a) The solubility product, Ksp, of Al(OH)3 is 1.3 x 10-33. Calculate its solubil
ID: 635996 • Letter: A
Question
a) The solubility product, Ksp, of Al(OH)3 is 1.3 x 10-33. Calculate its solubility (in g/L) in a solution of NaOH with a pH of 13.11.
b) At 25.0?C, in a sealed container that has a fixed volume of 30.0 L, 25.0g of CH4(g) reacts with O2(g) that has an initial partial pressure of 1.00 atm. The products of the combustion are CO2(g) and H2O(l). What is the mass of H2O(l) produced?
c) In a container of 40.0 L, we have a mixture of CO2 (g) and N2(g). The total pressure is 2.00 atm and the temperature is 25.0?C. If the mass of N2(g) is 55.5 g, what is the mass of CO2(g)?
Explanation / Answer
Ans a ) Balanced reaction : Al(OH)3 ------- Al 3+ + 3OH -
Solubility product equillibrium expression : Ksp = [Al3+ ] [OH- ]3
According to balanced equation , for every one mole of Al3+ formed, 3 moles of OH- are formed.
Hence, 1.3 * 10^-33 = x * (3x)^3
so, x^4 = 1.3 *10^-33/27
or, x = 4.68 * (10^-37)^1/4
or, x = 2.6 * 10^-9
Now 1 mol of Aluminium hydroxide = 78.003 grams.
Hence, SOLUBILITY = 78.003 * 2.6 * 10^-9 = 2.029 * 10 ^ -7 Answer.
Answer b)
CH4 (g) + O2 (g) ---------------- CO2(g) + 2 H2O
2 moles of H2O produced = 2* (2* 1 + 16) = 36 Grams.
Hence ,mass of water produced = 36 grams.
ANSWER c) 1 mol of N2 is 2 * 14 = 28 grams.
So, 55.5 grams of N2 is 1.98 moles.
1 mole of CO2 = 12 + 2*16 = 44 grams.
Hence, with 55.5 grams of N2 , Mass of CO2 = 44 * 1.98 = 87.12 grams.
Therefore, mass of CO2 (g) is 87.12 grams.
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