a. if a 0.1500M NaOH solution was used in this titration, how many milimoles of

Question

a. if a 0.1500M NaOH solution was used in this titration, how many milimoles of acid were involved in the first equivalence point.
b. if a 0.1500M NaOH solution was used in this titration, how many milimoles of acid were involved in the second equivalence point.
c. if the molecular weight of this acid was 104g/mol. what was thr sample size used in this analysis.

60%) Based on these plots provide the following infonnation: 4 00 o.sou 0.0 5.0 10.0 15.0 20.0 25 0 30.o aso a0o 450 500 s5o If a 0.1500 M NaOH selution was used in this titration, how many milimoles of acid wer involved in the first equivalence point. If a 0.1500 M NaOH solution was used in this titration, how many milimoles of acid were involved in the second equivalence point. If the molecular weight of this acid was 104 g/mol. What was the sample size used in this analysis?

Explanation / Answer

a.

Volume of NaOH required for first equivalence point is 20.00 mL.

So, milimoles of NaOH consumed upto first equivalence point is 20.00 * 0.1500 milimole = 3 milimole

b.

Volume of NaOH required for second equivalence point is 40.00 mL.

So, milimoles of NaOH consumed upto second equivalence point is 40.00 * 0.1500 milimole = 6 milimole

c.

Molar Mass of Acid is 104 g/mol.

So, amount of acid involved is (104 * 0.006) g = 0.624 g

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