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5
th
Nov , 2018 Muhammad Ismail
Roll no 55
Object : Accuracy and precession
Theory
KCl and CaCl2 are both very soluble salts. Adding KCl to a solution that already has CaCl2 in it would simply
add another cation K+ to the mixture, and increase the concentration of Cl- ions that are already present.
Now, if you’re starting with a saturated solution of CaCl2 (you didn’t specify, so I have to guess maybe), then
KCl won’t dissolve in it. That’s because the solubility is a function of both the Ca(2+) and Cl- ion
concentrations. And if it’s already saturated, then neither Ca(2+) nor Cl- can be added. What would probably
happen is some KCl would dissolve and force some dissolved CaCl2 to crystallize out.
Procedure
Take 0.3154 g of oxalic acid and make up in 100.0 ml .
Make a saturated solution of KCl by adding Ca(OH)
2
in a beaker and then filter this solution .
Similarly make saturated solution of CaCl
2
by adding Ca(OH)
2
in a beaker and then filter this
solution .
Take 0.2663 g of Na
2
CO
3
make up in 100.0 ml having concentration 0.1005 M.
Take 2.5 ml of 0.01005 M Na
2
CO
3
solution and titrate it against HCl .
Take 10.0 ml of saturated KCl solution & 20.0 ml of saturated CaCl
2
solution in two different beakers
and titrate with hcl by noting their ph on ph meter after 1.0 ml of hcl consumed .
Observations
Standardization of HCl with Na
2
CO
3
Sr #
initial volume of
hcl (ml)
final volumeof hcl
(ml)
difference
concordant
1
0.0
7.0
7.0
7.0
2
7.0
14.0
7.0
3
14.0
21.1
7.1
For saturated filtrate of KCL solution
Volume of KCL solution = 10.0 ml
sr #
PH
sr #
Volume of HCl
PH
1
13
7
6
10.2
2
12.97
8
7
4.5
3
12.82
9
8
2.15
4
12.64
10
9
1.41
5
12.38
11
10
1.12
6
11.95
12
11
1
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For saturated filtrate of CaCl
2
solution
sr #
Volume of HCl (ml )
PH
sr #
Volume of HCl (ml )
PH
1
0
12.8
9
8
10.4
2
1
12.75
10
9
7.5
3
2
12.66
11
10
2.4
4
3
12.55
12
11
1.7
5
4
12.44
13
12
1.44
6
5
12.27
14
13
1.3
7
6
12.03
15
14
1.26
8
7
11.7
16
15
1.25
Calculations
Volume of HCl used against saturated filtrate of KCL solution = 6.60 ml (from graph)
Volume of HCl used against saturated filtrate of CaCl
2
solution = 9.20 ml (from graph)
For solubility of KCl solution
Mile moles of HCl = conc of HCl x Volume of HCl
=0.07 x 6.60 = 0.642
Since mole ratio between H
+
and OH
-
ion is 1 : 1
Therefore mile moles of OH
-
= 0.642
Mole ratio between OH
-
and Ca
2+
is 2 : 1 ( two moles of OH
-
= one mole of Ca
2+
)
Mil moles of Ca
2+
= 0.642/2 = 0.231
Concentration or Molarity of Ca
2+
= Moles of Ca
2+
/ volume of Ca
2+
= 0.231 / 10.0 = 0.0231 M
Solubility of KCl solution = 0.0231
For solubility of CaCl
2
solution
Mile moles of HCl = conc of HCl x Volume of HCl
=0.07 x 9.20 = 0.644
Since mole ratio between H
+
and OH
-
ion is 1 : 1
Therefore mile moles of OH
-
= 0.644
Mole ratio between OH
-
and Ca
2+
is 2 : 1 ( two moles of OH
-
= one mole of Ca
2+
)
Mil moles of Ca
2+
= 0.644/2 = 0.322
Concentration or Molarity of Ca
2+
= Moles of Ca
2+
/ volume of Ca
2+
= 0.322 / 20.0 = 0.0161 M
Solubility of CaCl
2
solution = 0.0161

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5th Nov , 2018 Muhammad Ismail Roll no 55 Object : Accuracy and precession Theory KCl and CaCl2 are both very soluble salts. Adding KCl to a solution that already has CaCl2 in it would simply add another cation K+ to the mixture, and increase the concentration of Cl- ions that are already present. Now, if you’re starting with a saturated solution of CaCl2 (you didn’t specify, so I have to guess maybe), then KCl won’t dissolve in it. That’s because the solubility is a function of both the Ca(2+) and Cl- ion concentrations. And if it’s already saturated, then neither Ca(2+) nor Cl- can be added. What would probably happen is some KCl would dissolve and force some dissolved CaCl2 to crystallize out. Procedure • Take 0.3154 g of oxalic acid and make up in 100.0 ml . • Make a saturated solution of KCl by adding Ca(OH)2 in a beaker and then filter this solution . • Similarly make saturated solution of CaCl2 by adding Ca(OH)2 in a beaker and then filter this solution . • Take 0.2663 g of Na2 CO3 make up in 100.0 ml having concentration 0.1005 M. • Take 2.5 ml of 0.01005 M Na2 CO3 solution and titrate it against HCl . • Take 10.0 ml of saturated KCl solution & 20.0 ml of saturated CaCl 2 solution in two different beakers and titrate with hcl by noting their ph on ph meter after 1.0 ml of hcl consumed . Observations Standardization of HCl with Na2 CO3 initial volume of Sr # hcl (ml) final volumeof hcl (ml) difference 1 0.0 7.0 7.0 2 7.0 14.0 7.0 3 14.0 21.1 7.1 For saturated filtrate of KCL solution Volume of KCL solution = 10.0 ml Volume of sr # HCl PH concordant 7.0 sr # Volume of HCl PH 1 0 13 7 6 10.2 2 1 12.97 8 7 4.5 3 2 12.82 9 8 2.15 4 3 12.64 10 9 1.41 5 4 12.38 11 10 1.12 6 5 11.95 12 11 1 For saturated filtrate of CaCl 2 solution sr # Volume of HCl (ml ) PH 1 0 12.8 2 1 12.75 3 2 12.66 4 3 12.55 5 4 12.44 6 5 12.27 7 6 12.03 8 7 11.7 sr # 9 10 11 12 13 14 15 16 Volume of HCl (ml ) 8 9 10 11 12 13 14 15 Calculations Volume of HCl used against saturated filtrate of KCL solution = 6.60 ml (from graph) Volume of HCl used against saturated filtrate of CaCl2 solution = 9.20 ml (from graph) For solubility of KCl solution Mile moles of HCl = conc of HCl x Volume of HCl =0.07 x 6.60 = 0.642 Since mole ratio between H+ and OH- ion is 1 : 1 Therefore mile moles of OH- = 0.642 Mole ratio between OH- and Ca2+ is 2 : 1 ( two moles of OH- = one mole of Ca2+) Mil moles of Ca2+ = 0.642/2 = 0.231 Concentration or Molarity of Ca2+ = Moles of Ca2+ / volume of Ca2+ = 0.231 / 10.0 = 0.0231 M Solubility of KCl solution = 0.0231 For solubility of CaCl2 solution Mile moles of HCl = conc of HCl x Volume of HCl =0.07 x 9.20 = 0.644 Since mole ratio between H+ and OH- ion is 1 : 1 Therefore mile moles of OH- = 0.644 Mole ratio between OH- and Ca2+ is 2 : 1 ( two moles of OH- = one mole of Ca2+) Mil moles of Ca2+ = 0.644/2 = 0.322 Concentration or Molarity of Ca2+ = Moles of Ca2+ / volume of Ca2+ = 0.322 / 20.0 = 0.0161 M Solubility of CaCl2 solution = 0.0161 PH 10.4 7.5 2.4 1.7 1.44 1.3 1.26 1.25 Name: Description: ...
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