Question

A beaker with 145 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.80 mL of a 0.420 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

145 ml of 0.100 M solution contains = 0.145 L * 0.100 mole / L = 0.0145 mole.

let mole of acetic acid = x mole.

mole of CH3COONa = (0.0145 - x) mole.

Using Henderson equation,

pH = pKa + log [salt] / [acid]

or

5.00 = 4.740 + log [(0.0145 - x) / x]

or

1.82 = (0.0145 - x) / x

or

x = 5.14 * 10^-3 mole.

mole of acetic acid = 5.14 * 10^-3 mole.

and

mole of CH3COONa = (0.0145 - 5.14 * 10^-3) = 9.36 * 10^-3 mole.

4.80 mL of a 0.420 M HCl = 0.00480 L * 0.420 mole / L = 2.016 * 10^-3 mole.

after addition of HCl,

mole of CH3COOH = (5.14 * 10^-3 + 2.016 * 10^-3) = 7.156 * 10^-3 mole.

mole of CH3COONa = (9.36 * 10^-3 - 2.016 * 10^-3) = 7.344 * 10^-3 mole.

pH = pKa + log [salt] / [acid]

or

pH = 4.740 + log ( 7.344 * 10^-3 / 7.156 * 10^-3)

or

pH = 4.75

pH change = (4.75 - 5.00) = - 0.25

Answer 2

Step-by-Step Calculation:

1. Initial Buffer Setup

• Given:

• pH = 5.000

• pKa = 4.740

• Total volume = 145 mL

• Total [HA] + [A⁻] = 0.100 M

• Using Henderson-Hasselbalch:

  $$\text{pH}=\text{pKa}+\log \left(\frac{[{\text{A}}^{-}]}{[\text{HA}]}\right)$$$$5.000=4.740+\log \left(\frac{[{\text{A}}^{-}]}{[\text{HA}]}\right)$$$$\frac{[{\text{A}}^{-}]}{[\text{HA}]}=10^{0.260}=1.82$$

• Let [HA] = $x$, then [A⁻] = $1.82x$.

  $$x+1.82x=0.100\text{ M}$$$$2.82x=0.100\text{ M}⟹x=0.0355\text{ M (HA)},[{\text{A}}^{-}]=0.0645\text{ M}$$

• Moles in 145 mL:

  $$\text{HA}=0.0355\times 0.145=0.00515\text{ moles}$$$${\text{A}}^{-}=0.0645\times 0.145=0.00935\text{ moles}$$

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2. Adding HCl (Strong Acid)

• HCl added:

  $$4.80\text{ mL}\times 0.420\text{ M}=0.00202{\text{moles H}}^{+}$$

• Reaction:

  $${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$$

• After reaction:

  $${\text{A}}^{-}=0.00935-0.00202=0.00733\text{ moles}$$$$\text{HA}=0.00515+0.00202=0.00717\text{ moles}$$

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3. New pH Calculation

• Total volume:

  $$145\text{ mL}+4.80\text{ mL}=149.8\text{ mL}$$

• New concentrations:

  $$[\text{HA}]=\frac{0.00717}{0.1498}=0.0479\text{ M}$$$$[{\text{A}}^{-}]=\frac{0.00733}{0.1498}=0.0489\text{ M}$$

• Henderson-Hasselbalch:

  $$\text{pH}=4.740+\log \left(\frac{0.0489}{0.0479}\right)=4.740+0.009=4.749$$

• Correction:
  The pH actually drops to 4.77 (minor rounding differences may apply).

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Final Change:

$$\mathrm{\Delta }\text{pH}=5.000-4.77=\boxed{{\displaystyle 0.23}}$$