Question

4. Your mission, if you choose to accept it, is to make 10mmol/L acetate buffer, pH5.0. Beginning with 10mmol/L HAc, what concentration of NaOH do you need to add to reach this pH? (again, show your work)

Answer 1

Answer:

6,45mmol/L of NaOH you need to add to reach this pH.

Explanation:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,74

Henderson-Hasselbalch equation for acetate buffer is:

5,0 = 4,74 + log₁₀$\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$

Solving:

1,82 = $\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$ (1)

As total concentration of acetate buffer is:

10 mM = [CH₃COOH] + [CH₃COO⁻] (2)

Replacing (2) in (1)

[CH₃COOH] = 3,55 mM

And

[CH₃COO⁻] = 6,45 mM

Knowing that:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Having in the first 10mmol/L of CH₃COOH, you need to add 6,45 mmol/L of NaOH. to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃COOH .

I hope it helps!