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...
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Transcript of 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...
PROBLEM: You titrate 25.0 mL of 0.10 M NH3 with 0.10 M HCl. (a) What is the pH of NH3 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 M pOH = –log[OH–]
= 2.87
pH = 14.00 – pOH = 11.13
(b) NH3(aq) + H3O+(aq) NH⟺ 4+(aq) + H2O(l)
Total volume = 0.0250 + 0.0250 = 0.0500 L
[NH4+] =
= 0.050 M
NH4+(aq) + H2O(l) ⟺ H3O+(aq) + NH3(aq)
x = [H3O+] = 5.3 10–6 M pH = –
log[H3O+] = 5.28
(c) At titration midpoint, [NH3] = [NH4+] and pH = pKa = –log(5.6 10–10) =
9.25
(d) Methyl red would detect the equivalence point.
(e) mL HCl mol H3O+ mol NH4+ mol
NH3
added added producedremaining pH
5.00 0.00050 0.00050 0.0020 9.85
15.00 0.0015 0.0015 0.0010 9.08
20.00 0.0020 0.0020 0.0005 8.65
22.00 0.0022 0.0022 0.0003 8.39
30.00 0.00302.04
When 30.00 mL HCl is added, pH depends only on the excess H3O+
[H3O+] = (0.0030 mol – 0.0025 mol)/0.0550
L
= 0.009 M