acid base 2 titrations lecture 1617.pptx

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Acid-Base Part 2

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The process of adding a strong acid or a strong base

to a titrate of acid or base with the purpose of neutralizing the substance in the beaker

Looking to reach equivalence point  mol H+ = mol OH

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Generally done with a pH probe, which is used to

create a graph of pH vs. volume added

Acid-Base Titration

p

H

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Ways to find the equivalence point

1. Can find it stoichiometrically using stoichiometry for the

neutralization reaction

mol H+ = mol OH- stoichiometrically

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Ways to find the equivalence point

2. Can find it graphically based on pH vs. volume graph

 When the titrate is neutralized completely by the titrant, the pH

changes dramatically

 The midpoint of this dramatic change is the equivalence point – the middle of the area where neutralization is happening

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• If the pH at the equivalence point is known, an

indicator can be used

• Want the pKa of the indicator to match the pH at the

equivalence point

• Indicator will then change color as pH changes

dramatically, giving visual end to the titration

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They are weak acids and weak bases themselves!

Given Ka of possible indicators

RANGE of indicators = pKa ± 1 (10/1 or 1/10 ratio of A/B) Ex – bromothymol blue has Ka = 1.0 x 10-7

 HA side of indicator = yellow, A-_{side of indicator = blue}

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Need to be able to EXPLAIN the shape of the curve at

ALL points through the titration

What happens initially after a small amount of acid/base

added

When excess of the titrant still remains What happens near the equivalence point

What happens beyond the equivalence point – when excess

of the strong acid or strong base used as a titrate is added

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Calculate the initial pH of acid/base in the beaker

Normal acid/base calculation for strong or weak acid/base

Calculate the pH at the halfway point of the titration

Or use pH at halfway point to find K

a or Kb of substance titrated

Calculation of the pH at the equivalence point

OR using the equivalence point to find the concentration of unknown

acid/base you are titrating

Calculation of the pH at a point beyond the equivalence point

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1. Strong acid titrated with a strong base

2. Strong base titrated with a strong acid

3. Weak acid titrated with a strong base

4. Weak base titrated with a strong acid

5. Polyprotic acid titrated with a strong base

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Example – you have 50.0 mL of 0.200 M HNO

3 and

look to titrate it with 0.100 M NaOH

Find initial pH

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3 and

Strong Acid titrated with Strong

Base

p

H

Volume added

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3 and

Find pH at halfway point

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Upon adding strong base, there is a very SMALL

change as long as there is still excess acid.

Strong Acid titrated with Strong

Base

p

H

Volume added

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3 and

Find pH at equivalence point

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As the equivalence point is reached, there begins to

be a dramatic jump in pH as moles acid = moles base

Strong Acid titrated with Strong

Base

p

H

Volume added

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Even just beyond the equivalence point, the addition

of excess base leads to a basic pH

Strong Acid titrated with Strong

Base

p

H

Volume added

Excess base now = jump to high pH Midpoint =

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3 and

Find pH after 150.0 mL of NaOH has been added

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As more excess strong base is added, there is NOT a

dramatic change in pH

Strong Acid titrated with Strong

Base

p

H

Volume added

Same excess base = no dramatic

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OPPOSITE!!!!!

To find initial pH –

To find pH at halfway point –

To find pH at equivalence point –

To find pH beyond equivalence point -

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OPPOSITE!!!!!

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Create a titration curve

If molarity of original acid isn’t known, you can use

the curve to find its molarity. HOW??? That’s what you’re designing!!!

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Example – you have 50.0 mL of 0.100 M HC

2H3O2 with

0.100 M NaOH (K_a of acetic acid = 1.8 x 10-5)

Find initial pH

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Still starts at a low pH (though not quite as low) with

weak acid in beaker

WEAK Acid titrated with Strong

Base

p

H

Volume added

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The REACTION that occurs is now slightly different:

HC₂H₃O₂ + OH-  C

2H3O2- + H2O

WEAK Acid titrated with Strong Base

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The REACTION that occurs is now slightly different:

HC₂H₃O₂ + OH-  C

2H3O2- + H2O

WEAK Acid titrated with Strong Base

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Initial spike in pH due to formation of the conjugate base. Then

mostly levels off as with both weak acid and conjugate base you have a BUFFER.

WEAK Acid titrated with Strong

Base

p

H

Volume added

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2H3O2 with

Find pH at halfway point

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Ratio of weak acid to conjugate base is the same

Half of weak acid used up, equal amount of conjugate base

formed

pH = pKa!!!

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2H3O2 with

Find pH at equivalence point

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As the equivalence point is reached, there again is the point

where the base neutralizes all of the acid and there is a significant jump in pH. The pH at the equivalence point is GREATER than 7 since the GOOD conjugate base remains.

WEAK Acid titrated with Strong

Base

p

H

Volume added

Dramatic jump up as equivalence point is reached again

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Even just beyond the equivalence point, the addition

of excess base leads to a basic pH

WEAK Acid titrated with Strong Base

p

H

Volume added

Excess base now = jump to high pH Midpoint =

equivalence point =

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As more excess strong base is added, there is NOT a

dramatic change in pH

WEAK Acid titrated with Strong

Base

p

H

Volume added

Same excess base = no dramatic

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2H3O2 with

Find pH after 100.0 mL NaOH added

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OPPOSITE!!!!!

To find initial pH –

To find pH at halfway point –

To find pH at equivalence point –

To find pH after equivalence point -

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OPPOSITE!!!!!

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Create a titration curve

If the identity and the K

a / Kb of a weak acid or a weak

base are not known, you can determine it. HOW??? That’s what you’re designing!!!

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Acids that have more than one H+

That means there will be more than one equivalence point!!

 Each H+ reacted must be neutralized with a stoichiometric amount

of base

Example – H

2CO3 with NaOH

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Titration of a polyprotic acid with a strong base.

What would this look like? Why would it look that way?