Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent
1 answer:
Answer:
Following are the responses to the given question:
Explanation:
Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.
Please find the attached file.
D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.
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Answer:
Check the explanation
Explanation:
Answer – Given, acid and there are three Ka values
The transformation of is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.
Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L
First we need to calculate moles of each
Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1
= 0.162 moles
Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1
= 0.225 moles
[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M
[HPO42-] = 0.225 moles / 1.00 L = 0.225 M
Now we need to calculate the pKa2
pKa2 = -log Ka
= -log 6.2x10-8
= 7.21
We know Henderson-Hasselbalch equation
pH = pKa + log [conjugate base] / [acid]
pH = 7.21 + log 0.225 / 0.162
= 7.35
The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35