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1. The problems below concern CaCl2, (fw 110.98).
(a) 0.135 gram of CaCl2 is dissolved to 250 ml. What is the prepared molarity of the solution?
Molarity = grams/molar mass
Volume in litres
= 0.135/110.98
0.25
= 0.001216
0.25
= 0.0048
Molarity of CaCl2 solution = 0.0048 M
(b) What is the equili
ium molarity of chloride and the pCl value?
Calcium chloride dissociates as follows and forms the following ions at equili
ium:
CaCl2 ------------ Ca
+2
+ 2Cl
-
1 mole 1 mole 2 moles
Since molarity of CaCl2 solution = 0.0048 M, at equili
ium the concentrations of the ions
would be as follows:
0.0048 0.0048 2 * 0.0048 (= 0.0096)
Moles moles moles
P
Cl-
= -log [Cl
-
]
= - log (0.0096)
= 2.017
(c) A sample of CaCl2 is dissolved in 250 ml. A 5.00 ml aliquot of the solution is dissolved to
250 ml. Then a 5.00 ml of the second solution is dissolved in 500ml. The final solution is
analyzed and the chloride concentration is 3.60 x 10
-5
M. If all the chloride came from CaCl2,
how many grams CaCl2 were added to the first solution?
First step of initial dilution
M1 = Xmoles/litre M2 = ?
V1 = 5 ml V2 = 250 ml
M1 * V1 = M2 * V2
M2 = M1 * V1
V2
= X * 5
250
= 0.02 X moles/litre
Second step of initial dilution
M1 = 0.02x moles/litre M2 = 3.60 x 10
-5

V1 = 5 ml V2 = 500 ml
M1 * V1 = M2 * V2
M1 = M2 * V2
V1
0.02x = 3.60 x 10
-5
* 500
5
0.02x = 0.00360
X = 0.18 moles/litre
So, Initial concentration of CaCl2 = 0.18 moles/litre
-------------------------------------------------------------------
Molarity = grams/molar mass
Volume in litres
0.18 = x g /110.98 g/mole
1litre
X g = 0.0016
So, the amount of CaCl2 added to the first solution = 0.0016 g
2. 1.000 gram of lead sulfate, PbSO4 (FW 303.2), is mixed with 1.000g of potassium chromate,
K2CrO4 (FW 194.2) in 100.0 mL of water. The reaction produces the precipitate lead chromate
quantitatively, PbCrO4 (FW 323.2).
(a) Write the balanced chemical equation.
PbSO4 + K2CrO4 ---------- K2SO4 + PbCrO4
(b) Which reactant is the limiting reagent?
Limiting reagent is the reactant which is least concentration out of all the reactants, because of
which it decides the yield of the reaction
PbSO4 + K2CrO4 ---------- K2SO4 + PbCrO4
1 mole 1 mole
303.26 194.2
g/mole g/mole
[Considering molar masses]
Since, 303.26 g of PbSO4 requires only 194.2 of K2CrO4, indicating that the limiting reagent
[reagent whose concentration is low and thus determines the rate of reaction] is K2CrO4.
(c) How much lead chromate is generated in grams?
As per the balanced reaction, the amount of PbCrO4 produced would be same as number of
moles of K2CrO4 as it is the limiting reagent. Since 1 g of K2CrO4 is used, which is equal to 1 g/
194.2 g/mole = 0.00514 moles, the same amount of lead chromate [0.00514 moles] is produced.
Molarity = grams/molar mass
Litre
0.00514 = x g/ (323.2 g/mole)
1 L
X = 0.0000159 g
So, the amount of lead chromate produced = 0.0000159 g
(d) What are the concentrations of [CrO4
-2
] and [P
+2
] (Ksp = 1.8 x 10
-14
)?
Ksp = [P
+2
] [CrO4
-2
]
[PbCrO4]
1.8 x 10
-14
= x * x
0.00514
x
2
= 7.196 * 10
-17
x = 8.48 * 10
-9
The concentrations of [CrO4
-2
] and [P
+2
] would be 8.48 * 10
-9
M
3. 20.00 ml of 0.100 F Pb(NO3)2 (FW 331.2) are mixed with 15.00 ml of 0.125 F NaI (FW
149.9). Lead(II) and iodide form an insoluble salt PbI2 (FW 461.0). Calculate the
concentration of each of the remaining ions after mixing. ([P
+2
], [I
-
], [Na
+
] and [NO3
-
]). Ksp
PbI2 = 7.1 x 10
-9
2Pb(NO3)2 + 2NaI ----------- 2PbI2 + 2Na(NO3)2
1 mole 1 mole 1 mole
(0.1 * 0.020) (0.125 * 0.015)
Moles moles
0.002
0.001875
-----
Initial
concentrati
on
-x
-x
Change
(0.002 – x)
(0.001875 –
x)
x moles/litre
x moles/lire
----- At
equili
ium
Since NaI is the limiting reagent,

Ksp = [PbI2] [Na(NO3)2]
[Pb(NO3)2] [NaI]


7.1 x 10
-9
= x * x
(0.002) * (0.0018)
x
2
= 2.52 * 10
-14
x = 5.019 * 10
-6
concentration...