The Common-Ion Effect 공통이온효과

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Chapter 17

Lecture Presentation

Chapter 17

Additional Aspects (다른관점) of

Aqueous Equilibria 수용액평형(수용액평형)

John D. BookstaverSt. Charles Community College

Cottleville, MO© 2012 Pearson Education, Inc.

The Common-Ion Effect (공통이온효과)

• Consider a solution of acetic acid:

• If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO−(aq)

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shift to the left.

© 2012 Pearson Education, Inc.

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The Common-Ion Effect

“The extent of ionization of a weak l t l t i d d b ddi telectrolyte is decreased by adding to

the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

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Calculate the fluoride ion concentration and pH of a solution that is 0 20 M in HF and 0 10 MpH of a solution that is 0.20 M in HF and 0.10 Min HCl.

Ka for HF is 6.8 × 10−4.

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[H3O+] [F−][HF]Ka = = 6.8 × 10−4


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HF(aq) + H2O(l) H3O+(aq) + F−(aq)

Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

[HF], M [H3O+], M [F−], M

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Initially 0.20 0.10 0ChangeAt equilibrium





[HF], M [H3O+], M [F−], M

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Initially 0.20 0.10 0Change −x +x +xAt equilibrium


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[HF], M [H3O+], M [F−], M

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Initially 0.20 0.10 0Change −x +x +xAt equilibrium 0.20 − x ≈ 0.20 0.10 + x ≈ 0.10 x



(0 10) ( )

= x

3

(0.10) (x)(0.20)6.8 × 10−4 =

(0.20) (6.8 × 10−4)(0.10)

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1.4 × 10−3 = x


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• Therefore [F−] = x = 1 4 × 10−3• Therefore, [F ] = x = 1.4 × 10

[H3O+] = 0.10 + x = 0.10 + 1.4 × 10−3 = 0.10 M

• So,

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pH = −log (0.10)pH = 1.00


Buffers (완충용액)

• Buffers are solutions• Buffers are solutions of a weak conjugate acid–base pair.

• They are particularly resistant to pH changes, even when

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changes, even when strong acid or base is added.


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Buffers

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If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.


Buffers

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Similarly, if acid is added, the F− reacts with it to form HF and water.


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Buffer Calculations

Consider the equilibrium constant i f th di i ti fexpression for the dissociation of a

generic acid (일반적인산), HA:

HA + H2O H3O+ + A−

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[H3O+] [A−][HA]Ka =


Buffer Calculations

Rearranging slightly this becomesRearranging slightly, this becomes

[A−][HA]Ka = [H3O+]

Taking the negative log of both side, we get

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[A−][HA]−log Ka = −log [H3O+] + −log

pKa

pHacid

base


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Buffer Calculations

• So [base]pKa = pH − log [base][acid]

• Rearranging, this becomes

pH = pKa + log [base][acid]

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p p a g [acid]

• This is the Henderson–Hasselbalch equation.


Henderson–Hasselbalch Equation

What is the pH of a buffer that is 0.12 Mi l ti id CH CH(OH)COOH din lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 × 10−4.

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Henderson–Hasselbalch Equation

[base]

pH = 3.85 + (−0.08)

pH = pKa + log [base][acid]

pH = −log (1.4 × 10−4) + log(0.10)(0.12)

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pH 3.85 ( 0.08)

pH = 3.77


pH Range

• The pH range is the range of pH values hi h b ff t kover which a buffer system works

effectively.• It is best to choose an acid with a pKa

close to the desired pH.

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When Strong Acids or Bases Are Added to a Buffer

When strong acids or bases are added to a buffer, it is safe to assume that all of the strong acid or base issafe to assume that all of the strong acid or base is consumed in the reaction.

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Addition of Strong Acid or Base to a Buffer

1 Determine how the neutralization1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

2. Use the Henderson–Hasselbalch equation to determine the new

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equation to determine the new pH of the solution.


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Calculating pH Changes in Buffers

A buffer is made by adding 0.300 mol HC H O d 0 300 l N C H O tHC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

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Before the reaction sinceBefore the reaction, since

mol HC2H3O2 = mol C2H3O2−

pH = pK = log (1 8 10−5) = 4 74

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pH = pKa = −log (1.8 × 10−5) = 4.74


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The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:

HC2H3O2(aq) + OH−(aq) ⎯⎯→ C2H3O2−(aq) + H2O(l)

HC2H3O2 C2H3O2− OH−

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Before reaction 0.300 mol 0.300 mol 0.020 molAfter reaction



The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:

HC2H3O2(aq) + OH−(aq) ⎯⎯→ C2H3O2−(aq) + H2O(l)

HC2H3O2 C2H3O2− OH−

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Before reaction 0.300 mol 0.300 mol 0.020 molAfter reaction 0.280 mol 0.320 mol 0.000 mol


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Now use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log (0.320)(0.280)

pH = 4 74 + 0 06pH

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pH = 4.74 + 0.06pH

pH = 4.80


Titration (적정)

In this technique aIn this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base).

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Titration

A pH meter or indicators are used to determine when the solution has reached

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determine when the solution has reachedthe equivalence point (당량점), at which the stoichiometric amount of acid equals that of base.


Titration of a Strong Acid with a Strong Base

From the start of the tit ti t thtitration to near the equivalence point (당량점), the pH goes up slowly.

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Just before (and after) th i l i tthe equivalence point, the pH increases rapidly.

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At the equivalence i t l idpoint, moles acid =

moles base, and the solution contains only water and the salt from the cation of the base and the anion of

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base and the anion of the acid.


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

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Titration of a Weak Acid (약산) with a Strong Base (강염기)

• Unlike in the previous case the conjugatecase, the conjugate base of the acid affects the pH when it is formed.

• At the equivalence point the pH is >7.Ph l hth l i i

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• Phenolphthalein is commonly used as an indicator(지시약) in these titrations.


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Titration of a Weak Acid with a Strong Base

At each point below the equivalence point, the pH of the solution during titration is determined

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pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.


Titration of a Weak Acid with a Strong Base

With weaker acids, the initial pH isthe initial pH is higher and pH changes near the equivalence point are more subtle (감지하기힘들다).

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( 지하기 다)


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Titration of a Weak Base with a Strong Acid

• The pH at the equivalence point in these

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• The pH at the equivalence point in these titrations is <7, so using phenolphthalein would not be a good idea.

• Methyl red is the indicator of choice.© 2012 Pearson Education, Inc.

Titrations of Polyprotic Acids (다양성자산)

When one titrates atitrates a polyprotic acid with a base there is an equivalence point for each

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pdissociation.


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Solubility Products (용해도곱)

Consider the equilibrium that exists in a t t d l ti f B SO i tsaturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42−(aq)

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Solubility Products

The equilibrium constant expression for thi ilib i ithis equilibrium is

Ksp = [Ba2+] [SO42−]

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where the equilibrium constant, Ksp, is called the solubility product (용해도곱).


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Solubility Products

• Ksp is not the same as solubilityas solubility.

• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL)

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100 mL (g/mL) of solution, or in mol/L (M).


Factors Affecting Solubility

• The Common-Ion Effect (공통이온효과)– 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:

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BaSO4(s) Ba2+(aq) + SO42−(aq)


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Factors Affecting Solubility• pH

– If a substance has a basic anion, it will be more soluble in an acidic solution.

– Substances with acidic cations are more soluble in basic solutions.

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Factors Affecting Solubility• Complex Ions (착이온)

– Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

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Factors Affecting Solubility

• Complex Ions• Complex Ions– The formation

of these complex ions increases the solubility of these salts

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these salts.


Factors Affecting Solubility• Amphoterism (양쪽성)

– Amphoteric metal oxides and hydroxides are l bl i id b b hsoluble in strong acid or base, because they can

act either as acids or bases.– Examples of such cations are Al3+, Zn2+, and Sn2+.

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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

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pQ = Ksp.


Selective Precipitation of Ions

One can use differences indifferences in solubilities of salts to separate ions in a mixture.

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The Common-Ion Effect 공통이온효과

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