When 5 mL of 0.10 M HCl are added to 10 mL of 0.10 M NH3, what is the approximate pOH?
Kb for NH3 is 1.8 x 10^-5.
I get pOH between 4 and 5. Is that right?
I don't think so.
NH3 + NaOH ==> NH4Cl
Initial:
NH3 = 10 mL x 0.1M = 1 mmol
HCl = 5 mL x 0.1 M = 0.5 mmol
This will form 0.5 mmole NH4Cl, all of the HCl will be used up and it will leave 0.5 mmole NH3 unreacted. What do we have? We have weak base (NH3) and its salt (NH4Cl) which is a buffer.
pH = pKa + log p(base)/(acid)] =
pH = 9.255 + ....... you can finish.