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28 Sep 2019
A 2.894-g sample of a solid, weak, monoprotic acid is used tomake 100.0 mL of solution. 30.0 mL of this solution was titratedwith 0.08017-M NaOH. The pH after the addition of 22.45 mL of basewas 2.96, and the equivalence point was reached with the additionof 47.96 mL of base.
a) How many millimoles of acid are in the original solid sample?Hint: Don't forget that you are not using all of the 2.894 g samplein the titration.
1 mmol acidb) What is the molar mass of the acid?
2 g/molc) What is the pKa of the acid?
pKa = 3
A 2.894-g sample of a solid, weak, monoprotic acid is used tomake 100.0 mL of solution. 30.0 mL of this solution was titratedwith 0.08017-M NaOH. The pH after the addition of 22.45 mL of basewas 2.96, and the equivalence point was reached with the additionof 47.96 mL of base.
a) How many millimoles of acid are in the original solid sample?Hint: Don't forget that you are not using all of the 2.894 g samplein the titration.
b) What is the molar mass of the acid?
2 g/molc) What is the pKa of the acid?
pKa = 3Jarrod RobelLv2
29 Sep 2019