# Question 60376

Nov 27, 2016

WARNING! Very long answer! Here are my suggestions.

#### Explanation:

$\text{NaOH} :$

This one is easy.

The molar mass of $\text{NaOH}$ is 40.00 g/mol.

Simply transfer 40.00 g of $\text{NaOH}$ to a 1 L volumetric flask.

Add enough water to dissolve the $\text{NaOH}$, then make up to the mark with distilled water.

If you need to know the concentration to more than 1 significant figure, you must standardize the solution by titrating an aliquot with an acid of known concentration.

Lisinopril:

The structure of Lisinopril is

Its molar mass is 405.49 g/mol.

Per the US Pharmacopoeia, Lisinopril tablets must contain between 90 % and 110 % of the labelled amount of ${\text{C"_21"H"_31"N"_3"O}}_{5}$.

Preparing the solution:

Calculate the mass of Lisinopril needed to prepare I L of a 1 nmol/L solution.

For easier writing, let's symbolize Lisinopril as $\text{Lis}$.

"Mass of Lis" = 1 color(red)(cancel(color(black)("L solution"))) × (1 ×10^"-9" color(red)(cancel(color(black)("mol Lis"))))/(1 color(red)(cancel(color(black)("L solution")))) × "405.49 g Lis"/(1 color(red)(cancel(color(black)("mol Lis"))))
= 4.05 × 10^"-7"color(white)(l) "g Lis"

So, we want a solution that contains 4.05 × 10^"-7" color(white)(l)"g of Lis" in 1 L of solution.

It looks like a serial dilution is in order.

Step 1: Dissolve 40 mg of $\text{Lis}$ in 100 mL of water.

$\text{[Lis]" = (40 × 10^"-3" color(white)(l)"g")/"0.100 L" = "0.40 g/L}$

Step 2: Dilute 1 mL of this solution to 1 L.

We can use the formula

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

Then c_2 = c_1 × V_1/V_2

$\text{[Lis]" = "0.40 g/L" × (1 color(red)(cancel(color(black)("mL"))))/(1000 color(red)(cancel(color(black)("mL")))) = 4.0 × 10^"-4"color(white)(l) "g/L}$

Step 3: Dilute 1 mL of this solution to 1 L.

$\text{[Lis]" = 4.0 × 10^"-4" color(white)(l)"g/L" × (1 color(red)(cancel(color(black)("mL"))))/(1000 color(red)(cancel(color(black)("mL")))) = 4.0 × 10^"-7"color(white)(l) "g/L}$

Check:

$\text{[Lis]" = (4.0 × 10^"-7" color(red)(cancel(color(black)("g Lis"))))/(1 "L") × "1 mol"/(405.4 color(red)(cancel(color(black)("g Lis")))) = 9.9 × 10^"-10" color(white)(l)"mol/L}$
= 0.99 × 10^"-9" color(white)(l)"mol/L" = "0.99 nmol/L"#

Close enough!

If you need more precision, you will have to analyze the solution according to the methods of the US Pharmacopoeia.