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.1 L of .20M silver nitrate is mixed with .1 L of .15M calcium chloride.

The net ionic equation is :
2Ag(+) + Cl (-) -------------> 2AgCl

What are the concentrations of each ion
remaining in solution after precipitation is complete?
THANK YOU!!!

2006-12-29 09:43:39 · 5 answers · asked by kyle c 1 in Science & Mathematics Chemistry

5 answers

This is a limiting reagent problem, so your first step to write out the full equation, then you should find the quantity in moles of both silver and chlorine.

2AgNO3 + CaCl2 -> 2AgCl + Ca(NO3)2

moles = liters * molarity

(.1 L) (.20 M) = .020 mol AgNO3
(.1 L) (.15 M) = .015 mol CaCl2

Now you know how much of each solution you have, so you can use factor-label to find the limiting reagent, which I am not going to write out.

.020 mol AgNO3 requires .010 mol CaCl2 -- this works
.015 mol CaCl2 requires .030 mol AgNO3 -- this doesn't

Your ions are Ag (+), Ca (2+), NO3 (-) and Cl (-)

Write out how much of each you have -- not the numbers you are using for limiting reagent. Those will come into play later.

.020 mol Ag (+)
.020 mol NO3 (-)
.015 mol Ca (2+)
.030 mol Cl (-)

Now subtract how much of each is used. This is why your net ionic equation is important (and incorrect -- it should be 2Cl) -- anything that is not the precipitate in the net equation remains in solution.

Therefore you're left with .015 mol Ca and .020 mol NO3 in solution

The more difficult part involves your limiting reagent calculations. Some of the Ag and Cl precipitate out, but some of each ion is not used in solution.

You began with .020 mol Ag and .030 mol Cl. Because of limiting reagent, you are going to end with .020 mol AgCl. Your Ag is used up -- 0 mol Ag (+) remains in solution. You are left with .010 mol Cl (-)

So left in solution you have:

.015 mol Ca (2+)
.020 mol NO3 (-)
0 mol Ag (+)
.010 mol Cl (-)

...oh right. Molarity.

To find molarity, you divide moles by liters. You can find your new volume by adding the original volumes.

.1 L + .1 L = .2 L

.015 mol Ca (2+) / .2 L = .075 M Ca
.020 mol NO3 (-) / .2 L = .10 M NO3
0 mol Ag (+) / .2 L = .000 M Ag
.010 mol Cl (-) / .2 L = .050 M Cl

...someone want to double-check that for me? ;) it's been a while since I've done precipitation stoichiometry.

2006-12-29 10:00:43 · answer #1 · answered by Shanny 2 · 0 0

First write the whole equation:

2AgNO3(aq) + CaCl2(aq) ---> 2AgCl (ppt) + Ca(NO3)2 (aq)
where (aq) means in solution in water and ppt means precipitate (solid). (Although we know there is some left in solution)
ALready we found a mistake.

You are told that you have .02 Moles of AgNO3 and 0.015 Moles of CaCl2 (which makes 0.030 moles of Cl).

So ya gotta have at least 0.010 moles of CL left, right?

I assume you have the Ksp?
A.
If not then you have your answer, above - except for change it to moles per liter - 0.2 L after mixing, right? This assumes zero Ag is left is solution, instead of ppm.

B.
If you do have Ksp, then given 0.2 L and the Ksp you can figure the Ag concentration [Ag+] right? Ksp/[CL-] =[Ag+]. And I presume that putting the [Ag+] back into the Ksp equation won't change the [Cl-] concentration significantly. (it will be way too small to be important, see?)

2006-12-29 10:17:26 · answer #2 · answered by Anonymous · 0 0

well you'll have to find the Ksp of AgCl, but the equation is

Ksp= [Ag+][Cl-]

and since the two concentrations will be equal, divide the Ksp by two to get the ion concentrations. Good luck!

2006-12-29 09:47:26 · answer #3 · answered by The Frontrunner 5 · 0 0

Multiply your Liters by Molarity by # of moles in compound to get # of moles:

0.020 mol Ag+
0.020 mol NO3-
0.015 mol Ca+2
0.030 mol Cl-

The net ionic eqn is wrong. It should be:

Ag+ (aq) + Cl-(aq) ---> AgCl (s)

You will ppt out 0.020 mol of AgCl and leave behind:

0.020 mol NO3-
0.015 mol Ca+2
0.010 mol Cl-

Now divide each by the new volume which is 0.2 L:

[NO3-] = 0.10 M
[Ca+2] = 0.075 M
[Cl-] = 0.050 M
[Ag+] = 0.000 M

Ag+ is your limiting reactant. In reality, there is a slight solubility of Ag+, but the excess Cl- will make it almost nothing.

2006-12-29 09:52:27 · answer #4 · answered by serf_tide 4 · 1 0

Assuming answer is largely water, 3.5 liters weighs 3500 g. you have 5.fifty 5 x 10^-5 g of Pb so fraction by utilising mass of Pb is 5.fifty 5*10-5/3500. to visit ppm, multiply by utilising a million (a million*10^6) so fifty 5.5/3500 ppm = 0.0159 ppm.

2016-10-28 16:19:04 · answer #5 · answered by Anonymous · 0 0

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