1. a.) Would boric acid, B(OH) 3 , be more soluble in ethanol (C 2 H 5 OH) or in
ID: 983408 • Letter: 1
Question
1.
a.) Would boric acid, B(OH)3, be more soluble in ethanol (C2H5OH) or in benzene (C6H6)? Explain your answer.
b.) Which of the following ions in each pair would be expected to have the greater energy of hydration? Explain each choice.
Mg+2 or Al3+ F-1 or Cl-1
c). Consider the three ionic compounds, NaCl, KCl, and LiCl.
Which compound would you expect to have the greatest lattice energy? Why?
Which compound would have the greatest energy of hydration? Why?
Are the three ionic compounds soluble in water?
What is the relationship between the lattice energy and the energy of hydration that will make an ionic compound soluble in water?
d.). Calculate the vapor pressure at 23°C of a solution made by dissolving 1.20 g of naphthalene, C10H8, in 25.6 g of benzene, C6H6. The vapor pressure of pure benzene at 23°C is 86.0 mmHg. What is the vapor-pressure lowering of the solution? (Assume that naphthalene is nonvolatile.)
Explanation / Answer
1. a) Boric acid would be more soluble in ethanol because this acid is polar and is more soluble in a more polar solvent. It can also hydrogen-bond to ethanol but not to benzene. because it can form hydrogen bonding with the OH of the alcohol, which make it more polar. In the case of benzene it can only form some wan der vals forces, but these forces are weaker than hydrogens bonding.
d) Use the following expression:
mols naphthalene = grams/molar mass
mols benzene = grams/molar mass
Xnaph = nnaph/total mols
Xbenzene = nbenzene/total mols
pbenzene = Xbenzene*Pobenzene
plowering = Xnaph*Pobenzene
So, using these expressions:
moles naph = 1.20 / 128.17 = 9.36x10-3 moles
moles benz = 25.6 / 78.11 = 0.3201 moles
moles totals = 0.3201 + 9.36x10-3 = 0.32946
Xa (naph) = 9.36x10-3 / 0.32946 = 0.0284
Xb (benz) = 1 - 0.0284 = 0.9716
pbenz = 0.9716*86 = 83.56 mmHg
plowering = 0.0284 * 86 = 2.4424 mmHg
Hope this helps. Post part b and c, in other question thread.
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