Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 17 Additional Aspects of Aqueous Equilibria

Effect of Acetate on the Acetic Acid Equilibrium
Acetic acid is a weak acid: CH3COOH(aq) H+(aq) + CH3COO–(aq) Sodium acetate is a strong electrolyte: NaCH3COO(aq) Na+(aq) + CH3COO–(aq) The presence of acetate from sodium acetate in an acetic acid solution will shift the equilibrium, according to Le Châtelier’s principle:

The Common-Ion Effect “Whenever a weak electrolyte and a strong electrolyte containing a common ion are together in solution, the weak electrolyte ionizes less than it would if it were alone in solution.” This affects acid–base equilibria.

An Acid–Base Example What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? CH3COOH(aq) H+(aq) + CH3COO–(aq) Ka = [H+][CH3COO–]/[CH3COOH] = 1.8 × 10–5

Example (completed) 1.8 × 10–5 = (x)(0.30 + x)/(0.30 – x)
1.8 × 10–5 = (x)( x)/(0.30 – x) Assume that adding or subtracting x from 0.30 will not change 0.30 enough to matter and the equation becomes 1.8 × 10–5 = (x)(0.30)/(0.30) which results in: x = 1.8 × 10–5 M = [H+] So: pH = –log[H+] = 4.74

Weak Bases Work the Same Way
The common-ion effect applies to weak bases the same way. If a solution contains a weak base and a salt of its conjugate acid, the base won’t dissociate as much as if it is alone in solution.

Buffers – solutions which resist changes in pH
1. A mixture of a weak acid + soluble salt of the conjugate base of that weak acid or 2. A mixture of a weak base + soluble salt of the conjugate acid of that weak base Most important buffer: In blood HCO3– and H2CO3 Absorb acids and bases produced by metabolism Maintains a remarkably constant pH Important as cells live only in a very narrow pH range. Buffer Capacity – the amount of acid or base that can be added before the buffer is overwhelmed and pH dramatically changes

Buffers If a small amount of hydroxide is added to an equimolar solution of HF and NaF, for example, the HF reacts with the OH to make F and water. Similarly, if acid is added, the F reacts with it to form HF and water. The net result is the pH doesn’t change much.

Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H2O H3O+ + A [H3O+] [A] [HA] Ka =

Buffer Calculations Rearranging slightly, this becomes [A] [HA] Ka =
[H3O+] Taking the negative log of both side, we get base [A] [HA] log Ka = log [H3O+] + log pKa pH acid

Buffer Calculations So [base] pKa = pH  log [acid]
Rearranging, this becomes pH = pKa + log [base] [acid] This is the Henderson–Hasselbalch Equation.

Henderson-Hasselbalch Equation
[base] [acid] pH = pKa + log pH = pH of the buffered solution pKa = pKa of the weak acid [base] and [acid] are initial [ ]’s of the conjugate acid/base pair

Steps to Working Any Buffer Problem
1. Determine how all species exist in solution 2. Write the equilibrium reaction describing the species 3. Leave out spectator ions such as Na+ and Cl- because they don’t affect the pH Ka for formic acid = 1.8 X 10-4 Suppose g NaCHO2 (s) is dissolved into 305 mL of 0.45 M HCHO2 (aq). What is the pH of the resulting solution ?

Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  104. These are equilibrium concentrations which are entered directly into the Henderson-Hasselbalch Equation.

Henderson–Hasselbalch Equation
pH = pKa + log [base] [acid] pH = log (1.4  104) + log (0.10) (0.12) pH = (0.08) pH = 3.77

pOH = pOH of the buffered solution
Henderson-Hasselbalch equation for base [acid] [base] pOH = pKb + log pOH = pOH of the buffered solution pKb = pKb of the weak base [acid] and [base] are initial [ ]’s of the conjugate acid/base pair

1. Work this problem using the “base” form of Henderson-Hasselbalch
Determine the pH of a solution of 0.30 M NH3 (aq) and 0.18 M NH4Cl (aq). [pKa (NH4Cl) = 9.25 pKb (NH3) = 4.74] 1. Work this problem using the “base” form of Henderson-Hasselbalch pKa (NH4Cl) = 9.25 pKb (NH3) = 4.74 Do you get the same answers? 2. Work this problem using the “acid” form of Henderson-Hasselbalch

Buffer Capacity The amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree. Using the Henderson–Hasselbalch equation, pH will be the same for a conjugate acid–base pair of 1 M each or 0.1 M each; however, the buffer that is 1 M can neutralize more acid or base before the pH changes.

pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pKa close to the desired pH. A suitable buffer is within pH = pKa  1 range. Ideally want a 1:1 mixture of salt and acid whose pKa is equal to the desired pH. A suitable buffer is within pH = pKa  1 range.

A solution buffered at pH 3. 90 is needed for a reaction
A solution buffered at pH 3.90 is needed for a reaction. Would formic acid and its salt, sodium formate, make a good choice for this buffer? If so, what ratio of moles of the conjugate base anion (HCOO–) to the acid (HCOOH) is needed? From table pKa of formic acid is 3.74. This is within pH = pKa  1 range 2.90 to 4.90 So we can use this buffer system

Titration In this technique, an acid (or base) solution of known concentration is slowly added to a base (or acid) solution of unknown concentration. A pH meter or indicators are used to determine when the solution has reached the equivalence point: The amount of acid equals that of base.

Titration of a Strong Acid with a Strong Base
From the start of the titration to near the equivalence point, the pH goes up slowly.

Just before (and after) the equivalence point, the pH increases rapidly.

At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

As more base is added, the increase in pH again levels off.

From the start of the titration to near the equivalence point, the pH goes up slowly. Just before (and after) the equivalence point, the pH rises rapidly. At the equivalence point, pH = 7. As more base is added, the pH again levels off.

Titration of a Strong Base with a Strong Acid
It looks like you “flipped over” the strong acid being titrated by a strong base. Start with a high pH (basic solution); the pH = 7 at the equivalence point; low pH to end.

Titration of a Weak Acid with a Strong Base
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. At the equivalence point the pH is >7. Phenolphthalein is commonly used as an indicator in these titrations.

Ways That a Weak Acid Titration Differs from a Strong Acid Titration
A solution of weak acid has a higher initial pH than a strong acid. The pH change near the equivalence point is smaller for a weak acid. (This is at least partly due to the buffer region.) The pH at the equivalence point is greater than 7 for a weak acid.

Which slide, this or the next slide, depicts the titration of the weaker acid?

Which slide, this or the previous slide, depicts the titration of the weaker acid?

Use of Indicators Indicators are weak acids that have a different color than their conjugate base form. Each indicator has its own pH range over which it changes color. An indicator can be used to find the equivalence point in a titration as long as it changes color in the small volume change region where the pH rapidly changes.

Indicator Choice Can Be Critical!

Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO4 in water: BaSO4(s) Ba2+(aq) + SO42(aq) A saturated solution is one in which no more solute can dissolve. Saturated solution – no more solute will dissolve

Solubility Product Constant
– equilibrium constant for ionic compounds that are only slightly soluble The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42] Ksp is the equilibrium constant for ionic compounds that are only slightly soluble.

Solubility Products Ksp is not the same as solubility.
Solubility generally expressed as (g/L), (g/mL), or in mol/L (M). Ksp - Only one value for a given solid at a given temperature Temperature dependence Solubility and thus Ksp changes with T The solubility of barium sulfate is how much of this material will dissolve in a given amount of solution. The solubility product, in the case of barium sulfate will be the square of that [Ba][SO4].

Solubility Equilibria
When ionic salt dissolves in water It dissociates into separate hydrated ions Initially, no ions in solution CaF2(s)  Ca2+(aq) + 2F–(aq) As dissolution occurs, ions build up and collide Ca2+(aq) + 2F–(aq)  CaF2(s) At equilibrium CaF2(s) Ca2+(aq) + 2F–(aq) We now have a saturated solution Ksp = [Ca+][F–]2

Writing Ksp Equilibrium Laws
AgCl(s) Ag+(aq) + Cl–(aq) Ksp = [Ag+][Cl–] PbI2(s) Pb2+(aq) + 2I–(aq) Ksp = [Pb2+][I–]2 Ag2CrO4(s) Ag+(aq) + CrO42–(aq) Ksp = [Ag+]2[CrO42–] AuCl3(s) Au3+(aq) + 3Cl–(aq) Ksp = [Au3+][Cl–]3 Note that the solid does not appear in the Ksp expression because it is has a constant concentration.

Given Solubilities, Calculate Ksp
At 25 °C, the solubility of AgCl is 1.34 × 10–5 M. Calculate the solubility product for AgCl. AgCl(s) Ag+(aq) + Cl–(aq) Ksp = [Ag+][Cl–] AgCl(s) Ag+(aq) + Cl–(aq) I 0.00 0. 00 C E 1.34 × 10–5 M 1.34 × 10–5 M 1.34 × 10–5 M 1.34 × 10–5 M Ksp = (1.34 × 10–5)(1.34 × 10–5) Ksp = 1.80 × 10–10

Learning Check The solubility of calcium fluoride, CaF2, in pure water is 2.15 × 10‑4 M. What is the value of Ksp? A × 10–11 B × 10–11 C × 10–12 D × 10-7 CaF2 Ca2+ + 2F– [Ca2+] = (2.15 × 10–4) [F-] = 2(2.15 × 10–4) Ksp = [Ca][F]2 = (2.15 × 10–4) (4.3 x 10-4)2 Ksp = 3.98 × 10–11

Given Ksp, Calculate Solubility
What is the molar solubility of CuI in water? Determine the equilibrium concentrations of Cu+ and I– CuI(s) Cu+(aq) + I–(aq) Ksp = [Cu+][I–] Ksp = 1.3  10–12

Molar Solubilities from Ksp
Solve Ksp expression Ksp = 1.3 × 10–12 = (x)(x) x2 = 1.3 × 10–12 x = 1.1 × 10–6 M = calculated molar solubility of CuI = [Cu+] = [I– ] Conc (M) CuI(s) Cu+(aq) + I–(aq) Initial Conc. 0.00 Change Equilibrium Conc. No entries No entries No entries +x +x x x

Given Ksp, Calculate Solubilities
Calculate the solubility of CaF2 in water at 25 °C, if Ksp = 3.4 × 10–11. CaF2(s) Ca2+(aq) + 2F– (aq) 1. Write equilibrium law Ksp = [Ca2+][F–]2 2. Construct concentration table Conc (M) CaF2(s) Ca2+(aq) 2F–(aq) Initial Conc. (No entries 0.00 Change in this Equilibrium Conc. column) +x +2x x 2x

Molar Solubilities from Ksp
3. Solve the Ksp expression Ksp = [Ca2+][F–]2 = (x) (2x)2 3.4 × 10–11 = 4x3 x = 2.0 × 10–4 M = molar solubility of CaF2 [Ca2+] = x = 2.0 × 10–4 M [F–] = 2x = 2(2.0 × 10–4 M) = 4.0 × 10–4 M

Factors Affecting Solubility
The Common-Ion Effect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease: BaSO4(s) Ba2+(aq) + SO42(aq)

pH and Solubility Mg(OH)2(s) Mg2+(aq) + 2OH–(aq)
Increase OH– shift equilibrium to left Add H+ shift equilibrium to right Le Châtelier’s Principle Ag3PO4(s) Ag+(aq) + PO43–(aq) Add H+ increase solubility H+(aq) + PO43–(aq)  HPO42–(aq) AgCl(s) Ag+(aq) + Cl–(aq) Adding H+ has no effect on solubility Why? Cl– is very, very weak base, so neutral anion So adding H+ doesn’t effect Cl– concentration End Lecture:

Common Ion Effect What happens if another salt, containing one of the ions in our insoluble salt, is added to a solution? Consider PbI2(s) Pb2+(aq) + 2I–(aq) Saturated solution of PbI2 in water Add KI PbI2 (yellow solid) precipitates out Why? Le Chatelier’s Principle Add product I– Equilibrium moves to left and solid PbI2 forms

Learning Check What effect would adding copper(II) nitrate have on the solubility of CuS? A. The solubility would increase B. The solubility would decrease C. The solubility would not change

Common Ion Effect Consider three cases
What is the molar solubility of Ag2CrO4 in pure water? What is the molar solubility of Ag2CrO4 in 0.10 M AgNO3? What is the molar solubility of Ag2CrO4 in 0.10 M Na2CrO4? Ag2CrO4(s) Ag+(aq) + CrO42–(aq) Ksp = [Ag+]2[CrO42–] = 1.1 × 10–12 x = 9.0 10–10 M

Common Ion Effect A. What is the solubility of Ag2CrO4 in pure water?
Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries 0.00 M C in this E column) +2x +x 2x x Ksp = [Ag+]2[CrO42–] = (2x)2(x) = 1.1  10–12 = 4x3 x = 9.0 10–10 M x = Solubility of Ag2CrO4 = 6.5 x 10-5 M [CrO42–] = x = 6.5 x 10-5 M [Ag+] = 2x = 1.3 × 10–4 M

Common Ion Effect B. What is the molar solubility of Ag2CrO4 in 0.10 M AgNO3 solution? Ksp = 1.1 × 10–12 Ksp = 1.1 × 10–12 = (0.10 M)2[x] x = Solubility of Ag2CrO4 = 1.1 × 10–10 M [Ag+] = 0.10 M [CrO42–] = 1.1 × 10–10 M Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries 0.10 M 0.00 C in this E column) +2x +x ≈0.10 x

Common Ion Effect Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries
C. What is the solubility of Ag2CrO4 in M Na2CrO4? Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries 0.00 M 0.10 M C in this E column) +2x +x 2x ≈0.10 Ksp = (2x)2(0.10) = 1.1  10–12 = 4x2(0.10) x = Solubility of Ag2CrO4 = 1.66 × 10–6 M [CrO42–] = 0.10 M [Ag+] = 2x = 3.3 × 10–6 M

Common Ion Effect What have we learned about the solubility of silver chromate? Dissolving it in pure water the solubility was 6.5 × 10–5 M Dissolving it in AgNO3 solution solubility was 1.1 × 10–10 M Dissolving it in Na2CrO4 solution solubility was 1.66 × 10–6 M Common ion appearing the most in the formula of the precipitate decreases the solubility the most

Will a Precipitate Form?
In a solution, If Q = Ksp, the system is at equilibrium and the solution is saturated. If Q < Ksp, more solid can dissolve until Q = Ksp. If Q > Ksp, the salt will precipitate until Q = Ksp.

Predicting Precipitation
Will a precipitate of PbI2 form if mL of M Pb(NO3)2 are mixed with mL of M NaI? PbI2(s) Pb2+(aq) + 2I–(aq) Ksp = [Pb2+][I–]2 = 9.8 × 10–9 Strategy for solving Calculate concentrations in mixture prepared Calculate Qsp = [Pb2+][I–]2 Compare Qsp to Ksp

Step 1. Calculate concentrations Vtotal = mL mL = mL [Pb2+] = 1.67 × 10–2 M [I–] = 6.67 × 10–2 M

Step 2. Calculate Qsp Qsp = [Pb2+][I–]2 = (1.67 × 10–2 M)(6.67 × 10–2 M)2 Qsp =7.43 × 10–5 Step 3. Compare Qsp and Ksp Qsp = 7.43 × 10–5 Ksp = 9.8 × 109 Qsp > Ksp so precipitation will occur

Chapter 17 Additional Aspects of Aqueous Equilibria

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Chapter 17 Additional Aspects of Aqueous Equilibria

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