A 5.079 g sample of a solid, weak, monoprotic acid is used to make a 100.0...

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Chemistry

A 5.079 g sample of a solid, weak, monoprotic acid is used tomake a 100.0 mL solution. 26.00 mL of the resulting acid solutionis then titrated with 0.09454 M NaOH. The pH after the addition of18.00 mL of the base is 5.54, and the endpoint is reached after theaddition of 47.60 mL of the base. Please see Titration to DetermineMolecular Weight for assistance.

(a) How many moles of acid were present in the 26.00 mL sample?mol

(b) What is the molar mass of the acid? g/mol

(c) What is the pKa of the acid?

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Answer We are given mass of sample of solid weak monoprotic acid 5079 g volume of acid used for the titration 2600 mL NaOH 009454 M Volume of NaOH used for titration 4760 mL when volume of 180 mL of NaOH    See Answer
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