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Unformatted text preview: -3 ) 2 Mg 2+ + 2Na + ========= 2(RSO-3 ) 2 Na + + Mg + Number of moles of Na = 2* number of moles of Mg At the equivalence point Moles of analyte = moles of titrant Moles of analyte (Mg) = 0.05 M * V (ml) EDTA Substituting for Mg Number of moles of Na = 2 * 0.05 M * V (ml) EDTA Weight of Na (mg) per 10 ml of unknown Number of mg of Na = 2 * 0.05 M * V (ml) EDTA * MW Na Weight of Na (mg) for 1 liter of unknown (ppm) Na = 2 * 0.05 M * V (ml) EDTA * MW Na * 1000ml / 10 ml MW Na = 22.99 g/mole After titration V (ml) EDTA = (15-26) ml =11 ml (ppm) Na = 2 * 0.05 M * 11 ml * 22.99 g/mole * 1000ml / 10 ml (ppm) Na = 2530 ppm 2 CONCLUSION: -From this experiment we learned the use of ion exchange resins in column chromatography and its use in quantitative analysis. After running the experiment we found that the concentration of sodium in the unknown sample is equal to 2530 ppm. 3...
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