Please show all your work! Thank you :) 3) standards used generate curve are usu

Question

Please show all your work! Thank you :)

3) standards used generate curve are usually made by firs preparing a stock to concentration. then diluting it to the required solution of known concentration solution that is 1.00 x 102 M in phosphate ion. How many g of Na3PO4 irst prepare a stock would you add to a 250 volumetric flask to prepare this solution Mass Na3PO4 What is this concentration in ppm of phosphate? (That is, as PO4 no Na3PO4) PO4 concentration hat volume of this ppm solution would you dilute with water into a 1.000 L volumetric flask to make a solution of 10.0 ppm This much of the work will be done for you by the laboratory assistant. You will be given a To preserve stock solution of 10.0 ppm phosphate to use in calibrating your phosphate assay ne ammonium molybdate reagent, you will create the calibrating standards by doing a serial dilution, using an aliquot of each diluted solution to create the next dilution After adding 1 mL of ammonium molybdate reagent and 2 drops of stannous chloride to 25 mL of the 10.0 ppm solution to develop the color, use a 20 mL aliquot to dilute with 5 mL of water to make 25 mL of a solution of 8.0 ppm Using the dilution equation: V1 x Concl V2 xConc2 and rearranging 20mL 25mL 10 ppm o ate 25 mL of a 6.0 ppm solution, what volume of the 8.0 ppm solution and what volume of water should be mixed? o create 25 mL of a 4.0 ppm solution, what volume of the 6.0 ppm solution and what volume of water should be mixed? To create 25 mL of a 2.0 ppm solution, what volume of the 4.0 ppm solution and what volume of water should be mixed?

Explanation / Answer

We know that Molarity , M = ( mass/Molar mass) / Volume of solution in L

Given Molarity = 1.00x10-2 M

    Volume of the solution = 250 mL = 0.250 L

Molar mass of Na3PO4 = ( 3x23) + 31+ (4x16)

                                     = 136 g/mol

So mass of Na3PO4 is = Molarity x Volume of solution in L x Molar mass

                                  = 1.00x10-2 x 0.250 x 136

                                  = 0.34 g

                                 = 340 mg

So 340 mg of Na3PO4 present in 250 mL of solution

     M mg of Na3PO4 present in 1000mL = 1 L of solution

   M = ( 340x1000) / 250

       = 1360 mg

So concentration of solution is 1360 mg /L

                                               = 1360 ppm   since 1 ppm = 1mg/L

Therefore the concentration of phosphate = [Na3PO4 ] = 1360 ppm

According to law of dilution   MV = M'V'

Where M = Molarity of stock = 1360 ppm

V = Volume of the stock = ?

M' = Molarity of dilute solution = 10.0 ppm

V' = Volume of the dilute solution = 1.000L

Plug the values we get , V = ( M'V') / M

                                        = 7.35x10-3 L

                                          = 7.35 mL

So the volume of the stock solution is 7.35 mL

-----------------------------------------------------------------------------------------

According to law of dilution   MV = M'V'

Where M = Molarity of stock = 8.0 ppm

V = Volume of the stock = ?

M' = Molarity of dilute solution = 6.0 ppm

V' = Volume of the dilute solution = 25 mL

Plug the values we get , V = (M'V')/ M

                                         = 18.75 mL

Simillarly do the remaining

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.