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
<span>CH</span>₃<span>CH</span>₂<span>COOH + H</span>₂<span>O </span>↔ <span> CH</span>₃<span>CH</span>₂<span>COO</span>⁻<span> + H</span>₃<span>O</span>⁺<span>
</span>
pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
= 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴
where Ca is the acid concentration
To convert boiling water to steam, that would involve heat of vaporization. The heat of vaporization for water at atmospheric conditions is: ΔHvap = <span>2260 J/g.
Molar mass of water = 18 g/mol
Q = m</span>ΔHvap = (1.50 mol water)(18 g/mol)(<span>2260 J/g) = 61,020 J
Time = Q/Rate = (61,020 J)(1 s/20 J) = 3051 seconds
In order to express the answer in three significant units, let's convert that to minutes.
Time = 3051 s * 1min/30 s = <em>102 min</em></span>
Oxygen = 16
Iron = 55.8
16 x 27.6% = 4.4 /4 = 1.1
55.8 x 72.4% = 40.4 /4 = 10.1
1 oxygen and 10 iron, so Fe10 O
The rate constant for 1st order reaction is
K = (2.303 /t) log (A0 /A)
Where, k is rate constant
t is time in sec
A0 is initial concentration
(6.82 * 10-3) * 240 = log (0.02 /A)
1.63 = log (0.02 /A)
-1.69 – log A = 1.63
Log A = - 0.069
A = 0.82
Hence, 0.82 mol of A remain after 4 minutes.
