Asked by Taylor
A solution is prepared by adding 300 mL of 0.500 M NH3 and 100 mL of 0.500 M HCl. Assuming that the volumes are additive, what is the pH of the resulting mixture? Kb for ammonia is 1.8 x10 -5
Answers
Answered by
DrBob222
You've added a base to an acid; which one is in excess. The equation is
NH3 + HCl ==> NH4Cl
mmoles NH3 = 300 mL x 0.500 M = 150
mmoles HCl = 100 mL x 0.500 M = 50
It should be obvious that the reaction produces 50 mmoles NH4Cl, all of the HCl is used and there are 100 mmoles NH3 remaining unreacted. This is a solution of a weak base (NH3) and its salt (NH4Cl) and that is a buffered solution. So use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid))]
Post your work if you get stuck.
NH3 + HCl ==> NH4Cl
mmoles NH3 = 300 mL x 0.500 M = 150
mmoles HCl = 100 mL x 0.500 M = 50
It should be obvious that the reaction produces 50 mmoles NH4Cl, all of the HCl is used and there are 100 mmoles NH3 remaining unreacted. This is a solution of a weak base (NH3) and its salt (NH4Cl) and that is a buffered solution. So use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid))]
Post your work if you get stuck.
Answered by
stephen
8.9
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