The solubility of in water is 0.02 g in 1.0 L of water at 25 degree C. What is t
ID: 503874 • Letter: T
Question
Explanation / Answer
28.First, this is the dissociation equation for PbI2:
PbI2 <--------> Pb2+ + 2I-
This means that the Ksp is [Pb2+][I-]^2
Remember that the coefficients in the dissociation equation are the exponents in the Ksp equation.
Next, find the molarity of the PbI2:
(0.62 g / 1 L) * (1 mole / 461 g) = 1.3 * 10 ^-3 molar PbI2
Next, find the concentrations of the Pb2+ and the I-. Note that there are 2 I- for every Pb2+.
1.3 * 10^-3 M PbI2 * (1 mole Pb2+ / 1 mole PbI2) = 1.3 * 10^-3 M Pb2+
1.3 * 10^-3 M PbI2 * (2 mole I- / 1 mole PbI2) = 2.6 * 10^-3 M I-
Lastly, plug in these concentrations into the Ksp equation:
Ksp = (1.3 * 10^-3)(2.6 * 10^-3)^2 = 9.7 * 10 ^ -9
29. The hydroxide ion concentration can be found from the solubility, using Ksp for Ca(OH)2.
Ca(OH)2(s)
Ca2+(aq)+2OH^-
Ksp = [Ca2+] [OH–]^2 = 5.5×10–6
(from the Table of Solubility Product Constants)
5.5×10–6 = [x][2x]^2 = 4x3
x = 0.011
[OH–] = 2x = 2(0.011) = 2.2×10^-2M appx.
Option c is correct.
30.BaSO4(s)Ba^2+(aq) + SO4^2-(aq)
Ksp=[Ba^2+][SO4^2-]
let x=[Ba^2+] then it follows that x is also equal to [SO4^2-] because we can see from the balanced equation that for every one Ba^2+ ion that is formed one SO4^2- ion is also formed.
So we have:
1.1x10^-10=x²
x=(1.1x10^-10)=1.0x10^-5 M Ba^2+ which means 1.0x10^-5 M of BaSO4 dissolved.
Solubility of BaSO4=1.0x10^-5 mol/L
1.0x10^-5 mol BaSO4/L x 233. 4g/1 mol BaSO4=2.3x10^-3 g/L appx.
Option e is correct.
31. In 0.033M KF ...concentration of F(-) will be (0.033+ 2s) moles/L or 0.033moles/L approximately..
so as Ksp = [Ba2+] [F-]^2
1.8X 10^-7= s X (0.033)^2
1.8 X 10^-7= s X 0.001089
s = 1.8X 10^-7/0.001089 = 1.7×10^-3 moles/L
Option e is correct.