10 x 0.1 = 1
millimols HBz initially = 18 x 0.1 = 1.8
........................HBz + NaOH ==> NaBz + H2O
I.......................1.8.........0................0...............
added..........................0.1.......................................
C....................-0.1.....-0.1..............+0.1........................
E......................0.8.........0...............0.1.....................
So (HBz) @ equilibrium = millimoles/mL = 0.8/28 = ?
(NaBz) @ equilibrium = 0.1/28 = ?
Plug this into the HH equation as
pH = pKa for HBz + log [(NaBz)/(HBz)]

2. I assume you meant NH4Cl in that first line.
Kb for NH3 = 1.75E-5 but use the Ka in your text/notes. Then pKb = 4.76 and pKa + pKb = pKw = 14 so pKa = 9.24
pH = pKa + log [(NH3)/(NH4Cl)]
8.5 = 9.24 + log [(0.18)/(NH4Cl)]
-0.74 = log 0.18/(NH4Cl)
0.182 = 0.18/(NH4Cl)
(NH4Cl) = 0.18/0.182 = about 1 M for NH4Cl
1M NH4Cl will be about 54 g/1000 mL or 54/5 = about 10.8 g NH4Cl in 200 mL. I've estimated here and there but you can clean that up. NOTE, There are so many places to go wrong in a problem like this that I N EVER let it go without checking. So let's check it by placing 10.8 g NH4Cl in 200 mL of 0.18 M NH3 and see if it give a pH of 8.5.
10.8 g NH4Cl in 200 mL.
10.8/54 = 0.2 mol NH4Cl and that in 0.200 L = 1 M
pH = pKa + log b/a
pH = 9.24 + log 0.18/1
pH = 9.24 - 0.74 = 8.5 YEA!!!

Post your work if you get stuck.

Answer 2

thank you very much🙏🏻☺️