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

(Difference between revisions)
 Revision as of 11:08, 7 December 2012 (view source) (→ADA Kinetic Assay Preparations)← Previous diff Revision as of 04:45, 9 December 2012 (view source) (→ADA Kinetic Assay Preparations)Next diff → Line 10: Line 10: .050 L of water × $\frac{0.05 mol}{1L}$ = .0025 mol of Na2HPO4 × $\frac{268.07 g}{1 mol}$ = 0.6702 g .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 +

## 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