Solution Preparation and Calculations Flashcards

How much NaCl (MW = 58.44 g/mol) is needed to prepare 500 mL of a 0.9% (w/v) saline solution?

Answer: 4.5 g

0.9% (w/v) means 0.9 g per 100 mL. For 500 mL: 0.9 g/100 mL x 500 mL = 4.5 g NaCl.

A laboratory stock solution of HCl is labeled '12 M'. How much stock is needed to make 100 mL of 1 M HCl?

Answer: 8.33 mL

Using C1V1 = C2V2: (12 M)(V1) = (1 M)(100 mL), so V1 = 100/12 = 8.33 mL of stock.

What does 'pH 7.4 with NaOH' typically mean when listed in a buffer preparation protocol?

Answer: After dissolving all buffer components, measure pH and adjust to 7.4 by adding NaOH (to raise pH) or HCl (to lower pH) dropwise

Buffer protocols require pH adjustment after dissolving all components, adding NaOH to raise pH to the target value, with HCl available to correct any overshoot below target.

A protocol requires a '10x PBS' stock solution. What concentration of NaCl would you expect in the stock, compared to the 1x working solution that contains 137 mM NaCl?

Answer: 1,370 mM (137 mM x 10)

A 10x stock is 10 times more concentrated than the 1x working solution; 137 mM x 10 = 1,370 mM NaCl in the stock.

When preparing a calibration standard curve for a Bradford assay, why must standards be prepared in the same buffer as the protein samples?

Answer: Because buffer components can affect the Bradford dye color reaction, potentially causing systematic error if standards and samples are in different buffers

Buffer components (detergents, reducing agents, chaotropes) can interfere with the Bradford dye reaction; using matched buffers for standards ensures that any interference is constant across all measurements.

What is the molar concentration of a glucose solution prepared by dissolving 18.02 g of glucose (MW = 180.2 g/mol) in water to a final volume of 1 liter?

Answer: 0.1 M

Moles of glucose = 18.02 g divided by 180.2 g/mol = 0.1 mol. Concentration = 0.1 mol / 1.0 L = 0.1 M.