User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/11/13

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Revision as of 04:45, 9 December 2012

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A weight of 0.6702 g of sodium phosphate dibasic was dissolved in 50 mL of water to obtain a molarity of 0.05 M. The pH of the solution was adjusted to pH 7.4.

.050 L of water × $\frac{0.05 mol}{1L}$ = .0025 mol of Na₂HPO₄ × $\frac{268.07 g}{1 mol}$ = 0.6702 g

The pH was was adjusted to 7.4 by the addition of 2 drops of 12 M HCl.
To obtain 1 mM inosine, 1.5 mg of the solid was dissolved in 1 mL buffer. This was further diluted by collecting 89.3 µL of the 1.5 mg/ml inosine into 5 mL of the sodium phosphate buffer.

.0015 g of inosine × $\frac{1 mol}{268.2 g}$ = .000005596 mol ÷ .001 L = .005596 M = 5.596 mM

5.596 mM (V₁)= 0.1 mM (5 mL)

V₁ = 0.08934 mL = 89.3 μL in 5 mL of buffer

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 .050 L of water × <math>\frac{0.05 mol}{1L}</math> = .0025 mol of Na<sub>2</sub>HPO<sub>4</sub> × <math>\frac{268.07 g}{1 mol}</math> = 0.6702 g
+* The pH was was adjusted to 7.4 by the addition of 2 drops of 12 M HCl.
+* To obtain 1 mM inosine, 1.5 mg of the solid was dissolved in 1 mL buffer. This was further diluted by collecting 89.3 µL of the 1.5 mg/ml inosine into 5 mL of the sodium phosphate buffer.
+.0015 g of inosine × <math>\frac{1 mol}{268.2 g}</math> = .000005596 mol ÷ .001 L = .005596 M = 5.596 mM
+.596 mM (V<sub>1</sub>)= 0.1 mM (5 mL)
+V<sub>1</sub> = 0.08934 mL = 89.3 μL in 5 mL of buffer