Ch18_Rxn - 1 Acid-Base Reactions 1 2 Stomach Acidity...

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Unformatted text preview: 1 Acid-Base Reactions 1 2 Stomach Acidity & Acid-Base Reactions Chapter 18 Acid-Base Reactions 3 What is relative pH before & after reaction? • Strong acid + strong base HCl + NaOH ----> • Strong acid + weak base HCl + NH 3 ---> • Weak acid + strong base HOAc + NaOH ---> • Weak acid + weak base HOAc + NH3 ---> 4 Acid-Base Reaction • Add 0.010 mol NaOH to 100. mL of Add of 0.10 M HOAc. 0.10 HOAc • What is pH before reaction? • What is Knet for reaction? What • What is pH when reaction is done? 5 pH before Acid-Base Reaction Add 0.010 mol NaOH to 100. mL of 0.10 M HOAc. (a) What is pH before rxn? (b) What is Knet for reaction? (c) pH when reaction is complete? HOAc + H2O HOAc K1 = 1.8 x ¸ OAc- + H3O+ 10 -5 [H3O+] = [Ka • 0.10]1/2 0.10] = 1.3 x 10-3 1.3 pH before adding NaOH = 2.87 pH NaOH Page 1 6 K for Acid-Base Reaction Add 0.010 mol NaOH to 100. mL of 0.10 M HOAc. (a) What is pH before rxn? (b) What is Knet for reaction? (c) pH when reaction is complete? HOAc + H2O ¸ OAc- + H3O+ K1 = 1.8 x 10 -5 HOAc OH- + H 3O+ ¸ 2 H2 O K2 = 1.0 x 10 14 HOAc + OH- ¸ OAc- + H2O OAc K1 • K2 = 1.8 x 10 9 Reaction proceeds from completely left to right! 2 7 pH after Acid-Base Reaction Add 0.010 mol NaOH to 100. mL of 0.10 M HOAc. What is Knet for reaction? What is pH before rxn? When reaction is complete? HOAc + OH- ¸ OAc- + H2O OAc 0.010 mol NaOH ? Mol HOAc (0.100 L)(0.10 M) = 0.010 mol HOAc Quantity of OAc = 0.010 mol [OAc-] = 0.010 mol/0.1 L = 0.10 M 8 pH after Acid-Base Reaction Add 0.010 mol NaOH to 100. mL of 0.10 M HOAc. What is Knet for reaction? What is pH when reaction is complete? Reaction: HOAc + OHOH ¸ OAc- + H2O OAc After reaction solution contains only OAcAfter OAc Therefore, solution is __________ OAc- + H2O ¸ HOAc + OHOH Calculate pH of 0.10 OAc Kb = 5.6 x 10 -10 Page 2 9 pH after Acid-Base Reaction Add 0.010 mol NaOH to 100. mL of 0.10 M HOAc. What is Knet for reaction? What is pH when reaction is complete? Calculate pH of 0.10 OAcKb = 5.6 x 10-10 OAc- + H2O ¸ HOAc + OH- 0.10 M - x x [OH-] = [Kb • 0.10]1/2 = 7.5 x 7.5 10-6 x pOH = 5.13 pH = 8.87 ...
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