How many grams of each component do I need in order to prepare a solution containing 75 mmol /L sodium; 65 mmol/L chloride; 20 mmol/L potassium; 10 mmol/L citrate; and 75 mmol /L glucose?

1 Answer
Feb 17, 2016

Answer:

You need to dissolve #"2.9 g Na"_3"C"_6"H"_5"O"_7"·2H"_2"O", "2.6 g NaCl", "1.5 g KCl", and "13.5 g glucose"# in enough water to make #"1 L of solution"#.

Explanation:

In 1 L of the solution, you have:

#"Na"^+ color(white)(ml)"75 mmol" color(white)(m)"Cl"^"⁻"color(white)(ml) "65 mmol"#
#"K"^+ color(white)(mm)"20 mmol" color(white)(m)"Cit"^"3-"color(white)(l) "10 mmol"#
#"glucose 75 mmol"#

You have 95 mmol of cations and 95 mmol of negative charge (since #"1 mol of Cit"^"3-" = "3 mol of negative charge"#).

Our problem is that the cations and anions do not match up evenly.

However, in terms of charge, we can re-distribute them as follows:

#"10 mmol Na"_3"Cit" = "30 mmol Na"^+ + "30 mmol (Cit)"^"-"#
#"45 mmol NaCl" = "45 mmol Na"^+ + "45 mmol Cl"^"-"#
#"20 mmol KCl" = "20 mmol K"^+ + "20 mmol Cl"^"-"#

Our solution must contain:

#"10 mmol Na"_3"Cit"#
#"45 mmol NaCl"#
#"20 mmol KCl"#
#"75 mmol glucose"#

Now, we must convert these quantities to grams.

Trisodium citrate is available as the dihydrate, #"Na"_3"C"_6"H"_5"O"_7"·2H"_2"O"#.

#10 color(red)(cancel(color(black)("mmol Na"_3"C"_6"H"_5"O"_7"·2H"_2"O"))) × ("294.10 mg Na"_3"C"_6"H"_5"O"_7"·2H"_2"O") /(1 color(red)(cancel(color(black)("mmol Na"_3"C"_6"H"_5"O"_7"·2H"_2"O")))) = "2900 mg Na"_3"C"_6"H"_5"O"_7"·2H"_2"O" = "2.9 g Na"_3"C"_6"H"_5"O"_7"·2H"_2"O"#

#45 color(red)(cancel(color(black)("mmol NaCl"))) × "58.44 mg NaCl"/(1 color(red)(cancel(color(black)("mmol NaCl")))) = "2600 mg NaCl" = "2.6 g NaCl"#

#20 "mmol KCl" × "74.55 mg KCl"/(1 "mmol KCl") = "1500 mg KCl" = "1.5 g KCl"#

#75 color(red)(cancel(color(black)("mmol glucose"))) × "180.16 mg glucose"/(1 color(red)(cancel(color(black)("mmol glucose")))) = "13 500 mg glucose"= "13.5 g glucose"#

You need to dissolve #"2.9 g Na"_3"C"_6"H"_5"O"_7"·2H"_2"O", "2.6 g NaCl", "1.5 g KCl", and "13.5 g glucose"# in enough water to make #"1 L of solution"#.