9 22/1 6. (3 pts.) Using the attached list of pK, values for acids/bases, select

Question

9 22/1 6. (3 pts.) Using the attached list of pK, values for acids/bases, select from the ist below which buffer would offer the largest buffer capacity at pH 3.8 (A)0.01 M Formic Acid/Formate, with a ratio of acid/base of 1 (B)O.15 M Lactic Acid/Lactate, with a ratio of acid/base of 1 (C)o.o5 M Formic Acid/Formate, with a ratio of acid/base of 2 (D)0. 10 M Hypolodous Acid/Hypoiodate, with a ratio of acid/base of 1 7. (3 pts.) How would you prepare 100 mL o.10 M acetic acid buffer at pH 4.2? a Weigh out o.o5 mols acetic acid and 0.05 mols of acetate, add to 80 ml Di water add HCI until pH reaches 4.2, fill to the mark of a 100 mL flask. (B) Weigh out o.010 mol s of Acetic acid, add to 80 ml DI water add HCI until pH reaches 4.2, fill to the mark of a 100 mL flask. (C)Weigh out 0.010 mols of Acetic acid, add to 80 ml DI water add NaOH until pH (D) None of the above. reaches 4.2, fill to the mark of a 100 mL flask. 8. (3 pts.) For a substance to be suitable as a primary standard in a volumetric analysis, it should have: (A)A low molar mass. (B)A very low melting point. (C)No tendency to absorb moisture from the air. (D) React with oxygen.

Explanation / Answer

6)

Answer

B) 0.15M lactic acid/lactate with ratio of acid /base of 1

Explanation

Because of following reason B is the answer

i) Lactic acid having pKa value near to target pH

ii) Lactic acid having higher concentration than formic acid

iii) the acid / base ratio is 1

7)

Answer

C) Weigh out 0.010mole of acetic acid, add 80ml DI water add NaOH untill pH reaches 4.2 , fill to the mark of 100ml flask

Explanation

Henderson - Hasselbalch equation is

pH = pKa + log([A-]/[HA])

4.2 = 4.75 + log([CH3COO-]/[CH3COOH])

[CH3COO-]/[CH3COOH] = 0.28

to get this concentration ratio , CH3COOH should partly get neutralized by NaOH until pH reaches 4.2

8)

Answer

C) No tentency to absorb moisture from the air

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