Hi all, this is the question: Question data: Reaction at STP: NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH3] = 0.0180 M [OH-] = 5.6E-4 M The answers I tried are in blue. a. Write Keq expression for reaction above. Keq = ([NH4][OH-])/([NH3][H2O]) b. Find pH of 0.0180 M NH3. Do I use the Henderson Hasselbach equation for this? pH = pKa + log(Base/Acid)? = c. Find Kb for NH3. Kb = Kw/Ka right? How do I find Ka then? d. Find % ionization of NH3 in 0.0180 M NH3. I guess this is [NH4+]/0.0180*100% right? How do I find NH4? e. 20.0 mL of 0.0180 M NH3 was titrated beyond equivalence point using 0.0120 M HCl. a. Find volume of 0.0120 M HCl added to reach equivalence point. So this happens when moles NH3 = mol HCl. So, 0.0180 M NH3 * .020 L = .00036 mol NH3. We would then need .00036 mol HCl, which could be obtained from doing .00036/.0120 = .030 L or 30.0 mL, yes? b. Find pH of total solution after 15.0 mL of 0.0120 M HCl was added. .015 L * .012 M HCl = .00018 mol HCl reacting with .00036 mol NH3 in 0.035 total L solution...do I use Henderson-Hasselbach for this too? c. Find pH of total solution after 40.0 mL of 0.0120 M HCl was added. Since you have .04 L * .012 M HCl = .00048 mol HCl reacting with .00036 mol NH3, you have .00012 mol worth of excess H+, divided by total solution of 0.06 L = .002 M, take -log(.002) = 2.70 = pH? Yes? Help please!