8.How many grams of NaCl are required to precipitate most of the Ag+ions from 2.50 × 102mL of 0.0113 M AgNO3solution? Write the net ionic equation for the reaction. (5 points)
9.
What volume of 0.416M Mg(NO
3
)
2
should be added to 255 mL of 0.102 M KNO
3
to
produce a solution with a concentration of 0.278 M NO
3
–
ions? Assume volumes are
additive.
(4 points)
M1V1 +M2V2= M3V3
(Reference: Chang 4.69)
(.832 mol/L)(x) + (.102mol/L)(255 mL)=b(.278mol/L) (255+x)
0.832x + 26.01 = 70.89 + 0.278x
.554x = 44.88
X= 81.0L
3
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University
. All rights reserved.
10.Describe how to prepare 1.00 L of 0.646 MHCl solution, starting with 2.00 MHCl
11.
Calculate the volume in mL of a 1.420
M
NaOH solution required to titrate the
following solutions:
(9 points)
a.
25.00 mL of a 2.430
M
HCl solution
Moles HCl = 0.02500 x 2.430 = 0.06075
NaOH= 0.06075 mol / 1.420 M =0.04278 L =
42.78 ml
b.
25.00 mL of a 4.500
M
H
2
SO
4
solution
Moles H2SO4 = 0.02500 x 4.500 = 0.1125
When balanced 2 moles to 1 NaOH.
2 x .1125 = .2250
.2250/1.42 = .1585L =
158.8 ml
c.
25.00 mL of a 1.500
M
H
3
PO
4
solution
Moles H3PO4 = 0.02500 x 1.500 = 0.03750
3 to 1 ratio so, .03750 x 3 = .1125
0.1125 / 1.420 = 0.07923 L =
79.23 mL
(Reference: Chang 4.91)
12.
A 325 mL sample of solution contains 25.3 g of CaCl
2
.
(4 points)
a.
Calculate the molar concentration of Cl
–
in the solution.
25.3g x (1 mol CaCl2/110.9g CaCl2) x (2 mol Cl-/1 mol CaCl2)=.456 mol Cl-
Molarity = mol of sol/ L of sol
.456 mol Cl / .325 L =
1.4M Cl
b.
How many grams of Cl
–
are in 0.100 L of this solution?
1.40m x 35.45g Cl = 49.63g Cl
49.63g Cl x .100 L =
4.96g in .100 L of solution
(Reference: Chang 4.148)
13.
A student mixes 100.0 mL of 0.500
M
AgNO
3
with 100.0 mL of 0.500
M
CaCl
2
.
(8
points)
a.
Write the balanced molecular equation for the reaction.
