How to make 1000 ppm solution of NaCl

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In order to determine how much sodium chloride must be dissolved in 1 L of water, start from the definition of parts per million, ppm.

A concentration of 1 ppm is equivalent to 1 part solute, in your case sodium chloride, for every 1 million parts solvent, in your case water.

To get a solution's concentration in ppm, you multiply the ratio that exists between the mass of the solute and the mass of the water by 1 million, or #10^6#.

This is the exact same approach you use when calculating percentage, the only difference being the fact that you need to multiply the ratio by 1 million, instead of by 100.

So, you can safely assume the density of water to be equal to #"1 g/mL"#. This would make the mass of water equal to

#1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1 g"/(1cancel("mL")) = "1000 g"#

Sodium chloride will dissociate completely in aqueous solution to form sodium cations, #Na^(+)#, and chloride anions, #Cl^(-)#.

#NaCl_((aq)) -> Na_((aq))^(+) + Cl_((aq))^(-)#

The important thing to notice here is that you have a #1:1# mole ratio between sodium chloride and sodium cations. This mole ratio will help you determine the mass of sodium chloride you need to dissolve in order to get this particular solution.

The concentration of the target solution (in ppm) will be

#"ppm" = m_"solute"/m_"water" * 10^6#

Plug in your values and solve for #m_"solute"#.

#m_"solute" = ("ppm" * m_"water")/10^6#

#m_"solute" = (20 * "1000 g")/10^6 = "0.02 g"#

This means that your solution must contain 0.02 g of sodium cations, #Na^(+)#. Use sodium's molar mass to determine how many moles of sodium cations would be present

#0.02cancel("g") * "1 mole"/(23.0cancel("g")) = "0.000870 moles"# #Na^(+)#

The aforementioned mole ratio tells you that you must add the same number of moles of sodium chloride to the solution.

#0.000870cancel("moles"Na^(+)) * ("1 mole"NaCl)/(1cancel("mole"Na^(+))) = "0.000870 moles"# #NaCl#

Now use sodium chloride's molar mass to determine how much you need

#0.000870cancel("moles") * "58.44 g"/(1cancel("mole")) = color(green)("0.05 g NaCl")#

So, if you dissolve 0.05 g of sodium chloride in 1 L of water you'll get a 20-ppm #Na^(+)# solution.

Dilutions play a crucial role in quantitative estimations. The concentration of the analyte whenever high requires a single or multi stage dilution before estimation. Similarly dilutions are essential for preparing standard solutions for generation of calibration plots.

In this article critical concepts on dilutions are clarified with illustrative calculations, along with the explanation of “how to prepare 100 ppm solution.” Thus, you won’t have any doubts regarding the topic. 

But before moving on to the critical role of dilutions in quantitative estimations, you must first see what exactly dilution is and what is its general purpose. This information will further help you understand the concept of dilution better. 

What Is Dilution?

A mixture in chemistry consists of two parts: a solvent and a solute. Dilution simply means adding more solvent to the solution so that it decreases the concentration of the solute. This way, the amount of solute remains the same, while the total sample volume increases.

For example, you can add more water to sugar syrup to dilute it and reduce the strength of sugar in the entire mixture. The technique comes in handy when you have to deal with unknown substances. 

The concentration of a solution is usually stated in terms of mass percentage, molarity, molality, mole fraction, or normality. Hence, you have to take care of what unit you have taken, initially while diluting the mixture. The same has to be maintained throughout the process to obtain accurate results. 

The two significant benefits obtained by diluting a sample are:

• Keeping Solvent in Range: Sometimes, the solvent concentration goes beyond the specified range for the quantitative estimations. Therefore, the final result can’t be calculated correctly. That is why it becomes crucial to dilute the entire mixture. It brings down the range of solvent, making it easy to analyse the sample and come up with accurate results for the same.  
• Eliminate Interference: There can be a number of elements present in the mixture. These can interfere with the calculations when the overall concentration of the solvent is high. Thus, it becomes crucial to dilute the sample to eliminate the interference from these elements so that proper results can be obtained from the analysis. 

All in all, we can conclude that quantitative estimations become more comfortable and accurate with the help of dilution. It is incredibly crucial to perform dilution because most of the chemical components are found in high concentrations. Using them as they are won’t provide you with the correct results of the analysis. 

Now, let’s move on to ppm solution preparation, along with an example of the same. 

Volume or Weight Requirement

Before you start any quantitative estimation you have to do elementary calculations to arrive at weight or volume requirements of standards.For this purpose, you will first have to learn how to make ppm solution from the given sample.

Example – weight requirement

What weight of sodium chloride is required to make 100 ppm of sodium ion solution from NaCl standard?

Solution

Formula weight of NaCl = 23+35.5 = 58.5

23 g of Na is present in 58.5 g of NaCl

Therefore, 1 g of Na is present in 58.5/23 = 2.5 g of NaCl

1 ppm solution will contain 1 mg or 0.001 g Na/liter of solution

Therefore 100 ppm solution will contain 100 mg or 0.1 g Na / per litre of solution

1 g of Na is present in 2.5 g NaCl

Therefore 0.1 g of Na is present in 2.5/1.0 X 0.1 = 0.25 g of NaCl

Thus 0.25 g of NaCl dissolved in 1 L volume will give rise to 100 ppm Na solution. Now, you know how to prepare a 100 ppm solution for Na. 

Example – volume requirements

You have 1000 ppm Ca standard solution. How would you proceed to make 1, 2, 5 and 10 ppm solutions

Solution

The basic relationship for dilutions is

C1 x V1 = C2 x V2

where

C1 is initial concentration of standard solution (1000 ppm)

V1 is the volume of stock solution(1ml) to be diluted to get 100 ml of 10 ppm solution

C2  will be 1, 2, 5 or 10 ppm for respective dilution standards

V2 will be final volume of standards obtained after completion of dilutions

The individual dilutions calculations are illustrated below:

Stage – 1 1000 ppm standard to 100 ppm (10 fold dilution)

1.0 X 1000 ppm = V2  X 100 ppm

V2 = 1 X 1000 /100 = 10 ml

It means, for 10 ppm solution preparation from the given 1000 ppm, you will have to increase the volume to 10 ml. 

Stage – 2 dilutions of 100 ppm to 1, 2, 5 and 10 ppm concentrations using 10 ml volumetric flasks

100X V (10 ppm) = 10 X 10

V (10 ppm) = 10 X 10 / 100 = 1ml

100 X V (5ppm ) = 5 X 10

V (5ppm) = 5 X10 / 100 = 0.5 ml

100X V(2 ppm) = 2 X10

V (2 ppm) = 2X10 / 100 = 0.2 ml

100X V(1 ppm) = 1 X 10

V (1 ppm) = 1 X10 /100 = 0.1 ml

The above dilutions stages are illustrated in the video. So you can conveniently understand the process and perform it according to the requirements.

Suggested precautions for dilutions. These will help you perform the process without any mistakes. This way, you will be able to conduct the desired ppm solution preparation calculation. 

• Before proceeding with dilutions ensure that the standard solution is within its validity period and has been stored under prescribed conditions.  If the mixture is in a bad state initially, it won’t be of any use for you after dilution. 
• Make use of micropipette for dilutions so that accurate measurement can be obtained. 
• Wear disposable gloves to prevent contamination of the sample
• Use clean and dry volumetric  glassware for dilutions
• It is advisable to carry diluted sample flasks in trays to avoid breakages
• Never return unused standard back into the standard bottle

Remember that the quality of your estimations in terms of accuracy and precision is in direct proportion to the care taken at each stage of serial dilution.

Error at any stage gets multiplied in subsequent stages so utmost care should be exercised during dilutions operations. Otherwise, the results won’t be correct, and the process will have to be conducted again.