A 0.204-g sample of a CO3 antacid is dissolved with 25.0 mL of 0.0981 M HCL....

A 0.204-g sample of a CO3 antacid is dissolved with 25.0 mL of 0.0981 M HCL. The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol blue endpoint with 5.83 mL of 0.104 M NaOH.

a) Determine the moles of base per gram in the sample.

Expert Solution

Moles of HCl initially present = M x V ( in L)

= ( 0.0981) x ( 25/1000) = 0.0024525

Moles of HCl titrated with NaOH = NaOH moles used ( since HCl and NaOH react in 1:1 ratio)

= M x V ( NaoH) = 0.104 x ( 5.83/1000) = 0.0006063

Moles of HCl used in titrating antacid = Initial HCl moles - HCl moles used in titratiing NaOH

= 0.0024525 - 0.0006063 = 0.0018462

Moles of antacid = Moles of HCl used = 0.0018462

0.204 g sample of ant acid has 0.0018462 moles

Moles of base per gram = ( 0.0018462 /0.204) = 0.00905 moles / g

queen_honey_blossom answered 1 year ago

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