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Offline Runner19

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Mixing aqueous solutions and moles of ions
« on: October 10, 2017, 09:01:43 AM »
The following aqueous solutions are mixed: 90.0 mL of 0.100 M HCL and 80.0 mL of 0.100 AgNO3.

a) Before any reaction occurs, identify each ion and determine the moles of that ion.
          - I think the ions would be Ag+, NO3-, H+, and Cl- . Not sure how I determine the moles of each ion. Do I use the equation n(moles) = M(molarity) * V(volume) ?

b) Write the net ionic equation for the reaction that occurs.
          - Think I've got this one. Ag+ + Cl-  :rarrow: AgCl

c) After the reaction has occurred and assuming that it goes to completion, determine the moles of each ion still present in solution.
          - Running into the same problem here as part "a"

d) obtain 80 mL of 0.100 K2CO3 and identify new ions and the number of moles of any new ion.
          - the ions would be K+ and CO32- , right?

e) Add the solution from "d" to the mixture in step "c" . Write the net ionic equation for the next reaction that occurs
          - I don't understand how to know what the products will be if I add K2CO3 to the original     solution.

f) After the reaction has occurred and assuming it goes to completion, determine the moles of each ion whose amount in solution has changed.


I apologize for the all the questions. My professor is not having office hours today due to an unusual circumstance. I appreciate any help.

Offline Arkcon

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Re: Mixing aqueous solutions and moles of ions
« Reply #1 on: October 10, 2017, 09:06:43 AM »
The following aqueous solutions are mixed: 90.0 mL of 0.100 M HCL and 80.0 mL of 0.100 AgNO3.

a) Before any reaction occurs, identify each ion and determine the moles of that ion.
          - I think the ions would be Ag+, NO3-, H+, and Cl- . Not sure how I determine the moles of each ion. Do I use the equation n(moles) = M(molarity) * V(volume) ?

Yes.  You've given molarity and volume, they want you to work the rest of the problem with moles.  You have to convert, and you have a formula that relates the three variables.

Consider a different problem: if you were using H2SO4 instead of HCl, of the same concentration, would mole of ions be the same?  That's a trick they'll spring on an exam.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Runner19

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Re: Mixing aqueous solutions and moles of ions
« Reply #2 on: October 10, 2017, 09:16:50 AM »
Ok. How does this look?

0.09 L * 0.100 M = 0.009 mol HCl

(0.009 mol HCL ) * (1 mol H / 1 mol HCl) = 0.009 mol H+ ion


Offline Runner19

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Re: Mixing aqueous solutions and moles of ions
« Reply #3 on: October 10, 2017, 12:04:42 PM »
So you would have 0.009 moles of H+ ions and Cl- ions.
Likewise, you'd have 0.008 moles of Ag+ ions and NO3- ions.

For part "c" (determine moles of each ion still present in sol. after reaction has occurred), wouldn't you just be left with 0.009 moles of H ions and 0.008 of NO3 ions? After all, the products for this reaction would be AgCl + H+ + NO3- ...

Offline XeLa.

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Re: Mixing aqueous solutions and moles of ions
« Reply #4 on: October 10, 2017, 06:09:50 PM »
Just check your equation for part (b) - what is a "net ionic equation"? Where did the silver and chloride ions come from? What happened to the other species?

XeLa

Offline Runner19

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Re: Mixing aqueous solutions and moles of ions
« Reply #5 on: October 10, 2017, 07:07:03 PM »
Are you saying my net ionic equation for part "b" is wrong...?

for "d" I got K+ ions and CO32- ions. Did a little stoich and got 0.016 moles of K+ ions and 0.008 moles of CO32- ions.

for part "e" I have gotten 2H+ + CO32-  :rarrow: H2O + CO2
I got this by first setting up this equation: HNO3 + K2CO3  :rarrow: KNO3 + H2CO3
I balanced it and broke it up into ions, canceled what was on both sides, and ended with what I said above.

So this brings me to part "f" (after reaction has occurred, determine moles of each ion whose amount in solution has changed). I'm pretty lost here. The only ions I have in my products are potassium (K) ions and Nitrate (NO3) ions.

Offline Borek

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Re: Mixing aqueous solutions and moles of ions
« Reply #6 on: October 11, 2017, 02:58:07 AM »
For part "c" (determine moles of each ion still present in sol. after reaction has occurred), wouldn't you just be left with 0.009 moles of H ions and 0.008 of NO3 ions? After all, the products for this reaction would be AgCl + H+ + NO3- ...

Do you know what a limiting reagent is?

Can 8 mmoles of Cl- fully precipitate 9 mmoles of Cl-?
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