The volume of the base was not provided. The flask shown here contains 0.693 g o

Question

The volume of the base was not provided.

The flask shown here contains 0.693 g of acid and a few drops of phenolphthalein indicator dissolved in water. The buret contains 0.250 M NaOH.

What volume of base is needed to reach the end point of the titration?

What is the molar mass of the acid (assuming it is diprotic and that the end point corresponds to the second equivalence point)?

Part 2.

The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.580 moles of a monoprotic weak acid (Ka = 2.4 × 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?

Explanation / Answer

moles = conc * vol (in litres)

moles of NaOH = 0.250*(30.0/1000)

moles = 7.5* 10^-3)

Since the stoic ratio of acid : NaOH is 1 : 2 there are half as many moles of acid as of NaOH.

(7.5 * 10^-3) / 2 = 3.75* 10^-3 moles of acid

moles = mass / Molar mass

Molar mass = mass / moles

Molar mass = 0.693/3.75*10^-3

Molar mass = 184.8 (atomic mass unit)

Part 2

At the half equivalence point , pH = pKa
pKa = -log Ka= -log (2.4*10^-5)
pKa = 4.44

At half equivalence point pH = 4.61

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