Can someone please help me with my homework? The solubility product of PbF_2 is
ID: 952783 • Letter: C
Question
Can someone please help me with my homework? The solubility product of PbF_2 is 4.1 times 10^-8. How many grams of PbF_2 can be dissolved in 250 mL of water? What is the solubility of BaSO_4 (K_sp =1.1 times 10^-10) in a 0.11 M solution of BaCl_2? A 300 mL solution of 5 times 10^-6M Ca(NO_3)_2 is added to 250 mL of 3.5 times 10^-4 M NaF. Will CaF_2 (K_sp = 4 times 10^-11) precipitate? Name each of the following, give the coordination number of the central metal cation, the geometry of the complex, and show any possible geometric isomers. The complex (Co(NH_3)_6]^2+ absorbs light at a wavelength of 465 nm. What is the ligand field splitting energy for the complex in J/mole? Which complex is more paramagnetic, [Cr(OH)_4]^2+ or [Cr(NH_3)_4]^2+? Prove your answer.Explanation / Answer
1) The dissociation of aluminium hdyroxide can be shown as
Al(OH)3 --> Al+3 + 3OH-
It is worth to assume that the hydroxid ion concentration is due to only al(OH)3
Given :
pH = 3.09
thereofre pOH = 14-pH = 10.91 = -log[OH-]
[OH-] = 1.23 X 10^-11
As per the stoichiometry of above reaction , for each mole of [OH-] 1/3 moles of Al+3 are present
so [Al+3] = 1.23 X 10^-11 / 3 = 0.41 X 10^-11
Solubility product = [Al+3] [ OH-] ^3 = 0.41 X 10^-11 X (1.23 X 10^-11)^3 = 0.762 X 10^-33
2) The solubility product of PbF2 = 4.1 X 10^-8
PbF2 --> Pb+2 + 2F-
Let [PbF2] = x (solubility)
so concentrations of Pb+2 = [F-] / 2 = x
We know that solubility product = Ksp = [Pb+2] [F-] ^2 = 4x^3
4.1 X 10^-8 = 4x^3
10.25 X 10^-9 = x^3
x = 2.17 X 10^-3 moles / L = [ PbF2]
so moles in 250mL should be = 2.17 X 10^-3 X 0.25 = 0.5425 X 10^-3
Mass = Moles X molecular weight of PbF2 = 0.5425 X 10^-3 X 245.2 = 0.133 grams
3) Ksp = 1.1 X 10^-10
BaSO4 --> Ba+2 + SO4-2
BaCl2 --> Ba+2 + 2Cl-
So the [Ba+2] will be = [BaCl2 = 0.11
Ksp = [Ba+2] [SO4-2]
1.1 X 10^-10 = 0.11 X [SO4-2]
[SO4-2] = 10^-9
So solubility of BaSO4 = 10^-9
4)
Volume of Ca(NO3)2 added = 300mL
Molarity of Ca(NO3)2 = 5 X 10^-6 M
Moles of Ca(NO3)2 = Molarity X volume in litres = 1.5 X 10^-6 moles
It will disscoiate as
Ca(NO3)2 --> Ca+2 + 2NO3-
Volume of NaF = 250mL
Molarity = 3.5 X 10^-4
Moles = Molarity X volume = 3.5 X 10^-4 X 0.25 = 0.875 X 10^-4 moles
NaF --> Na+ + F-
The concentration of F- = Moles / Total volume = 0.875 X 10^-4 moles / 0.55 L = 1.59 X 10^-4M
concentration of Ca+2 = 1.5 X 10^-6 / 0.55 = 2.72 X 10^-6 M
ionic product of CaF2 = Product of concentration of ions = [Ca+2] [F-]^2 = 2.72 X 10^-6 [ 1.59 X 10^-4]^2
Ionic product = 4.32 X 10^-14
so ionic product is less than Ksp, so it will not precipitate
5) a) tetraammine dichloro cobalt (III) ion
b) trisoxalatochromium (III) anion
c) diaquadicynonickel (II)
d) diammineaquathiocynatoiron(III) ion
6) The energy = hc/ lambda
lambda = wavelength
h = planck's constant
c= speed of light
E = (6.626 X 0^-34)(3 X 10^8)/ 465 X 10^-9 m
E = 0.0427 X 10^-17 Joules
7) The complex ion with more number of unpaired elecrons will be more paramagnetic as compared to other.
NH3 is a strong field ligand and will cause pairing of electrons and metal forms low spin complex
However in case of OH- , comparitevly low field ligand and metal will form high spin complex (more paramagentic, more unpaired electrons)