A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by addi

Question

A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding 30.00 mL of 4.912 x 10-1 M barium nitrate, Ba(NO3)2, to 50.00 mL of 9.004 x 10-1 M sodium Iodate, NaIO3.

Assume that, in the experiment, 125 mL of 25 C distilled water was used to wash and transfer the precipitate, rather than 20 mL of chilled distilled water (4 C).

The solubility of Barium iodate monohydrate in 25 C water is 0.028 g per 100 mL of water; in 4 C water, it is 0.010 g per 100 mL of water.

What mass of product would you expect to isolate?

Note: Please explain how and where the info for solubility is used while doing the calculations.

Explanation / Answer

Solubility if the student used 125 mL water at 25 oC is

( 0.028g / 100mL ) x (125/100 ) = 0.035 g lost due to solubility Ba(IO3)2.

Solubility if the student used 20 mL chilled water at 4oC is

( 0.01 g / 100 ml ) x ( 20 / 100 ) = 0.2 g lost due to solubility Ba(IO3)2.

mass of product isolated = 6.895 g - 0.035 = 6.86 g

mass of product isolated = 6.86 g ----------------------------------> at 25oC

mass of product isolated = 6.895 g - 0.2 = 6.692 g

mass of product isolated = 6.692 g ----------------------------------> at 4 oC

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