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Hey guys,
Just a quick question regarding the math of an ion exchange column question in TBR gen. chem. section 3 (equilibrium), passage VIII, question 55.
The problem asks (I'll paraphrase): Assuming a 100% exchange ratio (i.e., H+ that is initially bound to the column is fully exchanged for any Sr^2+ that passes). Based on this information, what is the Sr^2+ concentration?
I am well aware the pH in the beaker at the end of the analysis can be found, followed by calculating the [H+]. From there, since a +2 cation was exchanged for a +1 cation, that means 2 H+ were exchanged for every 1 Sr2+. Therefore, the final [H+] concentration need to be cut in half to find the Sr2+ concentration. I got the question correct.
In the answer explanation, the author expands on the question by claiming that had there not been a 100% exchange of ions, we have to alter the math I did above. For instance, if only 80% of the +2 cation were exchanged, then we would have to multiply the [H+] by 1.25 (i.e., 1/08 = 1.25).
I can understand why we have to account for this, but I am not understanding the math. Does this means that I still divide the [H+] by 2 and then further multiply it by 1.25 to account for this new exchange ratio?
Regards,
Lunasly.
Just a quick question regarding the math of an ion exchange column question in TBR gen. chem. section 3 (equilibrium), passage VIII, question 55.
The problem asks (I'll paraphrase): Assuming a 100% exchange ratio (i.e., H+ that is initially bound to the column is fully exchanged for any Sr^2+ that passes). Based on this information, what is the Sr^2+ concentration?
I am well aware the pH in the beaker at the end of the analysis can be found, followed by calculating the [H+]. From there, since a +2 cation was exchanged for a +1 cation, that means 2 H+ were exchanged for every 1 Sr2+. Therefore, the final [H+] concentration need to be cut in half to find the Sr2+ concentration. I got the question correct.
In the answer explanation, the author expands on the question by claiming that had there not been a 100% exchange of ions, we have to alter the math I did above. For instance, if only 80% of the +2 cation were exchanged, then we would have to multiply the [H+] by 1.25 (i.e., 1/08 = 1.25).
I can understand why we have to account for this, but I am not understanding the math. Does this means that I still divide the [H+] by 2 and then further multiply it by 1.25 to account for this new exchange ratio?
Regards,
Lunasly.
