–10. A 50mL of 0.20 M solution of AgNO3is mixed with 50mL solution of 0.010 M NaCl will AgCl (s) precipitate?

17-24
To calculate the new concentration…
New [Ag
+
]
=
M
0.10
(100)
(0.20)(50)
V
V
M
total
New [Cl
–
]
=
M
0.0050
(100)
)
(0.010)(50
V
V
M
total
[Ag
+
][Cl
–
]
=
(0.10)(0.0050)
=
5.0 x 10
–
4
>>
1.6 x 10
–
10
=
K
sp
Therefore…
AgCl precipitates.
Do at home…
a)
Would there be a precipitate when one mixes 200mL of 1.0 x 10
–
5
M AgNO
3
with 800mL of 1.8 x 10
–
5
M NaCl?
b)
What is the [Cl
–
] in 0.25 M MgCl
2
? ( strong electrolyte ).
c)
Would there be a precipitate when one mixes 100 mL of 1.00 x 10
–
4
M AgNO
3
with 100 mL of 1.5 x 10
–
4
M MgCl
2
?
Hint: calculate [Ag
+
][Cl
–
].

Consider the following: To a solution in which [Ca2+] is 0.0050 M, one adds sufficient sodium oxalate solid, Na2C2O4, to make the solution also having [C2O42-] =0.0100M. Will the precipitation of CaC2O4(s) be complete? Kspfor CaC2O4=2.7x10-9.
(s)
(aq)
s

Question: Is there a way to precipitate one ion (one salt –the one that has the lower solubility of the two salts ), leaving the other ion still in solution? Namely, can one have selective precipitation?
-17What should the [Ag+] be to start precipitating AgCl(s)?
M
M

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- Summer '09
- Marky
- pH, Solubility, buffer solution