1. PROBLEM: You titrate 25.0 mL of 0.10 M NH 3 with 0.10 M HCl. (a) What is the pH of NH 3 solution before the titration begins? (b) What is the pH at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) What indicator in Figure 18.10 could be used to detect the equivalence point? (e) Calculate the pH of the solution after adding 5.00, 15.0, 20.0, 22.0, and 30.0 mL of the acid? Combine this information with that in parts (a)-(c) and plot the titration curve (a) x = [OH – ] = 0.0013 MpOH = –log[OH – ] = 2.87 pH = 14.00 – pOH = 11.13

2. (b) NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O(l) Total volume = 0.0250 + 0.0250 = 0.0500 L [NH 4 + ] == 0.050 M NH 4 + (aq) + H 2 O(l) H 3 O + (aq) + NH 3 (aq) x = [H 3 O + ] = 5.3  10 –6 MpH = –log[H 3 O + ] = 5.28 (c)At titration midpoint, [NH 3 ] = [NH 4 + ] and pH = pK a = –log(5.6  10 –10 ) = 9.25 (d)Methyl red would detect the equivalence point.

3. (e)mL HClmol H 3 O + mol NH 4 + mol NH 3 addedaddedproducedremaining pH 5.000.000500.00050 0.00209.85 15.000.00150.0015 0.00109.08 20.000.00200.0020 0.00058.65 22.000.00220.0022 0.00038.39 30.000.00302.04 When 30.00 mL HCl is added, pH depends only on the excess H 3 O + [H 3 O + ] = (0.0030 mol – 0.0025 mol)/0.0550 L = 0.009 M