Asked by bert
Consider the titration of 50 mL of 0.15 M NH3 with 0.1 M HCl. What is the pH at the equivalence point of the titration? (Kb= 1.8X 10-5 Ka= 5.6X10-10)
Answers
Answered by
DrBob222
The salt present at the equivalence point is NH4Cl and there are 0.05 L x 0.15 M. That hydrolyzes at the equivalence point and that determines the pH.
NH4^+ + HOH ==> NH3^+ + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+)
Plub in (NH4^+) from mols NH4Cl/liters (don't forget in liters to add the volume of NH3 initially (0.050 L) to the volume of HCl used in the titration.) and solve for H3O^+.
pH = - log (H^+). Post your work if you get stuck.
NH4^+ + HOH ==> NH3^+ + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+)
Plub in (NH4^+) from mols NH4Cl/liters (don't forget in liters to add the volume of NH3 initially (0.050 L) to the volume of HCl used in the titration.) and solve for H3O^+.
pH = - log (H^+). Post your work if you get stuck.
Answered by
Anonymous
IM LOST ON THIS ONE
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.