# Question 4057d

Jun 13, 2016

You add 11 mL of distilled water to the tris buffer above.

#### Explanation:

An osmole is 1 mol of particles that contribute to the osmotic pressure of a solution.

For example, $\text{NaCl}$ dissociates completely in water to form ${\text{Na}}^{+}$ ions and $\text{Cl"^"-}$ ions.

Thus, each mole of $\text{NaCl}$ becomes two osmoles in solution: one mole of ${\text{Na}}^{+}$ and one mole of $\text{Cl"^"-}$.

A solution of 1 mol/L $\text{NaCl}$ has an osmolarity of 2 Osmol/L.

A solution of 1 mol/L ${\text{CaCl}}_{2}$ has an osmolarity of 3 Osmol/L (1 mol $\text{Ca"^"2+}$
and 2 mol $\text{Cl"^"-}$).

To calculate the osmolarity of a solution, we add up the osmolarities of all the solutes.

Tris osmolarity

Tris is 2-amino-2-(hydroxymethyl)propane-1,3-diol.

Its formula is ("HOCH"_2)_3"C-NH"_2, and its molar mass is 121.13 g/mol.

$\text{Tris osmolarity" = (2.4 color(red)(cancel(color(black)("g"))))/(100 color(red)(cancel(color(black)("mL")))) × (1000 color(red)(cancel(color(black)("mL"))))/"1 L" × {1 color(red)(cancel(color(black)("mol"))))/(121.13 color(red)(cancel(color(black)("g")))) × "1 Osmol"/(1 color(red)(cancel(color(black)("mol")))) = "0.1981 Osmol/L}$

Citric acid osmolarity

The formula of citric acid is "HOC(COOH)"("CH"_2"COOH")_2, and its molar mass is 192.12 g/mol.

$\text{Citric acid osmolarity" = (1.4 color(red)(cancel(color(black)("g"))))/(100 color(red)(cancel(color(black)("mL")))) × (1000 color(red)(cancel(color(black)("mL"))))/"1 L" × (1 color(red)(cancel(color(black)("mol"))))/(192.12 color(red)(cancel(color(black)("g")))) × "1 Osmol"/(1 color(red)(cancel(color(black)("mol")))) = "0.0729 Osmol/L}$

Glucose osmolarity

The formula of glucose is ${\text{C"_6"H"_12"O}}_{6}$, and its molar mass is 180.16 g/mol.

$\text{Glucose osmolarity" = (0.8 color(red)(cancel(color(black)("g"))))/(100color(red)(cancel(color(black)("mL")))) × (1000color(red)(cancel(color(black)("mL"))))/"1 L" × (1color(red)(cancel(color(black)("mol"))))/(180.16 color(red)(cancel(color(black)("g")))) × "1 Osmol"/(1color(red)(cancel(color(black)("mol")))) = "0.0444 Osmol/L}$

Na-benzylpenicillin osmolarity

The formula of sodium benzylpenicillin is ${\text{C"_16"H"_17"N"_2"O"_4"S"^"-", "Na}}^{+}$, and its molar mass is 356.37 g/mol.

Note that 1 mol of benzylpenicillin is 2 Osmol.

$\text{Na-benzylpenicillin osmolarity =}$

(0.06color(red)(cancel(color(black)("g"))))/(100color(red)(cancel(color(black)("mL")))) × (1000color(red)(cancel(color(black)("mL"))))/"1 L" × (1color(red)(cancel(color(black)("mol"))))/(356.37 color(red)(cancel(color(black)("g")))) × "2 Osmol"/(1color(red)(cancel(color(black)("mol")))) = "0.0034 Osmol/L"

Streptomycin sulfate osmolarity

The formula of streptomycin sulfate is ${\left(\text{C"_21"H"_42"N"_7"O"_12^"3+")_2, ("SO"_4^"2-}\right)}_{3}$, and its molar mass is 1361.32 g/mol.

Note that I mol of streptomycin sulfate is 5 Osmol.

$\text{Streptomycin sulfate osmolarity =}$

(0.1color(red)(cancel(color(black)("g"))))/(100color(red)(cancel(color(black)("mL")))) × (1000color(red)(cancel(color(black)("mL"))))/"1 L"× (1color(red)(cancel(color(black)("mol"))))/(1361.32 color(red)(cancel(color(black)("g")))) × "5 Osmol"/(1color(red)(cancel(color(black)("mol")))) = "0.0037 Osmol/L"

Osmolarity of the buffer

Now we add up the individual osmolarities to get the total osmolarity of the buffer.

$\text{Compound" color(white)(mmmmll)"Osmolarity}$
stackrel(——————————————)("Tris" color(white)(mmmmmmmmml)0.1981)
$\text{Citric Acid} \textcolor{w h i t e}{m m m m m m} 0.0729$
$\text{Glucose} \textcolor{w h i t e}{m m m m m m m l l} 0.0444$
$\text{Na-penicillin} \textcolor{w h i t e}{m m m m m l} 0.0034$
$\text{Streptomycin sulfate} \textcolor{w h i t e}{m l} 0.0037$
stackrel(——————————————)(color(white)(mmmmm)"TOTAL =" color(white)(ml)0.3225)

The osmolarity of your buffer is 0.3225 Osmol/L or 322.5 mOsmol/L.

Dilution calculation

You must now dilute this buffer to get a concentration of 290 mOsmol/L.

You can use the dilution fofrmula

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {c}_{1} {V}_{1} = {c}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

You can rearrange this to

V_2 = V_1 × c_1/c_2#

${V}_{2} = \text{100 mL" × (322.5 color(red)(cancel(color(black)("mOsmol/L"))))/(290 color(red)(cancel(color(black)("mOsmol/L")))) = "111 mL}$

You need to dilute the buffer to 111 mL.

∴ Add 11 mL of distilled, deionized water to the buffer, and you will have a 290 mOsmol/L buffer.