Help with pH, using Kb and buffers?

Question

To prepare exactly 650 mL of a buffer solution which has a pH of 9.36 and a final ammonium chloride salt concentration [NH4+] = 0.180 M, one would need to use <> of NH4Cl(s)
and <> of 3.00 M sodium hydroxide solution, plus enough water to make up the total volume.

Given that Kb = 1.8×10-5 for ammonia:



I've tried using the equation pH=pKa + log (mol base/mol acid) and i get 6.26 g NH4Cl but apparantly thats wrong...any suggestions?

Answer 1

Ka (NH4+) = 5.55 * 10^-10; pKa = 9.26

pH = pKa + log [NH3]/[NH4+]
9.36 = 9.26 + log [NH3]/[0.180]
0.10 = log [NH3]/[0.180]
1.26 = [NH3]/0.180
[NH3] = 0.227 M

So that means that your total concentration of "ammonia" species in solution will be 0.227 + 0.180 or 0.407 M, all of which will have to come from the NH4Cl.

0.407 M * 0.650 L = 0.265 moles

0.265 moles * 53.5 g/mol = 14.18 grams of NH4Cl.

To get the 0.227 M NH3 portion, you'll have to add enough NaOH to react with the NH4Cl to get there.

0.227 M * 0.650 L = 0.148 moles NH3 = 0.148 moles OH- added

0.148 moles OH-/ 3.00 M = 0.0493 L = 49.3 mL

So, disssolve 14.18 grams of NH4Cl in roughly 500 mL of water, add 49.3 mL of 3 M NaOH, and then dilute up to 650 mL.