Chemical Forums

Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: gdogga on February 26, 2009, 01:50:05 AM

Title: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: gdogga on February 26, 2009, 01:50:05 AM

1.   Accurately weigh out 0.03g of finely ground fertilizer into a 100ml beaker

2.   Add about 20ml of boiling water and stir to dissolve powder

3.   Transfer mixture into 250ml volumetric flask. Add water to make 250ml of solution. Transfer solution to 600ml beaker and stir to ensure concentration is uniform.

4.   Dilute the fertilizer solution by factor of 10. Pour 25ml of the solution into a 250ml volumetric flask; add water to make 250ml solution.

5.   Label 6 test tubes: 10mgL-1, 7.5 mgL-1, 5.0 mgL-1, 2.5 mgL-1, 0.0 mgL-1 and unknown.

6.   Place 20ml of the diluted fertilizer solution into the “unknown” test tube.

7.   Use a 10ml measuring cylinder to add standard phosphate solution and water to the labeled test tubes as per table below.

Test tube label (mgL-1)    Volume of 10mgL-1 Standard Phosphate solution (ml)    Volume of Water

10.0  20  0

7.5  15  5

5.0  10  10

2.5  5  15

0.0  0  20

8.   Add 2ml of ammonium molybate reagent and some crystals of ascorbic acid to each of the test tubes. Stir to dissolve the crystals.

9.   Place all six test tubes in a 600ml beaker containing 200ml boiling water. Heat for 5 minutes. On heating the solutions should turn blue. Remove all test tubes and allow time to cool.

10.   Compare the colour of the solution in the test tube labeled “unknown” with the colour standards to estimate the concentration of phosphate in the diluted fertilizer solution.

•   Colour looked like 2.5gl-1

QUESTIONS

1.The initial 250ml of fertilizer solution was diluted 10 times before being analyzed. What is the concentration of phosphate ions, in mg L-1, in the initial 250ml if fertilizer solution?

2.Use your answer to Q1 to find the mass of phosphorous ion the initial 250ml solution. This is the same as the mass of phosphorous in the sample of fertilizer


3.Calculate the percentage of phosphorous, by mass, in the fertilizer.

Sorry for the formatting, I'm a noob ;D

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: Borek on February 26, 2009, 03:07:24 AM

You have to try by yourself first.

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: gdogga on February 26, 2009, 05:58:29 AM

Quote from: Borek on February 26, 2009, 03:07:24 AM

You have to try by yourself first.

I have tried, that's why i posted it here, because i couldn't get the right answer.

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: Borek on February 26, 2009, 07:50:08 AM

Show what you did then, will be easier to help.

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: el13 on February 27, 2009, 12:17:20 AM

you wrote:"•   Colour looked like 2.5gl-"
Is that correct or it is 2.5 mg/l?

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: gdogga on February 27, 2009, 01:27:39 AM

Quote from: el13 on February 27, 2009, 12:17:20 AM

you wrote:"•   Colour looked like 2.5gl-"
Is that correct or it is 2.5 mg/l?

My mistake, it it 2.5 mg L-1

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: gdogga on February 27, 2009, 01:39:57 AM

Quote from: Borek on February 26, 2009, 07:50:08 AM

Show what you did then, will be easier to help.

1. c(PO43-) = 2.5 mg L-1

2. m(PO43-)= ??
   c(PO43-) = 2.5 mg L-1 = 0.0025 g L-1 =
   v(PO43-) = 0.25L
   n(PO43-)= THIS IS WHERE I GET STUCK
               
   m(PO43-) = nM

3. m(PO43-)/ 0.03g x 100 =

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: Borek on February 27, 2009, 03:31:05 AM

Quote from: gdogga on February 27, 2009, 01:39:57 AM

   c(PO43-) = 2.5 mg L-1 = 0.0025 g L-1 =
   v(PO43-) = 0.25L
   n(PO43-)= THIS IS WHERE I GET STUCK

By definition of molar concentration (http://www.chembuddy.com/?left=concentration&right=molarity):

C=n/V

You know C, you know V - just solve for n.

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: el13 on February 27, 2009, 01:16:08 PM

2.5mg/l is the concentration of the second dilution.The initial solution has 2.5*10=25mg/l.

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: gdogga on February 27, 2009, 04:32:01 PM

Quote from: Borek on February 27, 2009, 03:31:05 AM

Quote from: gdogga on February 27, 2009, 01:39:57 AM

   c(PO43-) = 2.5 mg L-1 = 0.0025 g L-1 =
   v(PO43-) = 0.25L
   n(PO43-)= THIS IS WHERE I GET STUCK

By definition of molar concentration (http://www.chembuddy.com/?left=concentration&right=molarity):

C=n/V

You know C, you know V - just solve for n.

n=cv

But when im getting the moles, i am using Litres times Grams per litre. so wouldnt the litres cancel out and all i am left with is the grams. so by finding c x v i am finging the mass.

n= 0.0025 x 0.25 = 0.000625 g

%= 0.000625/0.03 x 100 = 2.08%

Title: Re: Phosphate content of lawn fertilizer by colorimetric analysis
Post by: el13 on February 27, 2009, 11:43:22 PM

The second dilution has no importance?