Using Henderson Hasselbalch equation to calculate the ratio! Please help. Pictur

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Using Henderson Hasselbalch equation to calculate the ratio! Please help. Picture has questions! Pre-lab Questions You will prepare two of the buffer solutions used in this lab. They are the 0.1 M CHC0OH/CH,COONa buffer at pH = 4.74 and at pH-6. You will need to write your own procedure for the preparation of these buffers. To write the procedure, study the manual and follow the pre-lab questions. 1. Using Henderson-Hasselbach equation to calculate the ratio of [CH,COONa [CH COOH] for each buffer and write your answers in the table on the right. Show calculations: pH ICH.COONaVICH,COOH 4.74 You just calculated the ratios of [CH,COONay[CH,COOH). You also know that the total concentration for both buffers is 0.1M (ICH,COONa] + [CH,COOH]-0.1M). Now calculate the concentration of [CH,CO0ONa] and [CH,COOH] for each buffer and write your answer in the table below. Show calculations 2. ICH,COOH CH COONa 4.74 49

Explanation / Answer

1) We know that pKa for CH3COOH = 4.74

According to Henderson Hasselbach equation,

pH = pKa + log (Salt / Acid)

Salt = [CH3COONa]

Acid = [CH3COOH]

=> 4.74 = 4.74 + log (Salt / Acid)

=>  log (Salt / Acid) = 0

=>  (Salt / Acid) = 1 (for pH = 4.74)

FOR pH = 6

6 = 4.74 + log (Salt / Acid)

=>  log (Salt / Acid) = 1.26

=> (Salt / Acid) = 18.2 (for pH = 6)

2)

For pH = 4.75

Salt = Acid

=> Salt = 0.05 M = Acid

For pH = 6

Salt + Acid = 0.1

Salt = 18.2 x Acid

=> 19.2 x Acid = 0.1

=> Acid = 0.00521 M

=> Salt = 0.0945 M

3)

Moles of CH3COONa = 0.05 x 0.1 = 0.005 moles

=> Grams of CH3COONa = 0.005 x 82.03 = 0.41 g

Moles of CH3COOH = 0.05 x 0.1 = 0.005 moles

Volume = Moles / Molarity = 0.005 / 1 = 0.005 L = 5 mL

4)

Moles of Acetic Acid = 0.00521 x 0.1 = 5.21 x 10^-4 Moles

Volume = 5.21 x 10^-4 / 1 = 5.21 x 10^-4 L = 0.521 mL

Moles of NaOH = 0.0945 x 0.1 = 0.00945 moles

=> Mass of NaOH = 0.00945 x 40 = 0.378 g

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