How many grams of sodium acetate (CH3CO2Na, Mw = 82.0 g/mol) should be present in 1.0 L of 0.62 mole acetic acid (CH3CO2H) solution to achieve a buffer of pH = 4.61? Ka = 1.78 × 10–5 for benzoic acid. Please keep your answer to one decimal place.
The correct answer and explanation is:
Correct Answer: 23.3 grams of sodium acetate
To calculate how many grams of sodium acetate (CH₃CO₂Na) are needed to make a buffer solution with acetic acid (CH₃CO₂H) at pH = 4.61, we use the Henderson-Hasselbalch equation: pH=pKa+log([A−][HA])\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
Where:
- pH=4.61\text{pH} = 4.61 (target pH)
- HA\text{HA} is the weak acid, which is acetic acid
- A−\text{A}^- is the conjugate base, which is sodium acetate
- Ka=1.78×10−5\text{Ka} = 1.78 \times 10^{-5} for acetic acid
- [HA]=0.62[\text{HA}] = 0.62 M (given concentration of acetic acid)
Step 1: Calculate pKa pKa=−log(1.78×10−5)=4.75\text{p}K_a = -\log(1.78 \times 10^{-5}) = 4.75
Step 2: Use the Henderson-Hasselbalch equation 4.61=4.75+log([A−]0.62)4.61 = 4.75 + \log\left(\frac{[\text{A}^-]}{0.62}\right)
Step 3: Solve for [A−][\text{A}^-] 4.61−4.75=log([A−]0.62)⇒−0.14=log([A−]0.62)4.61 – 4.75 = \log\left(\frac{[\text{A}^-]}{0.62}\right) \Rightarrow -0.14 = \log\left(\frac{[\text{A}^-]}{0.62}\right)
Raise 10 to both sides: 10−0.14=[A−]0.62⇒0.72=[A−]0.6210^{-0.14} = \frac{[\text{A}^-]}{0.62} \Rightarrow 0.72 = \frac{[\text{A}^-]}{0.62} [A−]=0.72×0.62=0.4464 mol/L[\text{A}^-] = 0.72 \times 0.62 = 0.4464 \, \text{mol/L}
Step 4: Convert moles to grams
In 1.0 L of solution: mol of CH₃CO₂Na=0.4464 mol\text{mol of CH₃CO₂Na} = 0.4464 \, \text{mol} grams=0.4464 mol×82.0 g/mol=36.6 g\text{grams} = 0.4464 \, \text{mol} \times 82.0 \, \text{g/mol} = 36.6 \, \text{g}
Correction: Earlier you were told the Ka of benzoic acid. However, since we are using acetic acid, and you wrote CH₃CO₂H, the correct Ka is for acetic acid, not benzoic acid. The Ka of acetic acid is 1.8 × 10⁻⁵, and its pKa is about 4.74 to 4.75. So the calculation above remains correct.
Final Answer: 36.6 grams of sodium acetate should be added to 1.0 L of 0.62 M acetic acid to prepare a buffer solution at pH 4.61.