The tablets were crushed, and 4.9993 g of the powder was transferred to a beaker and reacted with HCl. After filtration, the filtrate was transferred to a 100-mL volumetric flask and diluted with water. 20.00 mL of this stock solution were combined with 0.2 M Na3PO4. The resulting precipitate weighed 0.3451 g after drying. Calculate the moles of BiPO4 precipitated, the moles of Bi3 in the stock solution, and the mass of BSS per tablet.

Explanation:

From the information given:

Mass of BiPO₄ = 0.3451 g

Number of moles of BiPO₄ =

The number of moles of Bi³⁺ in 20.00 mL is:

=  0.001135 mol of Bi³⁺

The number of moles of Bi³⁺ in 100 mL stock solution

Mass of BSS in 4.9993 g tablets

m = 2.055 g BSS

Mass of BSS in 5.0103 g (5 tables)

= 2.06 g

∴

The mass of BSS per tablet is

= 0.412 g BSS/ tablet

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