# 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?

Feb 17, 2016

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

#### Explanation:

In 1 L of the solution, you have:

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

You have 95 mmol of cations and 95 mmol of negative charge (since $\text{1 mol of Cit"^"3-" = "3 mol of negative charge}$).

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:

$\text{10 mmol Na"_3"Cit" = "30 mmol Na"^+ + "30 mmol (Cit)"^"-}$
$\text{45 mmol NaCl" = "45 mmol Na"^+ + "45 mmol Cl"^"-}$
$\text{20 mmol KCl" = "20 mmol K"^+ + "20 mmol Cl"^"-}$

Our solution must contain:

$\text{10 mmol Na"_3"Cit}$
$\text{45 mmol NaCl}$
$\text{20 mmol KCl}$
$\text{75 mmol glucose}$

Now, we must convert these quantities to grams.

Trisodium citrate is available as the dihydrate, $\text{Na"_3"C"_6"H"_5"O"_7"·2H"_2"O}$.

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"

$20 \text{mmol KCl" × "74.55 mg KCl"/(1 "mmol KCl") = "1500 mg KCl" = "1.5 g KCl}$

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 $\text{2.9 g Na"_3"C"_6"H"_5"O"_7"·2H"_2"O", "2.6 g NaCl", "1.5 g KCl", and "13.5 g glucose}$ in enough water to make $\text{1 L of solution}$.