1. A solution is obtained by shaking excess AgCl solid in 0.1M KI solution. Given that Ksp of AgCl is 2.0 x10*-10 mol*2 dm*-6
and Ksp of AgI is 8.0 x 10*-17 mol*2dm*-6
calculate
i the concentration of I ion in the solution
ii the concentration of Ag ion in the solution
2. Draw the shape of a molecule of BeCl2 and name the shape of the molecule
3. 1 g of solid CaCO3 is added into water to form 1 dm3 solution. the solubility product Ksp for CaCO3 is 4.57 x 10*-9 mol*2 dm*-6
CaCO3 ___________\ Ca2+ + CO3 2-
in reality , the following reactions also exist in the solution after CaCO3 is added
H2CO3 __________\ HCO3 - + H+ pKa = 6.35
HCO3- ___________\ H+ + CO3 2- pKa =10.33
(i) If the pH of the solution is adjusted to 8.0, and the H2CO3 is fixed at 1.49 x 10*-5 mol*-3 dm , what is the concentration of CO3 2- and Ca2+ in the solution?
(ii) If the pH of the solution is adjusted to 4.0, what is the concentration of Ca2+ in the solution? Assume the volume of the solution is still
1 dm3 in your calculation.
f.6 chem的問題
2009-07-12 5:17 am
回答 (2)
2009-07-14 3:26 am
1.
(i)
AgCl(s) ≒ Ag+(aq) + Cl-(aq) ...... (1)
Equilibrium constant = 2.0 x 10-10 M2
Ag+(aq) + I-(aq) ≒ AgI(s) ...... (2)
Equilibrium constant = 1/(8.0 x 10-17 M2)
Add equations (1) and (2) together.
AgCl(s) + I-(aq) ≒ AgI(s) + Cl-(aq)
Equilibrium constant:
K = (2.0 x 10-10) x [1/(8.0 x 10-17)] = 2.5 x 106
K is very large. At equilibrium, almost all I- ions are converted to Cl- ions.
At equilibrium: [Cl-] ≈ [I-]o = 0.1 M
K = [Cl-]/[I-]
2.5 x 106 = (0.1)/[I-]
Hence, [I-] = 4.0 x 10-8 M
(ii)
Consider reaction (*):
[Ag+][Cl-] = 2.0 x 10-10 M2
[Ag+](0.1 M) = 2.0 x 10-10 M2
[Ag+] = 2.0 x 10-9 M
2.
Cl-Be-Cl
The molecule is linear in shape.
3.
(i)
H2CO3(aq) ≒ HCO3-(aq) + H+(aq) .. pKa = 6.35
[HCO3-][H+]/[H2CO3] = 10-6.35
[HCO3-](10-8)/(1.49 x 10-5) = 10-6.35
[HCO3-] = 6.66 x 10-4 M
HCO3-(aq) ≒ CO32-(aq) + H+(aq) .. pKa = 10.33
[CO32-][H+]/[HCO3-] = 10-10.33
[CO32-](10-8)/(6.66 x 10-4) = 10-10.33
[CO32-] = 3.12 x 10-6 M
CaCO3(s) ≒ Ca2+(aq) + CO32-(aq)
[Ca2+][CO32-] = 4.57 x 10-9
[Ca2+](3.12 x 10-6) = 4.57 x 10-9
[Ca2+] = 1.46 x 10-3 M
(ii)
When the pH is adjusted to 4.0, the solution is acidic enough to dissolve all CaCO3.
CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
Mole ratio CaCO3 : Ca2+ = 1 : 1
Molar mass of CaCO3 = 40 + 12 + 16x3 = 100 g/mol
No. of moles of CaCO3 = 1/100 = 0.01 mol
[Ca2+] = 0.01/1 = 0.01 M
(i)
AgCl(s) ≒ Ag+(aq) + Cl-(aq) ...... (1)
Equilibrium constant = 2.0 x 10-10 M2
Ag+(aq) + I-(aq) ≒ AgI(s) ...... (2)
Equilibrium constant = 1/(8.0 x 10-17 M2)
Add equations (1) and (2) together.
AgCl(s) + I-(aq) ≒ AgI(s) + Cl-(aq)
Equilibrium constant:
K = (2.0 x 10-10) x [1/(8.0 x 10-17)] = 2.5 x 106
K is very large. At equilibrium, almost all I- ions are converted to Cl- ions.
At equilibrium: [Cl-] ≈ [I-]o = 0.1 M
K = [Cl-]/[I-]
2.5 x 106 = (0.1)/[I-]
Hence, [I-] = 4.0 x 10-8 M
(ii)
Consider reaction (*):
[Ag+][Cl-] = 2.0 x 10-10 M2
[Ag+](0.1 M) = 2.0 x 10-10 M2
[Ag+] = 2.0 x 10-9 M
2.
Cl-Be-Cl
The molecule is linear in shape.
3.
(i)
H2CO3(aq) ≒ HCO3-(aq) + H+(aq) .. pKa = 6.35
[HCO3-][H+]/[H2CO3] = 10-6.35
[HCO3-](10-8)/(1.49 x 10-5) = 10-6.35
[HCO3-] = 6.66 x 10-4 M
HCO3-(aq) ≒ CO32-(aq) + H+(aq) .. pKa = 10.33
[CO32-][H+]/[HCO3-] = 10-10.33
[CO32-](10-8)/(6.66 x 10-4) = 10-10.33
[CO32-] = 3.12 x 10-6 M
CaCO3(s) ≒ Ca2+(aq) + CO32-(aq)
[Ca2+][CO32-] = 4.57 x 10-9
[Ca2+](3.12 x 10-6) = 4.57 x 10-9
[Ca2+] = 1.46 x 10-3 M
(ii)
When the pH is adjusted to 4.0, the solution is acidic enough to dissolve all CaCO3.
CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
Mole ratio CaCO3 : Ca2+ = 1 : 1
Molar mass of CaCO3 = 40 + 12 + 16x3 = 100 g/mol
No. of moles of CaCO3 = 1/100 = 0.01 mol
[Ca2+] = 0.01/1 = 0.01 M
2009-07-12 8:53 pm
(1) With:
[Ag+][Cl-] = 2 x 10-10 and [Ag+][I-] = 8 x 10-17
We have:
[Cl-]/[I-] = 2.5 x 106
Therefore [Cl-] >> [I-] and so we can say that they are completely separated.
Hence [Cl-] = 0.1 M
Then [Ag+] = (2 x 10-10)/0.1 = 2 x 10-9 M and [I-] = (8 x 10-17)/(2 x 10-9) = 4 x 10-8 M
(2) With reference to the following link, BeCl2 has two different molecule shape:
http://i388.photobucket.com/albums/oo325/loyitak1990/Jul09/Crazyinorg1.jpg
The molecule shape under vapour state is linear.
(3)(i) From the given:
[H+][HCO3-]/[H2CO3] = 10-10.33 = 4.68 x 10-11
[HCO3-] = (4.68 x 10-11)(1.49 x 10-5)/(10-8) = 6.97 x 10-8 M
[H+][CO32-]/[HCO3-] = 10-6.35 = 4.47 x 10-7
[CO32-] = (4.47 x 10-7)(6.97 x 10-8)/(10-8) = 3.11 x 10-6 M
[Ca2+][CO32-] = 4.57 x 10-9
[Ca2+] = 1.47 x 10-3 M
(ii) Since a pH = 4.0 is in fact acidic, we can assume that all CaCO3 solids dissolve since CaCO3 can react with acidic solution.
Therefore, no. of moles of CaCO3 = 0.01 with [Ca2+] = 0.01M.
[Ag+][Cl-] = 2 x 10-10 and [Ag+][I-] = 8 x 10-17
We have:
[Cl-]/[I-] = 2.5 x 106
Therefore [Cl-] >> [I-] and so we can say that they are completely separated.
Hence [Cl-] = 0.1 M
Then [Ag+] = (2 x 10-10)/0.1 = 2 x 10-9 M and [I-] = (8 x 10-17)/(2 x 10-9) = 4 x 10-8 M
(2) With reference to the following link, BeCl2 has two different molecule shape:
http://i388.photobucket.com/albums/oo325/loyitak1990/Jul09/Crazyinorg1.jpg
The molecule shape under vapour state is linear.
(3)(i) From the given:
[H+][HCO3-]/[H2CO3] = 10-10.33 = 4.68 x 10-11
[HCO3-] = (4.68 x 10-11)(1.49 x 10-5)/(10-8) = 6.97 x 10-8 M
[H+][CO32-]/[HCO3-] = 10-6.35 = 4.47 x 10-7
[CO32-] = (4.47 x 10-7)(6.97 x 10-8)/(10-8) = 3.11 x 10-6 M
[Ca2+][CO32-] = 4.57 x 10-9
[Ca2+] = 1.47 x 10-3 M
(ii) Since a pH = 4.0 is in fact acidic, we can assume that all CaCO3 solids dissolve since CaCO3 can react with acidic solution.
Therefore, no. of moles of CaCO3 = 0.01 with [Ca2+] = 0.01M.
參考: Myself
收錄日期: 2021-04-30 12:59:43
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