pH, Ka, pKa & Kw Calculations (Cambridge International A-Level Chemistry)

Calculating pH, Ka, pKA & Kw

pH

• The pH indicates the acidity or basicity of an acid or alkali

• The pH scale goes from 0.0 to 14.0

  • Acids have a pH below 7.0
  • Pure water is neutral with a pH of 7.0
  • Bases and alkalis have a pH above 7.0

• pH can be calculated using:

pH = -log₁₀[H⁺]

• where [H⁺] = concentration of H⁺ ions (mol dm^-3)
• The pH can also be used to calculate the concentration of H⁺ ions in solution by rearranging the equation to:

[H⁺] = 10^-pH

Worked Example Calculating the pH of acids

Calculate the pH of ethanoic acid, at 298K, when the hydrogen ion concentration is 1.32 x 10^-3 mol dm^-3.

Answer

pH = -log [H⁺]

pH = -log 1.32 x 10^-3

pH = 2.9

K_a & pK_a

• The K_a is the acidic dissociation constant

  • It is the equilibrium constant for the dissociation of a weak acid at 298 K

• For the partial ionisation of a weak acid HA, the equilibrium expression to find K_a is:

HA (aq) ⇌ H⁺(aq) + A^-(aq)

K_a=$\frac{[ H + ] [ A^{-} ]}{[ \text{HA} ]}$

• When writing the equilibrium expression for weak acids, the following assumptions are made:

  • The concentration of hydrogen ions due to the ionisation of water is negligible

  • The dissociation of the weak acid is so small that the concentration of HA can be approximated by its initial value

    • This is calculated from the number of moles divided by the volume of solvent

• The value of K_a indicates the extent of dissociation

  • A high value of K_a means that:

    • The equilibrium position lies to the right
    • The acid is almost completely ionised
    • The acid is strongly acidic

  • A low value of K_a means that:

    • The equilibrium position lies to the left
    • The acid is only slightly ionised (there are mainly HA and only a few H⁺ and A^- ions)
    • The acid is weakly acidic

• Since K_a values of many weak acids are very low, pK_a* *values are used instead to compare the strengths of weak acids with each other

pK _a = -log ₁₀ K_a

• The less positive the pK_a value the more acidic the acid is

Worked Example Calculating the K_a & pK _a of weak acids **

Calculate the K_a* and pK*_a values of 0.100 mol dm^-3 ethanoic acid at 298K which forms 1.32 x 10^-3 of H⁺ ions in solution.

Answer

-** Step 1: **Write down the equation for the partial dissociation of ethanoic acid:

CH₃COOH (aq) ⇌ H⁺ (aq) + CH₃COO^- (aq)

-** Step 2: Write down the equilibrium expression to find K_a:

K_a = $\frac{[ \mathrm{H}^{+} ] [ CH_{3} COO^{-} ]}{[ CH_{3} \text{COOH} ]}$

• Step 3: Simplify the expression:

The ratio of H⁺ to CH₃COO^- is 1:1

The concentration of H⁺ and CH₃COO^- is, therefore, the same

The equilibrium expression can be simplified to:

K_a = $\frac{[ \mathrm{H}^{+} ]^{2}}{[ CH_{3} \text{COOH} ]}$

• Step 4: Substitute the values into the expression to find K_a:

K_a = $\frac{[ 1.32 \times 10^{- 3} ]^{2}}{[ 0.100 ]}$

K_a = 1.74 x 10^-5

• Step 5: Determine the units of K_a:

K_a = $\frac{[ \text{mol} dm^{- 3} ]^{2}}{[ \text{mol} dm^{- 3} ]}$= mol dm^-3

Therefore, K_a is 1.74 x 10^-5 mol dm^-3

• Step 6: Find pK_a:

pK_a = - log₁₀ K_a

pK_a* =* - log₁₀ (1.74 x 10^-5)

pK_a = 4.76

K_w

• The K_w is the ionic product of water

  • It is the equilibrium constant for the dissociation of water at 298 K
  • Its value is 1.00 x 10^-14 mol² dm^-6

• For the ionisation of water, the equilibrium expression to find K_w is:

H₂O (l) ⇌ H⁺(aq) + OH^-(aq)

K_w=$\frac{[ H^{+} ] [ OH^{-} ]}{[ H_{2} O ]}$

• As the extent of ionisation is very low, only small amounts of H⁺ and OH^- ions are formed
• The concentration of H₂O can therefore be regarded as constant and removed from the K_w expression
• The equilibrium expression therefore becomes:

K_w= [H⁺] [OH^-]

• As the [H⁺] = [OH⁺] in pure water, the equilibrium expression can be further simplified to:

K_w= [H⁺]²

Worked Example Calculating the concentration of H⁺of pure water

Calculate the concentration of H⁺ in pure water, using the ionic product of water

Answer

• **Step 1: **Write down the equation for the partial dissociation of water:

  • In pure water, the following equilibrium exists:

H₂O (l) ⇌ H⁺(aq) + OH^-(aq)

• **Step 2: **Write down the equilibrium expression to find K_w:

K_w** =**$\frac{[ H^{+} ] [ OH^{-} ]}{[ H_{2} O ]}$

• **Step 3: **Simplify the expression:

  • Since the concentration of H₂O is constant, this expression can be simplified to:

K_w** = [H**⁺] [OH^-]

• **Step 4: **Further simplify the expression:

  • The ratio of H⁺ to OH^- is 1:1
  • The concentration of H⁺ and OH^- is, therefore, the same and the equilibrium expression can be further simplified to:

K_w** = [H**⁺]²

• Step 5: Rearrange the equation to find [H⁺]:

[H⁺] = $\sqrt{K_{w}}$

• Step 6: Substitute the values into the expression to find K_w:

[H⁺] = $\sqrt{1.00\times10^{-14}}$

[H⁺] = 1.00 x 10^-7 mol dm^-3

Exam Hint

• The greater the K_a* *value, the **more strongly acidic **the acid is
• The greater the pK_a value, the **less strongly acidic **the acid is.
• Also, you should be able to rearrange the following expressions:

pH = -log₁₀** [H**⁺] ⇔ [H⁺] = 10^-pH

K_a=$\frac{[ H^{+} ] [ A^{-} ]}{[ \text{HA} ]}$⇔ [H⁺] =equation

pK _a = - log ₁₀ K_a ⇔ K_a = 10^-pKa