You need to prepare 1.0 L of a buffer with a pH of 9.15. The concentration of the acid in the buffer needs to be 0.100 M. You have available to you a 1.00 M NH4Cl solution, a 6.00 M NaOH solution, and a 6.00 M HCl solution. Determine how to make this buffer. Write all the components with amounts in the box below.

Question

* Do we use all the solutions available to us?

Answer 1

I would google "pKa table acids/bases" (without the parentheses) and look for pKa values in the 9.15 range. I looked and found:

HCN = 9.21
H3BO3 = 9.23
NH4^+ = 9.25

but anything within a + or - 1 will do. I wouldn't use HCN because of the toxicity and since you are given a 1.0 M solution of NH4^+, I would choose that. You can choose any acid in the table withing the appropriate range but you won't get a final solution containing 0.1M NH4Cl unless you start with NH4Cl.
Then, using NH4^+ do this.
pH = pKa + log b/a where b is base concn and a is acid concn. Plug in the number for the acid. Using NH4Cl that gives you b/a = approx 0.8 but you need a more accurate number and do the math yourself for the acid you choose. That makes b = 0.8*a and since you want a to be 0.1 M you will want b to be about 0.08 which means a+b = 0.1 + 0.08 = approx 0.18 for NH4Cl at the beginning.
You want 1 L of the solution and you have 1.0 M NH4Cl and 6 M NaOH. Calculate the volume of NH4Cl and NaOH needed. Mix those and add water to make 1 L.
Post your work if you get stuck.

I ran through the calculation and here are my approx numbers (not exact).
about 180 millimols (180 mL of the 1.0M stuff) NH4Cl to start.
Add about 80 mmols NaOH (about 13 mL of 6M). Mix and add H2O to make 1000 mL (about 800 mL. Mix thoroughly and label.

Answer 2

I don't understand why you're using NaOH? Isn't the buffer made by simply using NH4^+ and NH4Cl?
My work:
pH=pKa + log [A-]/[HA]
9.15=9.26 + log [NH4Cl]/[NH4^+] (got 9.26 from the table given by my prof.)
0.78 = [NH4Cl]/[NH4^+] (mole ratio)
Found mol of NH4^+ by multiplying volume and concentration: (1.0 L)(0.100 M NH4^+) = 0.10 mol NH4^+. Found the mol of NH4Cl to be 0.08.
Converted 0.10mol NH4^+ to ml : (0.10 mol NH4^+/1.0 mol NH4^+)(1000 ml)= 100.0ml
Converted 0.08 mol NH4Cl to ml: (0.08 mol NH4Cl/1.00 mol NH4Cl)(1000 ml)= 80.0 ml

My final answer is 100.0 ml of 0.100M NH4^+, 80.0 ml of 1.00 M NH4Cl, and approx 820 ml water to get 1.0 L.

Answer 3

How can you have a buffer with NH4^+ and NH4Cl. You have no base. Both NH4^+ by itself and NH4^+ from NH4Cl are acidic.
A buffer usually consists of a weak base and a salt of the weak base (NH3 and NH4Cl) or a weak acid and a salt of the weak acid (acetic acid for example and a salt of the acid such as sodium acetate).

Answer 4

For whatever it's worth, here is the calculation to determine the pH of the solution you have made.
mols NH4^+ = M x L = 0.1 M x 0.1 L = 0.01
mols NH4^+ from NH4Cl = 1.0 M x 0.08 L = 0.08 mols.
Total mols NH4^+ = 0.01 + 0.08 = 0.09 and that in 1 L is 0.09M. The pH of that solution is determined by the hydrolysis of the NH4^+ in the water.

.........NH4^+ + H2O ==> NH3 + H3O^+
I......0.09 M.............0......0
C......-x.................x.......x
E.....0.09-x..............x.......x

Ka for NH4^+ = (Kw/Kb for NH3) =
(x)(x)/(0.09-x) and solve for x.
I found it to be 7.07E-6 and pH = 5.15 which isn't even close to 9.15. In fact, most NH4Cl solutions are about pH = 5.They are ALWAYS < 7.0