Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
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Answer:
You need to add 19,5 mmol of acetates
Explanation:
Using the Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [base]/[acid]
For the buffer of acetates:
pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]
As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:
5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]
1,95 = [CH₃COO⁻]/[10]
<em>[CH₃COO⁻] = 19,5 milimoles</em>
Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.
I hope it helps!
