# HCO3-(aq) H+(aq) + CO32-(aq)

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HCO3-(aq) H+(aq) + CO32-(aq)
Unit 4-5: Acids and Bases 3 Be able to write the chemical equation for a chemical species acting as an acid: HX(aq) H+(aq) + X-(aq) Example Write the equation showing the bicarbonate ion acting as an acid. HCO3-(aq) H+(aq) + CO32-(aq)

X-(aq) + H2O(l) OH-(aq) + HX(aq)
Unit 4-5: Acids and Bases 3 Be able to write the chemical equation for a chemical species acting as a base: X-(aq) + H2O(l) OH-(aq) + HX(aq) Example Write the equation showing the bicarbonate ion acting as a base. HCO3-(aq) + H2O(l) OH-(aq) + H2CO3(aq)

Unit 4-5: Acids and Bases 3 The Common Ion Effect What is the pH of a solution made by adding 0.30 mol of acetic acid to enough water to make 250 mL of solution? Ka is 1.8 x 10-5 at 25°C. CH3COOH(aq) CH3COO-(aq) + H+(aq) initial /0.250 M 0 M 0 M change - x M x M x M equilibrium x M x M x M Ka = [CH3COO-][H+] = x = 1.8 x 10-5 [CH3COOH] (1.2 - x) assume x<< x2 = 2.16 x x = M (<<1.2) pH = -log(0.0046) = 2.33 verify your assumption

Unit 4-5: Acids and Bases 3 The Common Ion Effect What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 250 mL of solution? CH3COOH CH3COO-(aq) + H+(aq) initial /0.250 M 0.30/0.250 M M change - x M x M +x M equilibrium x M x M x M Ka = [CH3COO-][H+] = (1.2 + x)x = 1.8 x 10-5 [CH3COOH] (1.2 - x) (assume x<<1.2) x = 1.8 x 10-5 M (x<<1.2) pH = -log(1.8 x 10-5) = 4.74 verify your assumption

CH3COOH CH3COO-(aq) + H+(aq)
Unit 4-5: Acids and Bases 3 The Common Ion Effect Adding acetate ion shifted the equilibrium to the left, decreasing [H+] and increasing pH. CH3COOH CH3COO-(aq) + H+(aq) adding C2H3O2- Common Ion Effect The extent of ionization of any weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.

Unit 4-5: Acids and Bases 3 The Common Ion Effect Find the solubility of calcium phosphate in water at 25°C. Find the solubility of calcium phosphate in 0.100M sodium phosphate at 25°C.

Unit 4-5: Acids and Bases 3 The Common Ion Effect Find the solubility of calcium phosphate in water at 25°C. What information do you need? formula for calcium phosphate source: your head (I hope) the equation for the equilibrium source: your head the equilibrium constant source: Appendix D

Unit 4-5: Acids and Bases 3 The Common Ion Effect Find the solubility of calcium phosphate in water at 25°C. Ca3(PO4)2(s) Ca2+(aq) + 2PO43-(aq) Ksp = 2.0 x 10-29 I present C lose x mol +3x M +2x M E present 3x M 2x M 2.0 x = 108x5 x = 7.1 x 10-7 M

Unit 4-5: Acids and Bases 3 The Common Ion Effect b) Find the solubility of calcium phosphate in 0.100M sodium phosphate at 25°C. Ca3(PO4)2(s) Ca2+(aq) + 2PO43-(aq) Ksp = 2.0 x 10-29 I present M C lose x mol +3x M +2xM E present 3x M ( x) M Assume 2x << .100 2.0 x = .27x3 x = 4.2 x M Check: 8.4 x << .1 (100,000,000 times smaller)

Unit 4-5: Acids and Bases 3 The Common Ion Effect a) Find the solubility of calcium phosphate in water at 25°C. 7.1 x 10-7 M b) Find the solubility of calcium phosphate in 0.100M sodium phosphate at 25°C. 4.2 x M The presence of phosphate from sodium phosphate decreased the solubility of the calcium phosphate by a factor of a 1000.

Unit 4-5: Acids and Bases 3 Buffered Solutions What is the pH of our 1.2M acetic acid solution if 10.0 mL of 1.0 M NaOH is added to it? CH3COOH CH3COO-(aq) H+(aq) If NaOH is added to the solution, it reacts with some of the acetic acid: CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l) This has the effect of decreasing the acetic acid concentration and increasing the acetate ion concentration. PLUS, addition of the NaOH changes the volume of the system.

Unit 4-5: Acids and Bases 3 Buffered Solutions What is the pH of our 1.2M acetic acid solution if 10.0 mL of 1.0 M NaOH is added to it? CH3COOH CH3COO-(aq) H+(aq) If NaOH is added to the solution, it reacts with some of the acetic acid and produces acetate ion: CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l) This is a stoichiometry problem.

Unit 4-5: Acids and Bases 3 Buffered Solutions What is the pH of our 1.2M acetic acid solution if 10.0 mL of 1.0 M NaOH is added to it? CH3COOH CH3COO-(aq) H+(aq) initial: ( )/0.260 M /0.260 M 0 M change: - x M x M x M equilibrium: x M x M x M Ka = [CH3COO-][H+] = ( x)x = 1.8 x 10-5 [CH3COOH] ( x) (assume x<<0.0385) x x x10-5 = 0 x = 5.2 x 10-4 M (x<<0.0385) pH = -log(5.2 x 10-4) = 3.29 verify your assumption

Unit 4-5: Acids and Bases 3 Buffered Solutions What is the pH of our acetic acid/sodium acetate solution if 10.0 mL of 1.0 M NaOH is added to it? CH3COOH CH3COO-(aq) H+(aq) initial: ( )/0.260 M ( )/0.260 M 0 M change: - x M +x M x M equilibrium: x M x M x M Ka = [CH3COO-][H+] = ( x)x = 1.8 x 10-5 [CH3COOH] ( x) (assume x<<1.115, 1.192) x = 1.7 x 10-5 M (x<<1.115 and 1.192) pH = -log(1.7 x 10-5) = 4.77 verify your assumption

Unit 4-5: Acids and Bases 3 Buffered Solutions HAc is short for acetic acid. pH pH after addition pH after addition of 10.0 mL of 10.0 mL of 1.0 M NaOH of 1.0 M HCl 250 mL of 1.2 M HAc 1.2 M HAc/Ac- buffer soln DI water Ac- is short for acetate.

Unit 4-5: Acids and Bases 3 Buffered Solutions pH pH after addition pH after addition of 10 mL 1.0 M NaOH of 10 mL 1.0 M HCl 250 mL of 1.2 M HAc 1.2 M HAc/Ac DI water The pH of the HAc solution ranged with the addition of 10.0 mL of the acid or base.

Unit 4-5: Acids and Bases 3 Buffered Solutions pH pH after addition pH after addition of 10 mL 1.0 M NaOH of 10 mL 1.0 M HCl 250 mL of 1.2 M HAc 1.2 M HAc/Ac DI water When the same amount of acid or base was added to the HAc/Ac solution, its pH ranged The addition of the acetate ion buffers the solution against changes in pH.

Buffered Solutions contain a weak conjugate acid-base pair
Unit 4-5: Acids and Bases 3 Buffered Solutions contain a weak conjugate acid-base pair such as HAc/Ac-, NH3/NH4+, H2CO3/HCO3-, HCN/CN-. resist changes in pH because they contain an acid to react with OH- and a base to react with H+. have the ability to resist changes in pH, known as their buffer capacity, which depends on the concentrations of the members of the pair. The pH of a buffer solution may be determined by the Henderson-Hasselbalch equation.

Henderson-Hasselbalch Equation
Unit 4-5: Acids and Bases 3 Henderson-Hasselbalch Equation For the general equilibrium HX(aq) H+(aq) + X-(aq) the pH is given by 𝒑𝑯=𝒑 𝑲 𝒂 +𝒍𝒐𝒈( [𝒃𝒂𝒔𝒆] [𝒂𝒄𝒊𝒅] ) Henderson-Hasselbalch equation

Henderson-Hasselbalch Equation
Unit 4-5: Acids and Bases 3 Henderson-Hasselbalch Equation Let’s recalculate the pH of our 1.2 M HAc/Ac solution. pH = pKa + log ([base]/[acid]) Ka = 1.8 x 10-5, so pKa = 4.74 pH = log(1.2M/1.2M) = 4.74 After the addition of 10.0 mL 1.0 M NaOH pH = log(1.192M/1.115M) = = 4.77

Unit 4-5: Acids and Bases 3 Buffered Solutions You have been asked to prepare mL of a buffer with the pH of blood, What chemicals and amounts will you use? Buffers most effectively resist a change in pH in either direction if they contain equal concentrations of both members of the acid-base pair. Find a weak acid with a Ka ≈ 10-7 (so the pKa will be 7). Ka3(citric acid) = 4.0 x 10-7 at 25°C. pKa = 6.40 7.40 = log [base]/[acid] [base]/[acid] = 10

Unit 4-5: Acids and Bases 3 Buffered Solutions You have been asked to prepare mL of a buffer with the pH of blood, What chemicals and amounts will you use? We chose citric acid (abbreviated H3Cit) and found that we need a base-to-acid ratio of 10. For Ka3, the base is Cit3- and the acid is HCit2-. HCit2-(aq) Cit3-(aq) + H+(aq) Ka3 = 4.0 x 10-7 at 25°C pH = pKa3 + log [Cit3-] [HCit2-]

Buffered Solutions Solution #1:
Unit 4-5: Acids and Bases 3 Buffered Solutions Solution #1: Put 1.0 mol of Na3Cit and 0.10 mol Na2HCit in a 500-mL volumetric flask and dilute to the mark with DI water. Solution #2: Put mol of Na3Cit and mol Na2HCit in a 500-mL volumetric flask and dilute to the mark with DI water. Both solutions will produce the correct pH…what is different about them? Their BUFFERING CAPACITY. Solution #2 has smaller concentrations of the acid-base pair.

Titration of a Strong Acid with a Strong Base
If we neutralize an acid incrementally and monitor the pH of the solution as we add the base, the resulting data would give us a titration curve, a plot of pH versus volume of titrant added.

Titration of a Strong Acid with a Strong Base
For the neutralization of 50.0 mL of M HCl with M NaOH, the titration curve would be a plot of pH versus volume of NaOH added.

𝒎𝒎𝒐𝒍 𝒎𝑳 = 𝒎𝒐𝒍 𝑳 = molarity (M)
Titration of a Strong Acid with a Strong Base For the neutralization of 50.0 mL of M HCl with M NaOH, here is some useful information: 50.0 mL of M HCl contain 5.00 mmol H+ 1.0 mL of M NaOH contains 0.10 mmol of OH- 𝒎𝒎𝒐𝒍 𝒎𝑳 = 𝒎𝒐𝒍 𝑳 = molarity (M)

Titration of a Strong Acid with a Strong Base
For the neutralization of 50.0 mL of M HCl with M NaOH, the [H+] before the equivalence point is the concentration of unreacted HCl. After 25.0 mL of NaOH have been added, the total volume is ( ) mL, and [H+] =(5.00mmol HCl mmol NaOH)/75 mL = M pH = 1.48

Titration of a Strong Acid with a Strong Base
For the neutralization of 50.0 mL of M HCl with M NaOH, the [OH-] after the equivalence point is the concentration of excess NaOH. After 75.0 mL of NaOH have been added, the total volume is ( ) mL, and [OH-] = (7.50 mmol NaOH mmol HCl)/125 mL = M pOH = 1.70 pH = = 12.30

Titration of a Strong Acid with a Strong Base
The initial pH depends solely on the concentration of the acid. Here it is 1.00. The pH at the equivalence point is 7. This is true for the titration of any strong acid by any strong base…and vice versa.

Titration of a Weak Acid with a Strong Base
For the neutralization of 50.0 mL of 0.10 M HAc with M NaOH, the [H+] before the equivalence point is calculated from the equilibrium expression using the concentration of unreacted HAc. After 25.0 mL of NaOH have been added, the total volume is ( ) mL, and HAc Ac-(aq) H+(aq) ( )/75.0 M /75.0 M We can use the Henderson-Hasselbalch equation: Ka = 1.8 x 10-5  pKa = 4.74 [Ac-] = [HAc] = 2.50/75.0 M  [base]/[acid] = 1.00 pH = log 1.00 = 4.74

Titration of a Weak Acid with a Strong Base
The initial pH depends on the concentration of the weak acid and its Ka.

Titration of a Weak Acid with a Strong Base
For the neutralization of 50.0 mL of M HAc with M NaOH, the [H+] at the equivalence point is calculated from the equilibrium expression for the base. After 50.0 mL of NaOH have been added, the total volume is ( ) mL, and all of the HAc has been changed to Ac-. Ac-(aq) + H2O(l) HAc(aq) + OH-(aq) initial: /100.0 M M M change: x M x M x M equilibrium: x M x M x M Kb = Kw/Ka = 1.0 x 10-14/(1.8 x 10-5) = 5.6 x 10-10 [OH-][HAc] = 5.6 x ≈ x x = [OH-] = 5.3 x 10-6 [Ac-] pOH = 5.28 pH = = 8.72

Titration of a Weak Acid with a Strong Base
The pH at the equivalence point is 8.72.

Titration of a Weak Acid with a Strong Base
For the neutralization of 50.0 mL of M HAc with M NaOH, the [OH-] after the equivalence point is the concentration of excess NaOH. This is exactly the same as the post-equivalence section of the strong acid-strong base titration curve. After 75.0 mL of NaOH have been added, the total volume is ( ) mL, and [OH-] = (7.50 mmol NaOH mmol HCl)/125 mL = M pOH = 1.70 pH = = 12.30

Titration of a Weak Acid with a Strong Base
This part of the curve is identical to that of the HCl/NaOH curve, since in both cases the NaOH volume added is the same.

Titration of a Weak Base with a Strong Acid
The titration curve for a weak base with a strong acid is calculated similarly to that of the weak acid/strong base. You must be careful to write the equilibrium expression correctly and use the correct Kb. For the titration of aqueous ammonia, use NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kb = 1.8 x 10-5 at 25°C

Titration of a Weak Base with a Strong Acid

Region of Maximum Buffer Capacity 𝒑𝑯= 𝒑𝑲 𝒂 +𝒍𝒐𝒈( [𝒃𝒂𝒔𝒆] [𝒂𝒄𝒊𝒅] )
Buffering is best when [base] = [acid]. But when [base] = [acid], pH = pKa This occurs when the volume of NaOH added is ½ the NaOH volume needed to reach the equivalence point.

Region of Maximum Buffer Capacity
pH = pKa = Maximum buffering capacity is here.

Titration of a Polyprotic Acid with a Strong Base
2nd equivalence point pH = 9.9 H2A is completely neutralized at 2nd eq pt. 1st equivalence point pH = 4.5 HA- and A2- H2A and HA-

Titration of a Polyprotic Acid with a Strong Base
halfway between 40 and 80 mL pKa2= 7.2 halfway between 0 and 40 mL HA- and A2- pKa1= 2.0 H2A and HA-

Not All Equivalence Points are Obvious