Difference between revisions of "User:Karlena L. Brown/Notebook/PVOH Research/2012/09/28"
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< User:Karlena L. Brown  Notebook  PVOH Research  2012  09
(→Na+ and Cu2+ Ion Calculations in 20ppm SO42 Solutions) 
(→Na+ and Cu2+ Ion Calculations in 20ppm SO42 Solutions) 

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'''(971 ppm)(0.00495 L) = (M<sub>2</sub>)(0.5 L)'''  '''(971 ppm)(0.00495 L) = (M<sub>2</sub>)(0.5 L)'''  
'''M<sub>2</sub> = 9.61 ppm Na<sup>+</sup>'''  '''M<sub>2</sub> = 9.61 ppm Na<sup>+</sup>'''  
+  
+  '''Actual concentration of Cu<sup>+2</sup> in 2020 ppm SO<sub>4</sub><sup>2</sup> solution:'''  
+  '''(6.41 × 10<sup>4</sup> g Cu<sup>+2</sup>)/0.5 L × 10<sup>6</sup> = 1283 ppm Cu<sup>+2</sup>'''  
+  '''(1283 ppm)(5.16 mL) = (M<sub>2</sub>)(0.5 L)'''  
+  '''M<sub>2</sub> = 13.2 ppm Cu<sup>+2</sup>'''  
Revision as of 22:29, 15 October 2012
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OBJECTIVES
XRay Diffraction InstructionsNa^{+} and Cu^{2+} Ion Calculations in 2000ppm SO_{4}^{2} Solutions
MW Na_{2}SO_{4}: 142.04 g/mol MW Na^{+}: 22.99 g/mol Mass Na^{+}: 0.00150 g Na_{2}SO_{4} × (1 mol Na_{2}SO_{4}/142.04 g Na_{2}SO_{4}) × (2 mol Na^{+}/1 mol Na_{2}SO_{4}) × (22.99 g Na^{+}/1 mol Na^{+}) = 4.856 × 10^{4} g Na^{+} Actual concentration of Na^{+} in 2020 ppm SO_{4}^{2} solution: (4.856 × 10^{4} g Na^{+})/0.5 L × 10^{6} = 971 ppm
MW CuSO_{4}·5H_{2}O: 249.69 g/mol MW Cu^{+2}: 63.55 g/mol 0.00252 g CuSO_{4}·5H_{2}O × (1 mol CuSO_{4}·5H_{2}O/249.68 g CuSO_{4}·5H_{2}O) × (1 mol Cu^{+2}/1 mol CuSO_{4}·5H_{2}O) × (63.55 g Cu^{+2}/1 mol CuSO_{4}·5H_{2}O) = 6.41 × 10^{4} g Cu^{+2} Actual concentration of Cu^{+2} in 2020 ppm SO_{4}^{2} solution: (6.41 × 10^{4} g Cu^{+2})/0.5 L × 10^{6} = 1283 ppm Cu^{+2} (1283 ppm)(5.16 mL) = (M_{2})(0.5 L) M_{2} = 13.2 ppm Cu^{+2} Na^{+} and Cu^{2+} Ion Calculations in 20ppm SO_{4}^{2} SolutionsActual concentration of Na^{+} in 2020 ppm SO_{4}^{2} solution: (4.856 × 10^{4} g Na^{+})/0.5 L × 10^{6} = 971 ppm (971 ppm)(0.00495 L) = (M_{2})(0.5 L) M_{2} = 9.61 ppm Na^{+} Actual concentration of Cu^{+2} in 2020 ppm SO_{4}^{2} solution: (6.41 × 10^{4} g Cu^{+2})/0.5 L × 10^{6} = 1283 ppm Cu^{+2} (1283 ppm)(5.16 mL) = (M_{2})(0.5 L) M_{2} = 13.2 ppm Cu^{+2}
