IB DP Chemistry: HL复习笔记18.2.2 Acid & Base Problem Solving

pH

• The acidity of an aqueous solution depends on the number of H₃O⁺ ions in solution
• pH is defined as:

pH = -log [H₃O⁺]

• • Where [H₃O⁺] is the concentration of H₃O⁺ ions in mol dm^–3
• Similarly, the concentration of H⁺ of a solution can be calculated if the pH is known by rearranging the above equation to:

[H₃O⁺] = 10^-pH

• The pH scale is a logarithmic scale with base 10
• This means that each value is 10 times the value below it
  • For example, pH 5 is 10 times more acidic than pH 6
• pH values are usually given to 2 decimal places

pOH

• The basicity of an aqueous solution depends on the number of hydroxide ions, OH^-, in solution
• pOH is defined as:

pOH = -log [OH^-]

• • Where [OH^-] is the concentration of hydroxide ions in mol dm^–3
• Similarly, the concentration of OH^- of a solution can be calculated if the pH is known by rearranging the above equation to:

[OH^-] = 10^-pOH

• If you are given the concentration of a basic solution and need to find the pH, this can be done by:

[H₃O⁺] = K_w / [OH^-]

• Alternatively, if you are given the [OH^-] and calculate the pOH, the pH can be found by:

pH = 14 - pOH

Answers

Answer 1:

The pH of the solution is:

• • pH = -log [H₃O⁺]
    • pH = -log 1.6 x 10^-4
    • pH = 3.80

Answer 2:

The hydrogen concentration can be calculated by rearranging the equation for pH

• • pH = -log [H₃O⁺]
  • [H₃O⁺] = 10^-pH
    • [H₃O⁺] = 10^-3.10
    • [H₃O⁺] = 7.94 x 10^-4 mol dm^-3

Answers

Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na⁺ (aq) + OH^– (aq) 

Answer 1:

The pH of the solution is:

• • [H⁺] = K_w  ÷ [OH^–]
    • [H⁺] = (1 x 10^-14) ÷ 0.15 = 6.66 x 10^-14
  • pH = -log[H⁺]
    • pH = -log 6.66 x 10^-14  = 13.17

Answer 2

Step 1: Calculate hydrogen concentration by rearranging the equation for pH

• • pH = -log[H⁺]
  • [H⁺] = 10^-pH
    • [H⁺] = 10^-10.50
    • [H⁺] = 3.16 x 10^-11 mol dm^-3

Step 2: Rearrange the ionic product of water  to find the concentration of hydroxide ions

• • K_w = [H⁺] [OH^–]
  • [OH^–] = K_w  ÷  [H⁺]

Step 3: Substitute the values into the expression to find the concentration of hydroxide ions

• • Since K_w is 1 x 10^-14 mol² dm^-6
    • [OH^–] = (1 x 10^-14)  ÷  (3.16 x 10^-11)
    • [OH^–] = 3.16 x 10^-4 mol dm^-3

K_a, pK_a, K_b and pK_b

• In reactions of weak acids and bases, we cannot make the same assumptions as for the ionisation of strong acids and bases
• For a weak acid and its conjugate base, we can use the equation:

K_w = K_a K_b 

• By finding the -log of these, we can use:

pK_w = pK_a + pK_b 

• Remember, to convert these terms you need to use:

pK_a = -logK_a                 K_a= 10^–pK_a

pK_b = -logK_b                 K_b= 10^–pK_b

• The assumptions we must make when calculating values for K_a, pK_a, K_b and pK_b are:
  • The initial concentration of acid ≈ the equilibrium concentration of acid
  • [A^-] = [H₃O⁺]
  • There is negligible ionisation of the water, so [H₃O⁺] is not affected
  • The temperature is 25 °C

Answer

Step 1: Calculate [H₃O⁺] using

• • [H₃O⁺] = 10^-pH
    • [H₃O⁺] = 10^-4.88
    • [H₃O⁺] = 1.3182 x 10^-5

Step 2: Substitute values into K_a expression (include image)

• • K_a = [H₃O⁺]² / [CH₃CH₂COOH]
    • K_a = (1.3182 x 10^-5) ² / 0.2
    • K_a = 8.70 × 10^-10 mol dm^-3

Answer

Step 1: Calculate the value for K_b using

• • K_b = 10^–pK_b
    • K_b= 10^-3.35
    • K_b = 4.4668 x 10^-4

Step 2: Substitute values into K_b expression to calculate [OH^-]

• • K_b = [OH^-]² / [CH₃NH₂]
  • 4.4668 x 10^-4 = [OH^-]² / 0.035
  • [OH^-] = √(4.4668 x 10^-4 x 0.035)
  • [OH^-] = 3.9539 x 10^-3

Step 3: Calculate the pH

• • [H⁺] = K_w  ÷ [OH^–]
    • [H⁺] = (1 x 10^-14) ÷ 3.9539 x 10^-3
    • [H⁺] = 2.5290 x 10^-12
  • pH = -log [H⁺]
    • pH = 2.5290 x 10^-12
    • pH = 11.60 to 2 decimal places 

OR

Step 3: Calculate pOH and therefore pH

• • pOH = -log [OH^-]
    • pOH = -log 3.9539 x 10^-3
    • pOH = 2.4029

• • pH = 14 - pOH
    • pH = 14 - 2.4029
    • pH = 11.60 to 2 decimal places