General Chem Buffers

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CHM 11600 Daily Page – Lecture 14 Monday, February 25, 2013

LEARNING OBJECTIVES  Be able to define a buffer

(mixture of conjugate acid-base pair).

 Be able to calculate how to prepare a buffer.

 Be able to use Henderson-Hasselbalch equation to calculate pH of a buffer.

 Be able to describe how buffer solution can resist change of pH by adding strong acid/base or dilution.

ASSIGNMENTS  CONNECT Homework:

 HW13 due today, 11:59pm  HW14 due Thurs., Feb 28

 Reading for Wednesday:  Silberberg: 19.2

 Laboratory (Thurs/Fri):  Lab 8: Acid-base titrations

 Exam II:  Next Tuesday, March 5,

6:30pm-7:30pm, Elliott Hall  Review in class time this

Friday (March 1)

Acid-Base Buffers

 A Buffer Solution is one that resists a large change in pH upon the addition of a strong acid or strong base.

 Buffer solutions in water typically have a weak acid – weak base conjugate pair, both present in the solution at roughly equal concentrations.

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Two solutions:

 pure water  0.05 M HF, 0.05 M F–

add 0.001 mol HCl to 1.00 L of each solution

Is it a buffer?

 A solution made by mixing 500 mL of 0.10 M acetic acid (HC2H3O2) and 500 mL of 0.08 M sodium acetate (NaC2H3O2).

(A)  Yes (B)  No (C)  Maybe (D)  Need more information (E)   I don’t know

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Is it a buffer?

 A solution made by mixing 200 mL of 0.10 M nitrous acid (HNO2) and 120 mL of 0.10 M potassium hydroxide (KOH).


Is it a buffer?

 A solution made by mixing 250 mL of 0.10 M acetic acid (HC2H3O2) and 250 mL of water.


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Is it a buffer?

 A solution made by mixing 200 mL of 0.05 M perchloric acid (HClO4) and 200 mL of 0.08 M potassium fluoride (KF).


Is it a buffer?

 A solution made by mixing 250 mL of 0.30 M hydrochloric acid (HCl) and 100 mL of 0.75 M sodium acetate (NaC2H3O2).


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HA + H2O ⇔ H3O+ + A-

Henderson-Hasselbalch Equation

€

Ka =A−[ ] H3O

+[ ]HA[ ]

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logKa = logA−[ ]HA[ ]

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ + log H3O

+[ ]

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−log H3O+[ ] = −logKa + log

A−[ ]HA[ ]

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

€

pH = pKa + logA−[ ]HA[ ]

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟


 If [HA] = [A–], reduces to pH = pKa  Buffers work near the pKa of the conjugate

acid of the pair.  Can use to calculate the pH of a buffer.  Can use to calculate the desired ratio of

“component concentrations” to make a buffer of a desired pH.

€


⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

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€


⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

 What is the pH of a buffer that has equilibrium concentrations: [HCN] = 0.700 M [CN–] = 0.600 M

(the Ka for HCN is 6.2 x 10–10)

Practice Problem

 What is the pH of a solution made by mixing 200 mL of 0.05 M perchloric acid (HClO4) and 200 mL of 0.08 M potassium fluoride (KF)?  We said a few slides ago that this is a buffer.  The Ka of HF is 7.2 x 10–4.

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Buffer Range

 The buffer range is the pH range over which the buffer acts effectively. By convention, the “effective” range is within ±1 pH unit of the pKa of the weak acid component.

 Through Henderson-Hasselbalch, pH = pKa ± 1 translates to max 10:1 ratio between conjugate acid and conjugate base (a quantitative definition for our “roughly equal concentrations.”)

Buffer solutions have a limited capacity to keep the pH relatively constant. The buffering will be overcome if so much strong acid or base is added so that the ratio of conjugate base to acid (after reaction) leaves the 10:1 ratio either way.

If the conjugate acid and base components are more concentrated, the buffer capacity is greater.

Buffer Capacity

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Preparing a Buffer (in words)

1.  Mix together about equal moles of a weak acid and its conjugate base.

2.  Mix together a number of moles of a weak acid and about half that many moles of a strong base.

3.  Mix together a number of moles of a weak base and about half that many moles of a strong acid.

Practice Problem 2

 Calculate the pH of a solution that is 0.10 M HNO2 and 0.15 M NaNO2. The Ka for HNO2 is 4.0 x 10–4.

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Practice Problem 3

 Calculate the pH of a solution containing 0.50 M C2H5NH2 (aniline), and 0.25 M C2H5NH3

+. The Kb for aniline is 3.8 x 10–10.

Practice Problem 4

 How many moles of NaOH must be added to 1.00 L of a 2.00 M acetic acid (HC2H3O2) solution to create a buffer with pH = 5.0? The pKa for acetic acid is 4.74.

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Practice Problem 5

 A buffered solution is made by mixing 75.0g of solid sodium acetate into 500.0 mL of a 0.64 M solution of acetic acid. The Ka for acetic acid is 1.8 x 10–5. What is the final pH of the solution (assume no changes in volume)?

General Chem Buffers

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