Applications of Aqueous Equilibria Chapter 15. Titration Curves & Indicators Sections 4-5
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1 Applications of Aqueous Equilibria Chapter 15 Titration Curves & Indicators Sections 45
2 Strong Acid vs. Strong Base Titration Titrate 50.0 ml of M HNO 3 with M NaOH What is the ph when no NaOH has been added? 0.200M = [H + ] ph = log =.699 What is the ph when 10.0 ml of 0.10 M NaOH has been added? Species in solution: H +, NO 3, Na +, OH, H 2 O H + + OH H 2 O 50.0mL 10.0mL 0.200M 0.100M LIMITING HAVE 10.0mmol 1.00mmol NEED 1.00mmol 1.00mmol XS 9.00mmol 0mmol 9.00mmol H mL mL SPECTATORS = 0.15 M H + ph = log 0.15 = 0.82 What would it look like at the equivalence point? STOICHIOMETRY!.. BEFORE THE EQUIVALENCE POINT..
3 Strong Acid vs. Strong Base Titration Titrate 50.0 ml of M HNO 3 with M NaOH What is the ph when ml of 0.10 M NaOH has been added? H + + OH H 2 O 50.0mL 150.0mL 0.200M 0.100M LIMITING HAVE 10.0mmol 15.0mmol NEED 10.0mmol 10.0mmol XS 0 mmol 5.0mmol 5.0mmol OH = M OH 50.0mL mL [H + ] = 1.0 x /0.025 = 4.0 x ph = log 4.0 x = AFTER THE EQUIVALENCE POINT..
4 Strong Acid vs Strong Base Titration Curves Small volumes make large ph changes Figure 15.1 & 15.2 page 699
5 Weak Acid vs. Strong Base Titration Similar to buffer calculations Even though the acid is weak it reacts completely with the OH added Titrate 50.0 ml of M HC 2 H 3 O 2 (K a = 1.8 x 10 5 ) with 0.10 M NaOH What is the ph when no NaOH has been added? HC 2 H 3 O 2 H + + C 2 H 3 O x 10 5 = 0.10M 0 0 x +x +x 0.10x x x x ph = log 1.8 x 10 6 = 2.87 x = 1.8 x 10 6 = [H + ] EQUILIBRIUM!.. BEFORE THE EQUIVALENCE POINT..
6 Weak Acid vs. Strong Base Titration Titrate 50.0 ml of M HC 2 H 3 O 2 (K a = 1.8 x 10 5 ) with 0.10 M NaOH What is the ph when 10.0 ml of 0.10 M NaOH has been added? Species in solution before reaction: HC 2 H 3 O 2, Na +, OH, H 2 O HAVE NEED XS HC 2 H 3 O 2 + OH C 2 H 3 O 2 + H 2 O 50.0mL 10.0mL 0.100M 0.100M 1.00mmol 4.00mmol 4.00mmol HC 2 H 3 O mL mL 1.00mmol C 2 H 3 O mL mL 1.00mmol 1.00mmol 0mmol LIMITING = M HC 2 H 3 O 2 = M C 2 H 3 O 2 ACID 1.00mmol BASE STOICHIOMETRY! Use these for an equilibrium problem to find [H + ].. BEFORE THE EQUIVALENCE POINT..
7 Weak Acid vs. Strong Base Titration Titrate 50.0 ml of M HC 2 H 3 O 2 (K a = 1.8 x 10 5 ) with 0.10 M NaOH What is the ph when 10.0 ml of 0.10 M NaOH has been added? Use M HC 2 H 3 O 2 and M C 2 H 3 O 2 HC 2 H 3 O 2 H + + C 2 H 3 O M M x +x +x 0.067x x x 1.8 x 10 5 = 0.017x ph = log 7.1 x 10 5 = 4.15 x = 7.1 x 10 5 = [H + ] Halfway to the equivalence point [HC 2 H 3 O 2 ] 0 = [C 2 H 3 O 2 ] 0 so K a = [H + ] and pk a = ph EQUILIBRIUM!.. BEFORE THE EQUIVALENCE POINT..
8 Weak Acid vs. Strong Base Titration Titrate 50.0 ml of M HC 2 H 3 O 2 (K a = 1.8 x 10 5 ) with 0.10 M NaOH What is the ph when 50.0 ml of 0.10 M NaOH has been added? Species in solution before reaction: HC 2 H 3 O 2, Na +, OH, H 2 O HAVE NEED XS HC 2 H 3 O 2 + OH C 2 H 3 O 2 + H 2 O 50.0mL 50.0mL 0.100M 0.100M 0mmol C 2 H 3 O mL mL 0mmol ACID BASE STOICHIOMETRY! = M C 2 H 3 O 2 This is a base that will force water to be an acid!!.. AT THE EQUIVALENCE POINT..
9 Weak Acid vs. Strong Base Titration Titrate 50.0 ml of M HC 2 H 3 O 2 (K a = 1.8 x 10 5 ) with 0.10 M NaOH What is the ph when 50.0 ml of 0.10 M NaOH has been added? Use M C 2 H 3 O 2 that reacts with water C 2 H 3 O 2 + H 2 O HC 2 H 3 O 2 + OH 0.050M 0 0 x +x +x 0.050x x x Find K b from K w =K a x K b K b = 5.6 x = x [H + ] = 1.0 x / 5.3 x 10 6 = 1.9 x 10 9 ph = log 1.9 x 10 9 = 8.72 x = 5.3 x 10 6 = [OH ] weak acid & strong base equivalence point >7 because the anion left in solution is a (conjugate) base EQUILIBRIUM!.. AT THE EQUIVALENCE POINT..
10 Weak Acid vs. Strong Base Titration Titrate 50.0 ml of M HC 2 H 3 O 2 (K a = 1.8 x 10 5 ) with 0.10 M NaOH What is the ph when 60.0 ml of 0.10 M NaOH has been added? HAVE NEED XS HC 2 H 3 O 2 + OH C 2 H 3 O 2 + H 2 O 60.0mL 50.0mL 0.100M 0.100M LIMITING 0mmol 1.00mmol OH 60.0mL mL 6.00mmol 1.00mmol = 9.10 x 10 3 M OH weak base strong base USE ONLY THE STRONG. [H + ] = 1.0 x / 9.10 x 10 3 = 1.1 x M ph= STOICHIOMETRY!.. AFTER THE EQUIVALENCE POINT.. Try Sample Exercise 15.9 on page 705 or #55, 57, 58
11 Weak Acid vs Strong Base Titration Curves Weak acid increases in ph more rapidly before the equivalence point Levels off near the halfway point caused by buffering effects because [HA] = [A ]!!! Equivalence point is determined by stoichiometry not ph. SA & SB =7 WA & SB >7 SA & WB <7 Figure 15.3 page 704
12 Comparing Weak Acids The strength of a weak acid has no bearing on the amount of acid needed at the equivalence point. Equivalence point is determined by stoichiometry not ph! The strength of a weak acid does affect the ph of the equivalence point. The weaker the acid the higher the ph at the equivalence point. It also affects the shape of the curve. Figure 15.4 page 707
13 Calculating Ka Sample Exercise page 707 Given 100.0mL H 2 O, 2.0 mmol of HA with 20.0mL of M NaOH and a ph of HA + OH A + H 2 O 100.0mL 20.0mL 0.050M LIMITING HAVE 2.00mmol 1.00mmol 1.00mmol /120.0mL = M = [A ] 0 NEED 1.00mmol XS 1.00mmol /120.0mL = M = [HA] 0 ph = 6.00 [H + ] eq = 1.0 x 10 6 M HA H + + A x +x +x x 1.0x x approximate K a = [1.0x10 6 ] [ ] K a = 1.0x This was halfway to the equivalence point. Try #63
14 The reaction of NH 3 and HCl Weak Base vs Strong Acid Species in reaction NH 3, H +, Cl, H 2 O NH 3 + H + NH 4 + base acid from HCl conjugate acid goes to completion use stoichiometry NH H 2 0 NH 3 + H 3 O + acid water (base) conjugate base hydronium does not go to completion use equilibrium calculations The ph is less than 7 because the weak base produces a conjugate acid.
15 Endpoint when the indicator in a titration changes color not necessarily the same as equivalence point Indicators complex molecules that are weak acids, H(In) and their conjugate base, In, in equilibrium the conjugate base, In, is a different color than the H(In) HIn H + + In K a = 1.0 x 10 8 K a = [H+ ] [In ] [HIn] OR AcidBase Indicators K a = [H + ] [In ] [HIn] If an acid is added [H + ] = 1.0 x 10 1 M then 1.0 x [In ] = = There is more [HIn] 1.0 x x 10 7 [HIn] so you will see its color [In ] How much In must be present to see a change? [HIn] = 1 10
16 AcidBase Indicators Given K a, which ph will cause a change? Use [In ] / [HIn] = 1/10 K a = [H+ ] [In ] [HIn] 1.0x 10 7 = [H+ ] 1 10 [In Or use the HendersonHasselbalch ph = pka ] + log [HIn] [1] log = 1 ph = pk a 1 [10] Flipping [In ]/[HIn] to [HIn]/[In ] gives ph = pk a + 1 So now you have a range. The range for indicators is pk a ± 1. If K a = 1 x 10 7 then [HIn] [In ] yellow blue
17 When choosing an indicator make the endpoint and equivalence point as close as possible. AcidBase Indicators Figure 15.9 page 716
18 Figure 15.8 AcidBase Indicators
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