APPLICATIONS OF AQUEOUS EQUILIBRIA. Chapter 15

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APPLICATIONS OF AQUEOUS EQUILIBRIA Chapter 15

Acid-Base Equilibria Common Ion Effect in Acids and Bases Buffer SoluFons for Controlling ph Buffer Capacity ph-titrafon Curves Acid-Base TitraFon Indicators

Common Ion Effect ShiL in the equilibrium posifon due to the addifon of an ion already involved in the equilibrium process. An applicafon of Le Châtelier s principle.

A Common Ion Effect Consider the following equilibrium: HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) Adding NaC 2 H 3 O 2 to the solufon will shil the equilibrium to the lel because [C 2 H 3 O 2 - ] increases; C 2 H 3 O 2 - is part of the equilibrium system. This equilibrium shil causes [H 3 O + ] to decrease and raise the ph of the solufon. SoluFons containing a mixture of HC 2 H 3 O 2 and NaC 2 H 3 O 2 are less acidic than those solufons of HC 2 H 3 O 2 alone, and they are less basic than those of NaC 2 H 3 O 2 alone.

ph of weak acid and the Effect of Common Ions Consider the following solufons: Calculate the ph of 1.00 M HC 2 H 3 O 2 solufon. What is the ph of a solufon that contains 1.00 M HC 2 H 3 O 2 and 0.50 M NaC 2 H 3 O 2. SoluFon-1: Equilibrium: HC 2 H 3 O 2 (aq) H + - (aq) + C 2 H 3 O 2 (aq) IniFal [ ], M 1.00 0.00 0.00 Change, D[ ], M -x +x +x Equilm. [ ], M (1.00 x) x x

ph of AceFc Acid by itself. SoluFon-1: K a = [H + 2 3O ][CH 3CO2 ] x -5 = = 1.8 x 10 [CH 3 COOH] (1.00 - x) By approximafon, x = -5 ( 1.00 x 1.8 x 10 ) = 4.2 x 10-3 [H 3 O + ] = x = 4.2 x 10-3 M, è ph = 2.37

AceFc Acid-Acetate Equilibrium SoluFon-2: Equilibrium: HC 2 H 3 O 2 (aq) H + - (aq) + C 2 H 3 O 2 (aq) IniFal [ ], M 1.00 0.00 0.50 Change, D[ ], M -x +x +x Equilm. [ ], M (1.00 x) x (0.50 + x)

ph of AceFc Acid + Sodium Acetate SoluFon-2: K a + 3O ][CH 3CO2 ] ( x)(0.50 + x) -5 = = 1.8 x 10 [H = [CH 3 COOH] (1.00 - x) By approximafon, x = (1.00/0.50)(1.8 x 10-5 ) = 3.6 x 10-6 M [H + ] = x = 3.6 x 10-6 M, è ph = 4.44 Solu.on containing HC 2 H 3 O 2 and NaC 2 H 3 O 2 is less acidic than one containing only HC 2 H 3 O 2 at the same concentra.on.

Buffered SoluFons Buffered SoluFon resists a change in ph. They are weak acids or bases containing a common ion. ALer addifon of strong acid or base, deal with stoichiometry first, then the equilibrium.

Solving Problems with Buffered SoluFons

How does buffering work?

How does buffering work?

How does buffering work?

Henderson Hasselbalch EquaFon HA(aq) H + (aq) + A - (aq); K a = + [H ][A [HA] ] ph = pka + log([a ]/[HA]) For a parfcular buffer system, solufons with the same [A ]/[HA] rafo have same ph.

ph of Buffer SoluFon What is the ph of a buffer solufon that is 0.45 M acefc acid (HC 2 H 3 O 2 ) and 0.85 M sodium acetate (NaC 2 H 3 O 2 )? The K a for acefc acid is 1.8 10 5. SoluFon: ph = pk a + log([c 2 H 3 O 2 -]/[HC 2 H 3 O 2 ] ph = -log(1.8 10 5 ) + log(0.85/0.45) ph = 4.74 + 0.28 = 5.02

CharacterisFcs of Buffer SoluFons Contain weak acids or weak bases and their corresponding conjugate partners (common ions). Resist changes in ph. Buffering capacity depends on concentrafons of weak acid or weak base and their common ions. EffecFve ph buffering range ~ pk a ± 1

CharacterisFcs of Buffer SoluFons 1. Buffers contain relafvely large amounts of the weak acids (HA) and their conjugate base,( A ) (or weak bases and their conjugate acids) 2. Buffer ph is determined by the pk a of the acid HA and the molar rafo of the conjugate base to acid: [A ]/[HA].

CharacterisFcs of Buffer SoluFons 3. Buffer ph changes very limle because the rafo [A ]/[HA] changes very limle when a small amount of strong acid or strong base is added. 4. [H 3 O + ] in buffer solufons remains more or less constant: Most of H + from strong acid is absorbed by the conjugate base ; A most of OH added from strong base reacts with acid HA in the buffer to yield A and H 2 O.

Buffering Capacity How much H 3 O + or OH - the buffer can absorb without significantly changing its ph.. A Depends on the concentrafons of HA and High [HA] and [ A ] lead to large buffering capacity. ;[ A ] OpFmal buffering occurs when [HA] = RaFo [A ] / [HA] ~ 1 strong resist to change when either H 3 O + or OH is added.

Some Common Buffers Buffers pk a ph Range HCHO 2 NaCHO 2 3.74 2.74 4.74 CH 3 CO 2 H NaCH 3 CO 2 4.74 3.74 5.74 KH 2 PO 4 K 2 HPO 4 7.21 6.20 8.20 CO 2 /H 2 O NaHCO 3 6.37 5.40 7.40 NH 4 Cl NH 3 9.25 8.25 10.25

Choosing a Buffer System The weak acid in buffer has pk a close to target ph. For example, KH 2 PO 4 and K 2 HPO 4 may be used to buffer at ph ~ 7.5 (H 2 PO 4 has pk a = 7.20) Phosphate buffer is most effecfve in the ph range 6.20 8.20; it has the highest buffering capacity at about ph = 7.20.

Making Buffer SoluFon A phosphate buffer with ph = 7.40 is prepared using KH 2 PO 4 and K 2 HPO 4. (a) What is the molar rafo of [HPO 4 2- ] to [H 2 PO 4 - ] in the buffered solufon? (b) If [H 2 PO 4 - ] = 0.20 M, what is [HPO 4 2- ]? (c) How many grams of KH 2 PO 4 and K 2 HPO 4, respecfvely, are needed to make 500. ml of this solufon? (H 2 PO 4 - has K a = 6.2 x 10-8 )

Solutions to Buffer example #2 (a) Use Henderson-Hasselbalch equation: ph = pk a + log([hpo 2-4 ]/[H 2 PO - 4 ]) 7.40 = 7.21 + log([hpo 2-4 ]/[H 2 PO - 4 ]) log([hpo 2-4 ]/[H 2 PO - 4 ]) = 7.40 7.21 = 0.19 [HPO 2-4 ]/[H 2 PO - 4 ] = 10 0.19 = 1.55 (b) If [H 2 PO 4 - ] = 0.20 M, [HPO 4 2- ] = 1.55 x 0.20 M = 0.31 M

Solutions to Buffer example #2 (c) Moles of KH 2 PO 4 needed = 500. ml x (1 L/1000 ml) x 0.20 mol/l = 0.10 mole Moles of K 2 HPO 4 needed = 500. ml x (1 L/1000 ml) x 0.31 mol/l = 0.155 mole Grams of KH 2 PO 4 needed = 0.10 mol x (136.086 g/mol) = 14 g Grams of K 2 HPO 4 needed = 0.155 mol x (174.178 g/mol) = 27 g

Buffer Example An acetate buffer solufon is prepared by mixing 35.0 ml of 1.0 M acefc acid and 65.0 ml of 1.0 M sodium acetate. (a) What is the ph of this solufon? (b) If 0.010 mole of HCl is added to this solufon without altering its volume, what will be the ph of the resulfng solufon? (K a = 1.8 x 10-5 )

Buffer Example The K a values of some acids and base are given below: 1. AceFc acid, CH 3 CO 2 H, K a = 1.8 x 10-5 ; 2. Dihydrogen phosphate, H 2 PO 4, K a = 6.2 x 10-8 ; 3. Ammonia, NH 3, K b = 1.8 x 10-5; 4. Hydrogen carbonate, HCO 3, K b = 2.3 x 10-8. What solufons are used to make buffers with the following ph s? (i) ph = 7.00; (ii) ph = 4.50; (iii) ph = 9.00 (iv) ph = 9.50; (v) ph = 5.00

Buffer Example How many milliliters of each solufon of 0.50 M KH 2 PO 4 and 0.50 M K 2 HPO 4 are needed to make 100.0 ml solufon of phosphate buffer with ph = 7.50? What are the final concentrafons of K +, H 2 PO 4 - and HPO 4 2-, in the buffer solufon? (for H 2 PO 4 -, K a = 6.2 x 10-8 )

TITRATION CURVES

TitraFon and ph Curves Ploqng the ph of the solufon being analyzed as a funcfon of the amount of Ftrant added. From ph-ftrafon curve determine the equivalence point when enough Ftrant has been added to react exactly with the substance in solufon being Ftrated.

The ph Curve for the TitraFon of 50.0 ml of 0.200 M HNO 3 with 0.100 M NaOH

The ph Curve for the TitraFon of 100.0 ml of 0.50 M NaOH with 1.0 M HCI

The ph Curve for the TitraFon of 50.0 ml of 0.100 M HC 2 H 3 O 2 with 0.100 M NaOH

The ph Curves for the TitraFons of 50.0-mL Samples of 0.10 M Acids with Various K a Values with 0.10 M NaOH

The ph Curve for the TitraFon of 100.0mL of 0.050 M NH 3 with 0.10 M HCl

Acid-Base Indicators An indicator is a substance added to acid or base solufon to marks the end point of a FtraFon by the change of its color. For example, phenolphthalein changes from colorless to pink at the end point when an acid is Ftrated with a base. The end point of a FtraFon should correspond to the equivalence points of the acid-base reacfon.

The Acid and Base Forms of the Indicator Phenolphthalein

The Methyl Orange Indicator is Yellow in Basic SoluFon and Red in Acidic SoluFon

Choosing Indicators 1. The ph range for color changes should occur within the sharp verfcal rise (or drop) in the ph-ftrafon curves. 2. An indicator changes color at ph = pk a ± 1, where pk a is that of the indicator used.

ph Ranges for Indicators

CalculaFng the ph of solufon during FtraFon Strong Acid-Strong Base TitraFon; 1. Net reacfon: H 3 O + (aq) + OH - (aq) 2H 2 O 2. Determine the limifng reactant and calculate the final concentrafon of H 3 O + or OH - that is in excess. 3. Calculate ph using concentrafon of excess H 3 O + or OH -

TitraFon Problem: A 20.0 ml aliquot of 0.100 M HCl is Ftrated with 0.100 M NaOH solufon. What is the ph of the resulfng solufon aler 0.0 ml, 15.0 ml, and 30.0 ml of NaOH has been added? ReacFon: H 3 O + (aq) + OH - (aq) 2H 2 O

TitraFon Problem: Weak Acid-Strong Base A 20.0 ml aliquot of 0.100 M HNO 2 is Ftrated with 0.100 M NaOH. K a of HNO 2 = 4.0 x 10-4 (a) What is the ph of the solufon before FtraFon? simplified ICE calcula.on (b) What is the ph of the solufon aler 15.0 ml of NaOH has been added? stoichiometry & ICE (c) What is the ph of the solufon at equivalent point (aler 20.0 ml of 0.100 M NaOH is added)? ICE & K b

TitraFon Problem: Strong Acid-Weak Base A 20.0 ml aliquot of 0.100 M NH 3 is Ftrated with 0.100 M HCl. K b of NH 3 = 1.8 x 10-5 (a) What is the ph of the solufon before FtraFon? K b calculate [OH - ] (b) What is the ph of the solufon aler 10.0 ml of HCl has been added? ICE & HH (c) What is the ph of the solufon at equivalent point (aler 20.0 ml of 0.100 M HCl is added)? K a = K w /K b = [H + ] & ICE

TitraFon Exercises 1. 25.0 ml of 0.10 M HCl is Ftrated with 0.10 M NaOH solufon. (a) What is the ph of the acid before NaOH solufon is added? (b) What is the ph aler 15.0 ml of NaOH solufon is added? (c) What is the ph of the solufon aler 25.0 ml of NaOH is added? 2. 25.0 ml of 0.10 M acefc acid is Ftrated with 0.10 M NaOH solufon. (a) What is the ph of the acid solufon before NaOH is added? (b) What is the ph aler 15.0 ml of NaOH solufon is added? (c) What is the ph aler 25.0 ml of NaOH is added?