# Solubility question

#### acetylmandarin

2+ Year Member
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

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

#### BerkReviewTeach

##### Company Rep & Bad Singer
Vendor
10+ Year Member
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

2+ Year Member
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

##### Company Rep & Bad Singer
Vendor
10+ Year Member
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

2+ Year Member
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?

Vendor
10+ Year Member
Exactly!!!!

laczlacylaci
OP

#### acetylmandarin

2+ Year Member
Exactly!!!!
I'm using the most recent version of the gen chem II book, on page 21

OP

#### acetylmandarin

2+ Year Member
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

2+ Year Member