Edexcel A Level Chemistry:复习笔记5.2.7 Buffers

Action of a Buffer Solution

 

  • A buffer solution is a solution which resists changes in pH when small amounts of acids or alkalis are added
    • A buffer solution is used to keep the pH almost constant
    • A buffer can consists of weak acid - conjugate base or weak base - conjugate acid

     

Ethanoic acid & sodium ethanoate as a buffer

  • A common buffer solution is an aqueous mixture of ethanoic acid and sodium ethanoate
  • Ethanoic acid is a weak acid and partially ionises in solution to form a relatively low concentration of ethanoate ions

5.-Buffers-equation-1

 

  • Sodium ethanoate is a salt which fully ionises in solution

5.-Buffers-equation-2

  • There are reserve supplies of the acid (CH3COOH) and its conjugate base (CH3COO-)
    • The buffer solution contains relatively high concentrations of CH3COOH (due to partial ionisation of ethanoic acid) and CH3COO- (due to full ionisation of sodium ethanoate)
  • In the buffer solution, the ethanoic acid is in equilibrium with hydrogen and ethanoate ions

5.-Buffers-equation-3

  • When H+ ions are added:
    • The equilibrium position shifts to the left as H+ ions react with CH3COO- ions to form more CH3COOH until equilibrium is re-established
    • As there is a large reserve supply of CH3COO- the concentration of CH3COO- in solution doesn’t change much as it reacts with the added H+ ions
    • As there is a large reserve supply of CH3COOH the concentration of CH3COOH in solution doesn’t change much as CH3COOH is formed from the reaction of CH3COO- with H+
    • As a result, the pH remains reasonably constant

5.5-Equilibria-Effect-of-adding-H

When hydrogen ions are added to the solution the pH of the solution would decrease; However, the ethanoate ions in the buffer solution react with the hydrogen ions to prevent this and keep the pH constant

  • When OH- ions are added:
    • The OH- reacts with H+ to form water

OH- (aq) + H+  (aq) → H2O (l)

    • The H+ concentration decreases
    • The equilibrium position shifts to the right and more CH3COOH molecules ionise to form more H+and CH3COO- until equilibrium is re-established

CH3COOH (aq) → H+ (aq) + CH3COO- (aq)

    • As there is a large reserve supply of CH3COOH the concentration of CH3COOH in solution doesn’t change much when CH3COOH dissociates to form more H+ ions
    • As there is a large reserve supply of CH3COO- the concentration of CH3COO- in solution doesn’t change much
    • As a result, the pH remains reasonably constant

5.5-Equilibria-Effect-of-adding-OH

When hydroxide ions are added to the solution, the hydrogen ions react with them to form water; The decrease in hydrogen ions would mean that the pH would increase however the equilibrium moves to the right to replace the removed hydrogen ions and keep the pH constant

Buffers & pH Calculations

  • The pH of a buffer solution can be calculated using:
    • The Ka of the weak acid
    • The equilibrium concentration of the weak acid and its conjugate base (salt)

     

  • To determine the pH, the concentration of hydrogen ions is needed which can be found using the equilibrium expression

6.-Calculating-pH-of-Buffer-Solutions-equation-1

  • To simplify the calculations, logarithms are used such that the expression becomes:

6.-Calculating-pH-of-Buffer-Solutions-equation-2

  • Since -log10 [H+] = pH, the expression can also be rewritten as:

5.6.5-Hendersen-equation-3

  • This is known as the Hendersen-Hasselbalch equation

Worked Example

Calculate the pH of a buffer solution containing 0.305 mol dm-3 of ethanoic acid and 0.520 mol dm-3 sodium ethanoate.

The Ka of ethanoic acid  = 1.74 × 10-5 mol dm-3 at 298 K

Answer

Ethanoic acid is a weak acid that ionises as follows:

CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

Step 1: Write down the equilibrium expression to find Ka

6.-Calculating-pH-of-Buffer-Solutions-equation-4

Step 2: Rearrange the equation to find [H+]

 

6.-Calculating-pH-of-Buffer-Solutions-equation-5

 

Step 3: Substitute the values into the expression

5.6.5

   = 1.02 x 10-5 mol dm-3

Step 4: Calculate the pH

   pH = - log [H+]

   = -log 1.02 x 10-5

   = 4.99

How to make a buffer solution with a required pH

  • To make a buffer solution with a pH of less than 7, you need to use a mixture of a weak acid and its conjugate base
  • Conversely, you can make a buffer solution with a pH greater than 7 by using a mixture of a weak base and its conjugate acid
  • Imagine we want to make a buffer solution with a pH of 5.00 at a temperature of 298K
  • This would require a hydrogen ion concentration of

      [H+(aq)] = 1.00 x 10-5 mol dm-3

  • The hydrogen ion concentration of a buffer solution of a weak acid and its conjugate base is calculated using the formula

   

  • We will use ethanoic acid as our weak acid of choice

      Ka = 1.74 x 10-5 mol dm-3

  • Substituting our known values into the equation we get

   

  • This gives a value for the ratio of the concentrations of acid and base needed in our buffer solution

   

  • Mixing an equal volume of ethanoic acid with a concentration of 0.575 mol dm-3 and a sodium ethanoate solution of 1.00 mol dm-3 would allow us to make this buffer solution
  • This would give us a solution with an acid concentration of 0.2875 mol dm-3 and a salt concentration of 0.500 mol dm-3

   

Buffer Applications

Controlling the pH of blood

  • In humans, HCO3- ions act as a buffer to keep the blood pH between 7.35 and 7.45
  • Body cells produce CO2 during aerobic respiration
  • This CO2 will combine with water in blood to form a solution containing H+ ions

CO2 (g) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)

  • This equilibrium between CO2 and HCO3- is extremely important
  • If the concentration of H+ ions is not regulated, the blood pH would drop and cause ‘acidosis’
    • Acidosis refers to a condition in which there is too much acid in the body fluids such as blood
    • This could cause body malfunctioning and eventually lead to coma

     

  • If there is an increase in H+ ions
  • The equilibrium position shifts to the left until equilibrium is restored

H+ (aq) + HCO3- (aq) ⇌ CO2 (g) + H2O (l)

  • This reduces the concentration of H+ and keeps the pH of the blood constant
  • If there is a decrease in H+ ions
    • The equilibrium position shifts to the right until equilibrium is restored

    CO2 (g) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)

  • This increases the concentration of H+ and keeps the pH of the blood constant

转载自savemyexams

更多Alevel课程
翰林国际教育资讯二维码