Answer:
6,45mmol/L of NaOH you need to add to reach this pH.
Step-by-step explanation:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,74
Henderson-Hasselbalch equation for acetate buffer is:
5,0 = 4,74 + log₁₀
Solving:
1,82 =
(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!