18 Acid-Base Equilibria 1 18 Additional Aspects of Acid – Base Equilibria Common ion effect...

18 Acid-Base Equilibria 1

18 Additional Aspects of Acid – Base Equilibria

Common ion effect solutions containing acids, bases, salts, solvent

Buffer solutions solutions containing a weak acid and its salt

Indicators (acid-base) colored acids and bases

Titration curve chemistry and pH during titration

Solutions of salts and polyprotic acids (hydration) reaction of solvent with ions


Exam paper return

The exam papers are piled according to the last digit of your ID number from 0 to 9.

All papers are now in front of the Chem123 Lab on the shelf. You may retrieve yours from the hallway in ESC on the first floor.


Ka Kb and Kw

H+ + Base = Conjugate_acid of Base+

Acid = H+ + Conjugate_base of Acid-

For example:

NH3 + H2O = NH4+ + OH-

Ka for NH4+ = Kw / Kb for NH3

HA = H+ + A- Kb for A- = Kw / Ka for HA

Thus, Ka Kb = Kw for conjugate pairs.

A- + H2O = HA + OH-

[HA] [OH-] [H+]Kb = ————— ———

[A-] [H+]

[HA] = ———— [OH-] [H+] [A-] [H+]

1 = —— Kw Ka


Ka Kb and Kw - comparison

HA = H+ + A-

[H+] [A-] [OH-] Ka = ———— ———

[HA] [OH-]

[A-] = ————— [H+] [OH-] [HA] [OH-]

1 = —— Kw

Kb

A- + H2O = HA + OH-

[HA] [OH-] [H+]Kb = ————— ———

[A-] [H+]

[HA] = ———— [OH-] [H+] [A-] [H+]

1 = —— Kw Ka


Ka Kb and Kw – another way

HA + H2O (l) = H3O+ (aq) + A– (aq) Ka of HA +) A– + H2O (l) = OH–(aq) + HA (aq) Kb of A–

2 H2O (l) = H3O+ (aq) + OH– (aq) Kw = Ka Kb

Review qn:When you add two equations to get a third, what are the relationship between the Ks?


Ka, Kb of A- & Kw

A- + H2O = HA + OH-

[HA] [OH-] [H+]Kb = ————— ———

[A-] [H+]

[HA] = ———— [OH-] [H+] [A-] [H+]

1 = —— Kw Ka

HA = H+ + A- Ka

A- + H2O = HA + OH- Kb

-- add them together - -

H2O = H+ + OH- Kw = Ka Kb

Two ways to show their relationship


Properties of salt solutionsThe spectator ions in acid-base reactions form salts. Salts completely ionize in their aqueous solutions. A salt NH4A, the ionizes

NH4A = NH4+ + A-

The ions of salts interact with water (called hydration),A- + H2O = HA + OH-

If the anions are stronger base than H2O, the solution is basic.NH4

+ + H2O = NH3 + H3O+

If the cations are stronger acid than H2O, the solution is acidic.These competitive reactions make the solution acidic or basic depending on the strength of the acids and bases.

Are the solutions of the following salts neutral, basic or acidicNaCl? NaAc? NH4Cl? NH4Ac? KClO4? KNic? CH3NH3ClO4?

modified


Hydration problemsWhat is the pH and concentrations of various species of a 0.100 M KA salt solution if Ka for HA = 1.0e-5? (K = potassium, HA a general acid)

Solution: KA = K+ + A-

A– + H2O (l) = OH–(aq) + HA (aq) Kb = 1e–14/1.0e–5 = 1e–90.10-x x x

x2

Kb = ———— = 1.0e–90.10 – x (0.10 – x) 0.10 = [A–]

x = 1e-9*0.1 = 1e–5 = [OH–] = [HA]pOH = 5 (= pKb / 2=10/2); pH = 14 – 5 = 9 (basic)


Acidity of Salt SolutionsWhat is the pH of a 1.0-M NaHCO3 solution? For H2CO3, Ka1 = 4.3e-7 and Ka2 = 4.8e-11.

Solution: In the solution, [Na+] = 1.0 M and [HCO3–] = 1.0 M. Two

competitive reaction takes place:

H2O + HCO3– = H2CO3 + OH– Kb = Kw / Ka1 = 2.33e-8

HCO3–

= H+ + CO32– Ka2 = 4.8e-11

Thus, HCO3 – is a stronger base than an acid. The solution is basic.

Discussion: The overall dissociation constant, Koverall = Ka1* Ka2 = 2.1e-17

Generalization: if Ka2 < Kb or Ka1*Ka2 < Kw, the solution of NaHA is basic for the diprotic acid H2A.


Common ion effectSince salts, acids and bases ionize in their solutions, their common ions such as H+, OH–, other cations, anions, will affect the equilibria.

Calculate the degree of ionization of HA in a solution containing 0.10 M each of HCl and HA, if Ka = 1.0e–5 for HA.

Solution: (assume [A-] = x and work out the relations as shown)HCl = H+ + Cl–

0.10+x 0.10 (0.10+x) xHA = H+ + A– Ka = 1.0e–5 = ——————

0.10-x 0.10+x x 0.10-x (0.1-x=0.1+x=0.1)

x = 1.0e–5 Mdegree of ionization = 1.0e–5/0.1 = 1.0e–4 = 0.01%

ReDo if [HA] = 0.01 M and [HCl] = 0.10 M?


Mass and Charge Balance Equations

In a solution of electrolytes, the stoichiometry leads to mass or material balance (mbe) and charge balance equations (cbe).

The merit is to consider the various ion species and equilibria in the solution when we write these equations.

In a 0.10 M NaHCO3 solution,

mbe: 0.10 = [Na+] = [H2CO3] + [HCO3 –] + [CO3

2– ]

cbe: [Na+] + [H+] = [HCO3 –] + 2 [CO3

2– ] + [OH–]

New See Week 8 Part b problem #1.

Consider these equilibrium eqn for mbe and cbe:

NaHCO3 = Na+ + HCO3-

HCO3- + H2O = H2CO3 + OH-

HCO3- + H2O = CO3

2- + H3O+


BuffersA buffer contains a weak acid and a salt of the same acid.A buffer contains a weak base and a salt of the same base.

Concentrations of various species for a solution containing 0.10 M each of KA and HA, Ka = 1.0e–5.

KA = K+ + A– 0.10 0.10+x x (0.10+x)

HA = H+ + A– Ka = 1.0e–5 = —————— 0.10-x x 0.10+x 0.10-x (0.1-x=0.1+x=0.1)

x = 1.0e–5 = [H+]

pH = pKa in this case, because [HA] = [A–]


pH of buffersFor a weak acid, HA, we have

HA = H+ + A–

[H+] [A–]Ka = ————

[HA]

[A–] – log Ka = – log [H+] – log ———

[HA]

[A–] [A–] pKa = pH – log —— pH = pKa + log ——

[HA] [HA]

[salt]

[acid]

Adding acid or base only affect the ratio [A–] / [HA], the pH changes little.

Henderson-Hasselbach equationn


A buffer solution

A buffer solution is made up using 10.0 mL0.10 M acetic acid (HA, Ka = 1.7e-5) and 20.0 0.10 M sodium acetate (NaA). Evaluate its pH.

Hint: [A–] = 20.0*0.1 / (20.0+10.0) = 0.067 M

[HA] = 10.0*0.1 / (20.0+10.0) = 0.033 MHenderson-Hasselbach equationn

[A–] pH = pKa + log ——

[HA]

= – log (1.7e-5) + log (2/1) = 4.77 + 0.30 = 5.07

Find ways to see [A-] / [HA] = 2 / 1

pH of a sol’n by mixing 10.0 mL 0.1 M HCl and 10.0 mL 0.3 M NaA


Making a buffer of certain pH

Find a weak acid with pKa close to the desirable pH

Find out the desirable [A-] / [HA] ratio

Measure appropriate amount (moles) of acid and its salt

Dissolve in appropriate amount of water (volume does not matter)

Use a strong acid and a salt or a weak acid and a strong base instead of acid and salt.


Indicators

Indicators are substanceswhose solutions change color due to changes in pH.

HIn = H+ + In-,

and define the equilibrium constant as Kai,

[H+][In-] Kai [In-] Kai = ————— —— = ——

[HIn] [H+] [HIn]

The color varies according to the ratio Kai / [H+]

___


Common indicators

Name Acid color pH range Base color

Methyl violet Yellow 0 – 1.6 Blue-violet

Methy orange Red 3.2-4.4 Yellow

Litmus Red 5-8 Blue

Phenolphthalein Colorless 8.2 - 10.0 Pink

Thymolphthalein Colorless 9.4-10.6 Blue


Phenolphthalein – a indicator

C20H14O4 (MW = 318.33)

Formal name: 3,3-bis(4-hydroxyphenyl)-1(3H)-isobenzofuranone, 3,3-bis(4-hydroxyphenyl)phthalide)

Colorless in acidic solution

Pink in basic solution, C20H12O42–

pH ~ 10


Titration calculationWhen a strong acid is titrated with a strong base, consider:

The amount of acid present = Va * Ca

The amount of base NaOH added = Vb * Cb

The amount of acid left = Va * Ca - Vb * Cb

The concentration of acid and thus

Va * Ca - Vb * Cb [H+] = —————————

Va + Vb

Titration of 10.0 mL 1 M HCl using 1 M NaOH

Base add [H+] pH

0 1.0 05 5/15 0.489 1/19 1.289.5 .5/19.5 1.599.9 0.1/19.9 2.309.95 0.05/19.95 2.60

10 NaCl soln 7 [OH]10.5 0.05 / 20.05 11.4010.10 0.5 / 20.1 11.7011 1 / 21 12.6815 5 / 25 13.320 20 / 30 13.92

Please plot the curve from the data


Titration curve

Weak HAslide 7

Buffers–slide 10-11Hydration of saltsSlides 6 & 7


Acid-base equilibria - summaryKnow names, formula, & properties of some acids & bases: HA, NH3

Evaluate Ka and Kb and apply them to calculate [ ]’s of various speciesusing approximationusing quadratic equation

Theory of polyprotic acidsevaluate concentrations of various species

Derive and apply the relation Ka Kb = Kw

Explain why some salt solutions are acidic or basicEvaluate pH of salts (hydration reacting with water)Explain theory of buffer

evaluate pH of buffermake a buffer solution

Explain indicators as weak acids and basesPlot a titration curve (weak acid titrating with strong base)


ReviewIn solving x in problems involving equilibrium constant, how dowe know which method to use. When can we neglect x?Also, when can we use something like Successive Approxiamation?In the example given in class, K was quite small, so as donein other examples, I probably would have neglected x.Then there is Newton's method where we plug in trial values for x. How do we know where a good place is to start? Technically x could be any value!


Review 1 – pH of weak acid solutions

Solutions of a weak acid HA, Ka = 1.0e–6 with concentrations of 1.0 M, 0.10 M, 0.00010 M. What are their pH?Solution:

HA = H+ + A–, Ka = 1.0e-6C-x x x

x2

Ka = ——— C–x

C = 1.0 Mx = Ka*C = 1e–3pH = -log0.001 = 3

C = 0.10 Mx = Ka*C = 3.2e–4pH = -log3.2e-4=3.49

C = 0.00010 Mx = Ka*C = 1e–5, 10%of C

Do not approximate usex2 + Ka x – Ka*C = 0

– Ka + (Ka2 + 4 Ka*C)

x = ————————— = 1.9e–5 2

pH = -log1.9e-5 = 4.72 not 5


Review 2 – pH of buffer25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 5.00 mL NaOH has been added?

AnalysisThe solution containing 25.00 mL 0.10 M HAc and 5.00 mL 0.20 M NaOH actually contains 5.00 mL * 0.20 mol/L = 1.0 mmol NaAc,and (25.00*0.10 – 1.0) mol = 1.50 mmol HAc. This solution is a buffer

Solution: For a buffer, apply

[A-] 1.0 / total volume

pH = pKa + log ------- = 4.75 + log --------- = 4.75 – 0.18 = 4.57 [HA] 1.5/ total volume






pH = pKa + log ------- = 4.75 + log --------- = 4.75 – 0.00 = 4.75 [HA] 1.25 / total volume






pH = pKa + log ------- = 4.75 + log --------- = 4.75 + 0.10 = 4.85 [HA] 1.1/ total volume not 4.57


Review 5 – pH of salt a solution25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 12.50 mL NaOH has been added?

AnalysisThe solution containing 25.00 mL 0.10 M HAc and 12.5 mL 0.20 M NaOH actually contains 12.50 mL * 0.20 mol/L = 2.5 mmol NaAc,and (25.00*0.10 – 2.5) mol = 0.00 mmol HAc. [NaA] = 2.5 mmol / (25.0+12.5)mL = 0.067 M

Solution: hydration problemA- + H2O = HA + OH-

0.067-x x x x2 Kw 10-14 ------------ = Kb = ----- = -------- = 5.62e-10 0.067 – x Ka 10-4.75

x = (5.62e-10)*0.067 = 1.36e-6pOH = -log(1.36e-6) = 5.2


Review 6 – pH of salt and base25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 15.0 mL NaOH has been added?

Analysis The solution containing 25.0 mL * 0.10 mol/L = 2.5 mmol NaAc,[NaAc] = 2.5 mmol /40 mL = 0.0625 Mand (15.0 m* 0.20 – 2.5) = 0.5 mmol NaOH[NaOH] = 0.5 mmol / 40 mL = 0.0125 M

Solution: The pH is dictated by the NaOHNaOH = Na+ + OH-

0.0125 + xA- + H2O = HA + OH-

0.0625-x x 0.0125+xThe extend of this reaction is small, x << 0.0125, [OH] = 0.0125 M

pOH = -log(0.0125) = 1.90pH = 14.00 – 1.90 = 12.10

Review 5; x = 1.36e-6 M


Buffer solution sites

Suppliers:

postapplescientific.com/reagents/buffersoln.htmlcaledonlabs.com/cgi-bin/products.cgi?category=K

Buffer solution preparation

csudh.edu/oliver/chemdata/buffers.htmbiochem.mcw.edu/~simont/java/BufferMaker.html

18 Acid-Base Equilibria 1 18 Additional Aspects of Acid – Base Equilibria Common ion effect...

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