Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>
<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
<h2>pH = 8.34</h2>
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
This is the same as its freezing point
hope this helps
You need to look at the electronegativity and decide wheter the difference of both of the numbers are significant enough to form a polar bond
