Just for fun concentration problem.

[bjwojcik]

Hi All,

It looks like a lot of you enjoy answering calculation questions, just as I do. I was wondering if any of you would like to try some more challenging problems. These are problems that would never come up in a clinical setting, but I believe help hone our skills. Here is one I came up with. Let me know what you think and maybe post one of your own.

Brad Wojcik, PharmD

You place 6 g of NaCl and 4 g of KCl in a large glass container and let an IV of 0.9% NaCl, with a drop factor of 20, drip into the container at the rate of 3 gtt/min for 2 weeks. After exactly two weeks, you stop the drip and remove 1000 mL from the container and add 500 mL of 200 mEq/L KCl solution to the 1000 mL. What is the concentration of Cl^- in mg/mL in the final solution?

Assume:

No evaporation from the container.

No change in volume when adding NaCl and KCl crystals to water.

Use these values in the calculation:

[TABLE]

[TR]

[TD]Na⁺[/TD]

[TD]22.99 g/mol[/TD]

[/TR]

[TR]

[TD]Cl^-[/TD]

[TD]35.45 g/mol[/TD]

[/TR]

[TR]

[TD]NaCl[/TD]

[TD]58.44 g/mol[/TD]

[/TR]

[TR]

[TD]K⁺[/TD]

[TD]39.10 g/mol[/TD]

[/TR]

[TR]

[TD]KCl[/TD]

[TD]74.55 g/mol[/TD]

[/TR]

[/TABLE]

Good luck!

[chare]

Cl^-: 9.69 mg/mL

Na⁺: 8.21 mg/mL

K⁺: 3.49 mg/mL

[bjwojcik]

Hi Chare,

Thanks for working the problem!

I didn't calculate the concentrations of the K+ and Na+, but I got 7.23 mg/mL on the Cl-. Here is my work. Maybe I did it wrong.

Calculate weight of Cl- in the glass container.

6 g NaCl((35.45 g Cl⁻)/(58.44 g NaCl))=3.64 g Cl⁻

4 g KCl((35.45 g Cl⁻)/(74.55 g KCl))=1.90 g Cl⁻

Total = 3.64 g + 1.90 g = 5.54 g

Calculate volume of 0.9% NaCl solution which dripped into the glass container over 2 weeks.

14 days((24 h)/day)((60 min)/h)((3 gtts)/min)((1 mL)/(20 gtts))=3024 mL

Calculate weight of Cl- from the 3024 mL of 0.9% NaCl.

3024 mL((0.9 g NaCl)/(100 mL))((35.45 g Cl⁻)/(58.44 g NaCl))=16.51 g Cl⁻

Calculate total weight of Cl- in the 3024 mL of solution.

Total= 5.54 g + 16.51 g = 22.05 g

Calculate weight of Cl- in 1000 mL of the above solution.

1000 mL((22.05 g Cl⁻)/(3024 mL))=7.29 g Cl⁻

Calculate weight of Cl- in 500 mL of 200 mEq/L solution.

500 mL((200 mEq Cl⁻)/(1000 mL))((1 mmol)/mEq)((35.45 mg Cl⁻)/mmol)((1 g)/(1000 mg))=3.55 g Cl⁻

Total the weight of Cl- from the 1000 mL and 500 mL, divide by 1500 mL and change to mg/mL.

7.29 g + 3.55 g = 10.84 g. (10.84 g )/(1500 mL) ((1000 mg )/g)=7.23 mg/mL

[chare]

Hi Chare,

Thanks for working the problem!

I didn't calculate the concentrations of the K+ and Na+, but I got 7.23 mg/mL on the Cl-. Here is my work. Maybe I did it wrong.

[...]

No, you didn't, and I see where my errors were.

This is what happens when I try this before the first cup of coffee.

ETA: Thank you for posting this, it definitely made me think.