Buffer Calculations (Cambridge International A-Level Chemistry)

Buffer Solution pH Calculations

• The pH of a buffer solution can be calculated using:

  • The K_a* *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:

K_a= $\frac{[ \text{salt} ] [ H^{+} ]}{[ \text{acid} ]}$

• This can be rearranged to determine the concentration of hydrogen ions:

[H⁺] =$K_{ } a \times \frac{ \text{stretchy} [ \text{acid} \text{stretchy} ]}{ \text{stretchy} [ \text{salt} \text{stretchy} ]}$

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

–log₁₀[H⁺] =$- log_{ } 10 K_{ } a \times - log_{ } 10 \frac{ \text{stretchy} [ \text{acid} \text{stretchy} ]}{ \text{stretchy} [ \text{salt} \text{stretchy} ]}$

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

$pH = p K_{ } a a + log_{ } 10 \frac{ \text{stretchy} [ \text{salt} \text{stretchy} ]}{ \text{stretchy} [ \text{acid} \text{stretchy} ]}$

Worked Example Calculating the pH of a buffer solution

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

The K_a of ethanoic acid = 1.43 x 10^-5 mol dm^-3.

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 K_a:

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

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

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

• Step 3: Substitute the values into the expression:

[H⁺] = 1.43 x 10^-5 x $\frac{[ 0.305 ]}{[ 0.520 ]}$

[H⁺] = 8.39 x 10^-6 mol dm^-3

• Step 4: Calculate the pH:

pH = –log₁₀ [H⁺]

pH = –log 8.39 x 10^-6

pH = 5.08