Just for fun concentration problem.

Cl^-: 9.69 mg/mL

Na⁺: 8.21 mg/mL

K⁺: 3.49 mg/mL

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

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.