Experimental+procedure+for+determination+of+low+solubilities

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 * 1) Take two 2 cm 3 aliquots of 35 S-spiked strontium nitrate solution and place in separate 15 ml centrifuge tubes, add excess 2.5 M sulphuric acid (approx. 4-5 cm 3 ) – a white precipitate will rapidly form
 * 2) Centrifuge at 5000 rpm for 5 min and then pour off supernatant into the waste beaker provided
 * 3) Wash repeatedly with distilled water using at least 100 cm 3 – carefully pour off wash water into the waste beaker leaving the precipitate in the centrifuge tube each time
 * 4) Add 10 cm 3 of distilled water to the washed precipitate and leave to stand for 10-30 min, shaking periodically – this allows time for precipitate and water to equilibrate (see step 9)
 * 5) Meanwhile dilute the strontium nitrate solution by a factor of 100 in the volumetric flask with distilled water– use 0.25 cm 3 in 25 cm 3
 * 6) Add 10 cm 3 of scintillation fluid to 5 plastic scintillation vials
 * 7) Add 1 cm 3 of original undiluted solution to the1 st vial
 * 8) Add 1 ml of the 100x diluted solution to a 2 nd vial
 * 9) Once the strontium sulphate has equilibrated, centrifuge off the precipitate and add 1 cm 3 of the supernatant from 1 st precipitated sample to a 3 rd vial
 * 10) Add 1 cm 3 of the supernatant from the 2 nd precipitated sample to a 4 th vial
 * 11) The 5 th vial has no activity added and is the background sample
 * 12) Count all 5 vials on liquid scintillation counter using 14 C channel (Protocol 2). Press button F2 to start the run.

As the majority of the activity is in the precipitate, it is clear that a minute amount of this in the supernatant may cause a gross error in the determination.

From the results obtained the solubility of strontium sulphate may be determined as follows: Let count of 1 cm 3 original solution diluted 1/100 th =z cpm Let count of 1 cm 3 of SrSO 4 = y cpm Thus, 100z cpm is equivalent to 0.1 g Sr(NO 3 ) 2 math =\frac{1}{10}\cdot\frac{184}{212}g \, SrSO_{4} math y cpm is equivalent to: math \frac{1}{10}\cdot\frac{184}{212}\cdot\frac{y}{100z}\, g \, SrO_4 math Solution of SrSO 4 = math \frac{1}{10}\cdot\frac{184}{212}\cdot\frac{y}{100z}\cdot\frac{100\, g}{100 \, cm^3} = \frac{184}{212} \cdot \frac{y}{10z} \cdot g \, per 100 \, cm^3 math