Question

'A buffer is prepared by mixing a 100.omL of a 0.100 M NH3 solution with a 0.200M solution of NH4CL solution and making the total volume up to 1.000L of water. What is the volume of NH4Cl solution required to achieve a buffer at ph=9.5? Ka of NH4=5.6x10^-10. The textbook says the answer is 28.1mL.

Answer 1

mols NH3 = M x L = ?
mols NH4Cl = M x L = ?
pH = pKa + log(b/a)
9.5-9.25 + log (10/0./2x)
Solve for x and convert to mL.

Answer 2

thank you that definitely helps. i just have one question, line 5:
9.5-9.25 + log (10/0./2x) is that supposed to read
9.5-9.25 + log (10/0.1/2x)? just a bit confused.

Answer 3

thank you that definitely helps. i just have one question, line 5:
9.5-9.25 + log (10/0./2x) is that supposed to read
9.5-9.25 + log (10/0.1/2x)? just a bit confused., and also why is it divided by 2x? i have an exam for this on friday morning.

Answer 4

Sorry about that. I made several typos.
First mols = M x L (which is right BUT I used millimoles in the equation that follows).
100 mL x 0.100M = 10 millimols NH3.
x mL x 0.2M = 0.2x millimols NH4^+. Using millimoles, it should read
pH = pKa + log(base/acid)
9.5 = 9.25 + log(10/0.2x)
1.778 = 10/0.2x
0.3557x = 10
x = 10/0.3557 = 28.1 mL. Thank you for bringing this to my attention. Usually I check these fairly closely; I was in a hurry to leave today and didn't. I hope this clears things for you.