Find the OH- concentration of a solution that is 0.198 M NaOH and 0.262M NH4Cl.
I did the ice table,using OH- from NaOH for an initial, and NH4 from the NH4Cl molecule, and ended with .20M. But that's not right apparently...
3 answers
I redid it, and im pretty sure i did the ice table backwards. I redid it with oh- on the same side as nh4. Only issue is now i have 0 oh left in solution...
I would have liked to see some volumes. As it is I assumed a volume of 1L so
..........OH^- + NH4^+ ==> NH3 + H2O
initial..0.198..0.262......0.......0
change..-0.198..-0.198...+0.198..+0.198
equil......0......0.064....198....198
So what do you have. I see a weak base (NH3) and its salt(NH4^+) which is a buffered solution. So I substitute into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa + log[(base)/(acid)]
pH = 9.25 + log (0.198)/(0.064)
pH = about 9.74 and convert to pOH and OH^-.
You should look up pKa in your text tables and recalculate everything. That might change the answer slightly; especially if you are checking against an on-line data base.
..........OH^- + NH4^+ ==> NH3 + H2O
initial..0.198..0.262......0.......0
change..-0.198..-0.198...+0.198..+0.198
equil......0......0.064....198....198
So what do you have. I see a weak base (NH3) and its salt(NH4^+) which is a buffered solution. So I substitute into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa + log[(base)/(acid)]
pH = 9.25 + log (0.198)/(0.064)
pH = about 9.74 and convert to pOH and OH^-.
You should look up pKa in your text tables and recalculate everything. That might change the answer slightly; especially if you are checking against an on-line data base.
There weren't any volumes. That's all that was gven.
But it's correct, so thank you so much. :-)
But it's correct, so thank you so much. :-)