1) Give the expected hybridization of the central atom in the following molecule
ID: 780797 • Letter: 1
Question
1) Give the expected hybridization of the central atom in the following molecules
sulphur in SF6
oxygen in H2O
seleniurn SeO3
phosphorus in PBr5
tellurium in TeF4
2) if we trcat H2O as weak acid , and H2O+ as bronted acid, what would be the Equation for their ionization constants (Ka) ?
3) culculate the pH of the 0.25M NH3 0.30M NH4CL buffer system. what is the pH after the addition of (i) 40ml of 0.15M NaOH and (ii) 30ml of 0.20M HCl to 60ml of thebuffer solution ?
4) circle the molecules with a dipole moment ?
NH3 , CO2 , CHCl3 , CH4 , SO2
5) calculate the percentage ionization of benzoic acid at the following concentration : assume the approximation for both cases to be correct : a) 0.20M b) .0.00020M
Explanation / Answer
1) SF6 - S = sp3d2 , H2O - O = sp3 , SeO3 Se = sp3d2 , PBr5 , P=sp3d2 , TeF4 Te= sp3d
2) H2O(l) + H2O(l) ------> H3O +(aq) + OH-(aq)
Ka =[OH-][H3O+]
3) pOh = pkb + log[NH4Cl]/[NH3]
pOH = 4.75 + log ( 0.3/0.25) = 4.83 , pH = 14-4.82 = 9.17
when NaOH is added NH4Cl + NaOh <-> NH3
moles of NaOH added = ( 0.04x0.15) = 0.006,
hence net [NH3] = ( 0.006+0.25x0.06)/(0.06+0.04) = 0.21
[NH4Cl] = ( 0.3x0.06-0.006)/(0.06+0.04) = 0.12
pOH = 4.75+log(0.12/0.21) = 4.507 , pH = 14-4.507 = 9.493
when HCL is added NH3 + HCl <---> NH4Cl
HCl moles added = 0.03x0.2 = 0.006,
[NH3] = ( 0.25x0.06-0.006)/(0.06+0.03) = 0.1,
[NH4Cl] = ( 0.3x0.06+0.006)/(0.09) = 0.2677
pOH = 4.75 + log ( 0.2677/0.1) = 5.178, pH = 14-5.178 = 8.822
4) NH3, CHCl3, SO2 have dipole moment
5) a) 0.2 C6H5COOh <--> C6H5COO- + H+
Ka =[H+][C6H5COO-]/[C6H5COOH]
6.28 x10^ -5 = x^2/( 0.2-x)
x = 0.0035=[C6H5COO-].
% of ionization = [C6H5COO-]/[C6H5COOH] x 100 = ( 0.0035)/(0.2-0.0035) x 100
= 1.78 %
b) Ka = 6.28 x10^ -5 = x^2/( 0.0002-x)
x = 0.000085,
% ionization = ( 0.000085)/( 0.0002-0.000085) x 100 = 73.9 %
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