Buffers and Titrations SSS 3 Class 3. Materials 0.5 M Solutions VinegarNaOH HCl ammonia Sodium...

Buffers and Titrations

SSS 3 Class 3

Materials 0.5 M Solutions

Vinegar NaOHHCl

ammonia Sodium Acetate Ammonium Chloridewater

buffer

Buffers

• What is a Buffer?– A Solution made of a weak acid and salt of weak

acid ( ion is the acid’s conjugate base) or weak base and salt of weak base ( ion is the base’s conjugate acid)

• What are the advantages of having a Buffer?– Solution will not significantly change in pH when

an acid or base are added.

ExampleHCN(aq) + H2O(l) H3O+(aq) + CN-(aq) weak acid

• Addition of NaCN (salt contain ion that is the conjucate base, CN-) will shift the equilibrium to the left because of the addition of CN-, which is already involved in the equilibrium reaction.

• A solution of HCN and NaCN is less acidic than a solution of HCN alone and is a buffer solution

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Buffers demonstrate theCommon Ion Effect

• Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction.

• The salt dissociates to produce an ion that is the same as the ion that is a product in the weak acid reaction with water

• An application of Le Châtelier’s principle.

Identifying a Buffer

1. Write the net ionic equation for the reaction or solution

2. Look for a weak acid and conjugate base ion or weak base and conjugate acid ion

3. At equilibrium is there a concentration for the reactant( Weak Acid or Weak base) and product ( conjugate base or acid)

Henderson–Hasselbalch Equationthe easiest way to solve buffer problems

or [H3O+] = Ka x [acid]/[base]


a

ApH = p + log

HA

K

Buffering: How Does It Work?


What is the pH of a buffer solution that is 0.45 M acetic acid (HC2H3O2) and 0.85 M sodium acetate (NaC2H3O2)? The Ka for acetic acid is 1.8 × 10–5.

pH = 5.02


Problem 1

Buffered Solution Calculations

• Buffers contain relatively large concentrations of a weak acid and corresponding conjugate base.

Added H+ reacts to completion with the weak base.[H3O+] = Ka x [acid]+[H+ ]/[base] – [H+ ]

Added OH- reacts to completion with the weak acid.

[H3O+] = Ka x [acid]-[OH- ]/[base] + [OH- ]



• The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH.

• Determined by the magnitudes of [HA] and [A–].• A buffer with large capacity contains large

concentrations of the buffering components.• Optimal buffering occurs when [HA] is equal to [A–].• It is for this condition that the ratio [A–] / [HA] is

most resistant to change when H+ or OH– is added to the buffered solution.


Buffering Capacity

Choosing a Buffer

• pKa of the weak acid to be used in the buffer should be as close as possible to the desired pH.


Titration

• Laboratory procedure used to determine the concentration of a solution or molar mass of a unknown substance by reacting unknown with known concentration of a solution

• Can only be used if the stoichiometric endpoint can be monitored (color change, conductivity or pH graph)

• Plotting the pH of the solution being analyzed as a function of the amount of titrant added.

• Equivalence (Stoichiometric) Point – point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated.

• Endpoint is point of color change with indicator


The pH Curve for the Titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH


Strong Acid/Strong Base Titration

Features:1. Low initial pH. (Excess Acid)2. Very little change until endpoint 3. Sharp rise in pH at endpoint

(neutral salt)4. High pH after endpoint (Excess Base)

Strong Acid/Strong Base Titration

A stoichiometry problem (reaction is assumed to run to completion) then determine concentration of acid or base remaining. The

salt formed is neutral. HCl + NaOH H2O + NaCl

I CEWork in Moles! Then calculate concentration (Molarity)!

You will need this for pH.

The pH Curve for the Titration of 100.0 mL of 0.50 M NaOH with 1.0 M HCI


The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH


Weak Acid /Strong Base

Features:1. Higher initial pH (Weak Acid)2. As base is added the pH climbs slightly

(Buffering)3. Equivalence point is greater than 7.0 (basic salt)4. pH rises rapidly after equivalence point (Excess base)5. pH = pKa ½ distance to equivalence point

Weak Acid–Strong Base Titration

Step 1: A stoichiometry problem (reaction is assumed to run to completion) then

determine concentration of acid remaining, conjugate base formed, or excess base added.

At the end you will have : weak acid, buffer, salt or, excess base

Step 2: An equilibrium problem is next if you have weak acid, buffer, or a salt.

Determine Molarity of Hydronium ions or hydroxide ions and then calculate pH


Calculate the pH of a solution at the equivalence point when 100.0 mL of a 0.100 M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–5, is titrated with a 0.10 M NaOH solution.

pH = 8.72


Problem 2

Calculate the pH of a solution made by mixing 100.0 mL of a 0.100 M solution of acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–5, and 50.0 mL of a 0.10 M NaOH solution.

pH = 4.74


Problem 3

• Weak acids or bases that change color when equilibrium shifts

• Marks the end point of a titration by changing color.• The equivalence point is not necessarily the same as

the end point (but they are ideally as close as possible).

• Choose a indicator that has a pKa close to pH at the endpoint in the titration.


Indicators

The Acid and Base Forms of the Indicator Phenolphthalein


Acid Base

Multiple Choice 1• In the titration of a weak acid of unknown concentration

with a standard solution of a strong base, a pH meter was used to follow the progress of the titration. Which of the following is true for this experiment?

• (A) The pH is 7 at the equivalence point.(B) The pH at the equivalence point depends on the indicator used.(C) The graph of pH versus volume of base added rises gradually at first and then much more rapidly.(D) The graph of pH versus volume of base added shows no sharp rise.(E) The [H+] at the equivalence point equals the ionization constant of the acid.

Multiple Choice 2Questions 1-5 refer to aqueous solutions containing 1:1 mole ratios of the following pairs of substances. Assume all concentrations are 1 M. (A) NH3 and NH4Cl(B) H3PO4 and NaH2PO4

(C) HCl and NaCl(D) NaOH and NH3

(E) NH3 and HC2H3O2 (acetic acid)

1. The solution with the lowest pH 2. The most nearly neutral solution3. A buffer at a pH > 84. A buffer at a pH < 6

1. C2. E3. A4. B

Multiple Choice 3

• Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.0 milliliters of the vinegar 26.7 milliliters of 0.600-molar NaOH solution was required. What was the concentration of acetic acid in the vinegar if no other acid was present?

(A) 1.60 M (B) 0.800 M (C) 0.600 M (D) 0.450 M (E) 0.200 M

The graph below shows the titration curve that results when 100. mL of 0.0250 M acetic acid is titrated with 0.100 M NaOH.

Multiple Choice 4Which of the following indicators is the best choice for this titration? indicator pH of color change

(A) Methyl orange 3.2 - 4.4(B) Methyl red4.8 - 6.0(C) Bromothymol blue 6.1 - 7.6(D) Phenolphthalein 8.2 - 10.0(E) Alizarin 11.0 - 12.4

Multiple Choice 5

What part of the curve corresponds to the opti mum buffer action for the acetic acid/acetate ion pair?(A) Point V (B) Point X(C) Point Z (D) Along all of section WY

Multiple Choice 6

Which of the following represents acceptable lab oratory practice?(A) Placing a hot object on a balance pan(B) Using distilled water for the final rinse of a buret before filling it with standardized solu tion(C) Adding a weighed quantity of solid acid to a titration flask wet with distilled water(D) Using 10 mL of standard strength phenolph thalein indicator solution for titration of 25 mL of acid solution(E) Diluting a solution in a volumetric flask to its final concentration with hot water

Multiple Choice 7

Mixtures that would be considered buffers in clude which of the following?

I. 0.10 M HCl + 0.10 M NaClII. 0.10 M HF + 0.10 M NaFIII. 0.10 M HBr + 0.10 M NaBr

(A) I only (B)II only (C) III only (D) I and II (E) II and III

Free Response 11972

Given a solution of ammonium chloride. What additional reagent or reagents are needed to prepare a buffer from the ammonium chloride solution?Explain how this buffer solution resists a change in pH when:(a) Moderate amounts of strong acid are added.(b) Moderate amounts of strong base are added.(c) A portion of the buffer solution is diluted with an equal volume of water.

• Answer:• Since ammonium chloride is a salt of a weak base, the weak base is needed,

ammonia, NH3.• (a) When moderate amounts of a strong acid, H+, are added, the ammonia

reacts with it. The concentration of the hydrogen ion remains essentially the same and therefore only a very small change in pH.

• NH3 + H+ <=> NH4+

• (b) When moderate amounts of a strong base, OH-, are added, the ammonium ion reacts with it. The concentration of the hydrogen ion remains essentially the same and therefore only a very small change in pH.

• NH4+ + OH- <=> NH3 + H2O

• (c) By diluting with water the relative concentration ratio of [NH4+]/[NH3]

does not change, therefore there should be no change in pH.•

Free Response 2A sample of 40.0 milliliters of a 0.100 molar HC2H3O2 solution is titrated with a 0.150 molar NaOH solution. Ka for acetic acid = 1.8x10-5

(a) What volume of NaOH is used in the titration in order to reach the equivalence point?(b) What is the molar concentration of C2H3O2

- at the equivalence point?(c) What is the pH of the solution at the equivalence point?

Answer:(a) MaVa=MbVb

(0.100M)(40.0 mL) = (0.150M)(Vb)

Vb = 26.7 mL

(b) acetate ion is a weak base withKb=Kw/Ka = 1.010-14/1.810-5 = 5.610-10

[CH3COO-]eq = 0.600M -X

[OH-] = [CH3COOH] = X

0.0600M-9.6610-5M = 0.0599M [CH3COO-]eq

(c) pH = -log [H+] = -log(1.0410-10) = 9.98

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Buffers and Titrations SSS 3 Class 3. Materials 0.5 M Solutions VinegarNaOH HCl ammonia Sodium...

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