If you are given the following solids: Na 3 PO 4 MW = 163.94 g/mol K 3 PO 4 MW =

Question

If you are given the following solids:

Na3PO4             MW = 163.94 g/mol

K3PO4                  MW = 212.26 g/mol   

NaH2PO4           MW = 119.98 g/mol

K2HPO4             MW = 174.17 g/mol

Na2HPO4           MW = 141.96 g/mol

KH2PO4             MW = 136.08 g/mol

Which compounds would you use to prepare your stock solutions? Why?

I picked KH2PO4 and K2HPO4 because potassium salts dissolve more readily in H2O which would make the lab go more smoothly. I would like to know if I'm missing any valuable pieces of information in this answer.

If you are given deionized water, the solid you selected in the previous question and the following volumetric glassware:

100.00 ml volumetric flask(s)

250.00 ml volumetric flask(s)

   25.00 ml volumetric transfer pipet(s)

   10.00 ml volumetric transfer pipet(s)

     5.00 ml volumetric transfer pipet(s)

            Describe how you would prepare the required stock solutions (two). Your explanation should be detailed and quantitative.  

Explanation / Answer

When making solutions for use in the biology lab, controlling the concentration of solutes in the solution is often critical. The control of the concentration of solutes in solutions used to culture cells can cause the cells die. Inappropriate concentrations of solutes in solutions that contain bio molecules may cause the bio molecules to denature or aggregate, thus making them inactive. Concentrations of solutes can also affect the speed and types of chemical reactions that take place within a solution. Dissolving solid solutes usually granules or powder in the solvent usually dH20. It cannot measure directly because we have no way to count atoms or molecules. We use molecular weights of the molecules.

Weight of solute (g) = M.wt of solute (g/mole) X molarity (mole/L) X final volume (L)

For example all listed compounds molarity 1M in 1L then their weights of the solutes are

Na3PO4 = 163.94 g/mol X 1 M X 1L = 163.94 gm

Na2HPO4 = 141.96 g/mol X 1 M X 1L = 141.96 gm

KH2PO4 = 136.08 g/mol X 1 M X 1L = 136.08 gm

K3PO4 = 212.26 g/mol X 1 M X 1 L = 212.26 gm

NaH2PO4 = 119.98 g/mol X 1M X 1L = 119.98 gm

K2HPO4 = 174.178 g/mol X 1M X 1L = 174.178 gm

When preparing solutions, it is often easier to dilute a stock solution than it is to weigh out a solid solute and then dissolve it in the solvent. This is especially true if you want to prepare several solutions that each have a different concentration of the same solute.

Using C1V1 = C2V2

KH2PO4 and K2HPO4 salts used as buffers, but they contain two primary plant macronutrients: Potassium and Phosphorus. These salts are mono-, di-, and tri-potassium phosphates, solid in the crystalline form, they are creating them by adding phosphoric acid to potassium hydroxide:

H3PO4 + KOH --> H2O + KH2PO4

KH2PO4 + KOH --> H2O + K2HPO4

K2HPO4 + KOH --> H2O + K3PO4

The buffering capacity from mixing those two products seems to yield very strong buffers. These salts are using pH up to 8 (6- 8).

I m already described above

Basing on Molarity equation both volumetric flask and volumetric transfer pipette having exact volumes. On which marker is available we add salts (weighed by formula) add distilled water. Take upper meniscus volume make up with distilled water which is used as a stock solution.

Suppose all chemicals prepared for 1litre

Now calculated for 100 ml

Na3PO4 = 163.94 g/mol X 1 M X 100/1000 = 16.394 gm

Na2HPO4 = 141.96 g/mol X 1 M X 100/1000 = 14.196 gm

KH2PO4 = 136.08 g/mol X 1 M X 100/1000 = 13.608 gm

K3PO4 = 212.26 g/mol X 1 M X 100/1000 = 21.226 gm

NaH2PO4 = 119.98 g/mol X 1M X 100/1000 = 11.998 gm

K2HPO4 = 174.178 g/mol X 1M X 100/1000 = 17.4178 gm

Now calculated for 250 ml

Na3PO4 = 163.94 g/mol X 1 M X 250/1000 = 40.985 gm

Na2HPO4 = 141.96 g/mol X 1 M X 250/1000 = 35.49 gm

KH2PO4 = 136.08 g/mol X 1 M X 250/1000 = 34.02 gm

K3PO4 = 212.26 g/mol X 1 M X 250/1000 = 53.065 gm

NaH2PO4 = 119.98 g/mol X 1M X 250/1000 = 29.995 gm

K2HPO4 = 174.178 g/mol X 1M X 250/1000 = 43.5445 gm

Now calculated for 25 ml

Na3PO4 = 163.94 g/mol X 1 M X 25/1000 = 4.0985 gm

Na2HPO4 = 141.96 g/mol X 1 M X 25/1000 = 3.549 gm

KH2PO4 = 136.08 g/mol X 1 M X 25/1000 = 3.402 gm

K3PO4 = 212.26 g/mol X 1 M X 25/1000 = 5.3065 gm

NaH2PO4 = 119.98 g/mol X 1M X 25/1000 = 2.9995 gm

K2HPO4 = 174.178 g/mol X 1M X 25/1000 = 4.35445 gm

Now calculated for 10 ml

Na3PO4 = 163.94 g/mol X 1 M X 10/1000 = 1.6394 gm

Na2HPO4 = 141.96 g/mol X 1 M X 10/1000 = 1.4196 gm

KH2PO4 = 136.08 g/mol X 1 M X 10/1000 = 1.3608 gm

K3PO4 = 212.26 g/mol X 1 M X 10/1000 = 2.1226 gm

NaH2PO4 = 119.98 g/mol X 1M X 10/1000 = 1.1998 gm

K2HPO4 = 174.178 g/mol X 1M X 10/1000 = 1.74178 gm

Now calculated for 5 ml

Na3PO4 = 163.94 g/mol X 1 M X 5/1000 = 0.8197 gm

Na2HPO4 = 141.96 g/mol X 1 M X 5/1000 = 0.7098 gm

KH2PO4 = 136.08 g/mol X 1 M X 5/1000 = 0.6804 gm

K3PO4 = 212.26 g/mol X 1 M X 5/1000 = 1.0613 gm

NaH2PO4 = 119.98 g/mol X 1M X 5/1000 = 0.5999 gm

K2HPO4 = 174.178 g/mol X 1M X 5/1000 = 0.87089 gm

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