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...
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.
There weren't any volumes. That's all that was gven.
But it's correct, so thank you so much. :-)
To find the OH- concentration of a solution containing both sodium hydroxide (NaOH) and ammonium chloride (NH4Cl), you need to consider the reaction that occurs between NaOH and water (H2O) and the reaction between NH4Cl and water.
The balanced chemical equation for the reaction between NaOH and water is:
NaOH + H2O -> Na+ + OH-
The balanced chemical equation for the reaction between NH4Cl and water is:
NH4Cl + H2O -> NH4+ + Cl-
Now let's write down the initial concentration of each species:
NaOH: [NaOH] = 0.198 M
NH4Cl: [NH4Cl] = 0.262 M
From the reaction between NaOH and water, we know that for every 1 mol of NaOH that dissolves, 1 mol of OH- is produced. Therefore, the initial concentration of OH- is also 0.198 M.
From the reaction between NH4Cl and water, we know that for every 1 mol of NH4Cl that dissolves, 1 mol of NH4+ is produced. However, NH4+ is not OH-, so it does not contribute to the OH- concentration.
Hence, the OH- concentration of the solution is 0.198 M. It seems like you obtained the correct answer in your initial calculation. If you received a different answer, please double-check your calculations and make sure all values were entered correctly.