How to make 5 ml of a NaBH4 (3.42 M in 1.0 M NaOH) I don\'t know how to make the

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How to make 5 ml of a NaBH4 (3.42 M in 1.0 M NaOH)

I don't know how to make the above solution provided the molarity. Please see below for the reaction scheme, we need to figure out step #5.

Scheme 2. Reaction Scheme OH 1. NaBH HO 2. H30 workup HO 4-hydroxy-3-methoxybenzyl alcohol 4-hydroxy-3-methoxybenzaldehyde a.k.a. vanillyl alcohol a.k.a. vanillin m.p. 110-117°C m.p. 81-83°C C8H10O3 154.16 3 152.15 CeHaO Reaction Table Formula Name Weight densi eg mmol wt/ vol 1.00 17.1 2.60 g vanillin 152.15 NaBH4 (3.42 M in 1.0 M NaOH) 37.83 1.00 17.1 5.0 mL note. use YOUR weight and YOUR volume for your reaction table Safety Precautions Hydrochloric acid: Hydrochloric acid is corrosive and a chemical burn hazard. Avoid contact. If contact is made with the skin, wash the affected area with cold water for 15 minutes NaB solution: NaBHL solution is caustic and will decompose violently with acid. Handle with care. Experimental Procedure Synthesis of Vanillyl Alcohol 1. Weigh between 2.5 and 2.6 g of vanillin into a 50 mL beaker. Record the weight directly into your notebook to the maximum accuracy of the balance 2. Fill out your reaction table by calculating the mmol and equivalents of reagents used based on YOUR mass of vanillin. 3. Transfer the vanillin to a 25 mL round bottom flask using a solid addition funnel. Rinse your beaker and funnel with 5.0 mL of ethanol. This ethanol will be the solvent for the reaction. 4. Add stir bar and dissolve the vanillin. If the vanillin does not dissolve, gently warm the flask with your hand. Read thoroughly before starting this step! Set up the reaction as follows: a. Add 5.0 mL of NaBH. solution to a dropping funnel. b. Clamp the 25 mL round bottom flask in an ice bath as depicted in Figure 2, and immediately procedure to the next step

Explanation / Answer

5mL of NaBH4 solution can be made as per the calculations given below:

Molarity of NaBH4 = moles of NaBH4 / L of solvent

3.42M = moles of NaBH4 / 0.005L of 1.0M NaOH

Moles of NaBH4 = 3.42 x 0.005

Moles of NaBH4 = 0.0171mol

Weight of 0.0171 mol of NaBH4 = (37.83 g of NaBH4 / 1 mol of NaBH4) x 0.0171mol

Weight of 0.0171 mol of NaBH4 = 0.647 g of NaBH4

Thus, 3.42M NaBH4 in 1.0M NaOH can be prepared by dissolving 0.647 g of NaBH4 in 5 mL of 1.0M NaOH.

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