Solubility question

[deleted647690]

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Would anyone like to give me some direction on this question?

"For a column that has a 90% exchange efficiency when releasing a +1 cation in place of Ca2+, what is the concentration of the +1 cation that elutes from a column to which 0.010 M Ca(NO3)2 has been added?"

 

[lpp06]

The column will exchange 9 +1 cations for every 10 Ca2+ (90% efficiency) - how much Ca2+ is in solution?

 

[BerkReviewTeach]

Would anyone like to give me some direction on this question?

"For a column that has a 90% exchange efficiency when releasing a +1 cation in place of Ca2+, what is the concentration of the +1 cation that elutes from a column to which 0.010 M Ca(NO3)2 has been added?"

What version of our book are you using and what page is that question on? I can help explain the solution step-by-step if you'd like.

The column will exchange 9 +1 cations for every 10 Ca2+ (90% efficiency) - how much Ca2+ is in solution?

You're missing a stoichiometry factor of 2 : 1.

If 90% of the Ca2+ cation bind and release the corresponding charge equivalent of a +1 cation, there would need to be 180% of the original [Ca2+] in solution.

 

[laczlacylaci]

You have 0.001mol/L of Ca(NO3)2, meaning you will have 0.001moles of Ca2+ and 0.002moles of NO3-. When the column elutes 9 +1 cations for every 10 Ca2+ cation (like lpp06 said), you will get 0.001mol Ca2+ *({9mol of +1}/{10 mol of Ca2+})=0.09mol of +1. Since we never changed volume, it would be 0.09M of +1.

@lpp06 would this be correct?

The column will exchange 9 +1 cations for every 10 Ca2+ (90% efficiency) - how much Ca2+ is in solution?

 

[BerkReviewTeach]

You have 0.001mol/L of Ca(NO3)2, meaning you will have 0.001moles of Ca2+ and 0.002moles of NO3-. When the column elutes 9 +1 cations for every 10 Ca2+ cation (like lpp06 said), you will get 0.001mol Ca2+ *({9mol of +1}/{10 mol of Ca2+})=0.09mol of +1. Since we never changed volume, it would be 0.09M of +1.

@lpp06 would this be correct?

No, because you need to keep in mind that for every +2 that binds, TWO +1 cations must be released. Charge must be balanced.

 

[laczlacylaci]

No, because you need to keep in mind that for every +2 that binds, TWO +1 cations must be released. Charge must be balanced.

So 0.018 M of +1 then?

 

[BerkReviewTeach]

Exactly!!!!

 

[deleted647690]

Exactly!!!!

I'm using the most recent version of the gen chem II book, on page 21

 

[deleted647690]

Sorry, answer choices are
A. [+1 cation] = 0.010 * .9 * 2
B. [+1 cation] = 0.010 * 1/0.9 * 2
C. [+1 cation] = 0.010 * 0.9 * 1/2
D. [+1 cation] = 0.010 * 1/0.9 * 1/2

 

[laczlacylaci]

Sorry, answer choices are
A. [+1 cation] = 0.010 * .9 * 2
B. [+1 cation] = 0.010 * 1/0.9 * 2
C. [+1 cation] = 0.010 * 0.9 * 1/2
D. [+1 cation] = 0.010 * 1/0.9 * 1/2

So 0.018 M of +1 then?

Its A
0.01 M of Ca2+ * 90% (0.9) * 2 +1 cations/for every 1 +2 cation (2)
0.01*0.9*2=0.018

Let me know if you still have questions! 🙂