Question

Does any solid Ba(IO₃)₂ form when 7.5 mg of BaCl₂ is dissolved in 500. mL of 0.023 M NaIO₃?

Answer 1

17.55 mg of Ba(IO3)2 will be formed.

What is equilibrium reaction ?

In a chemical reaction, chemical equilibrium is the state in which both the reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.

BaCl2 + 2 NaIO3 = Ba(IO3)2 + 2 NaCl

500ml of 0.023M sodium iodate =   500 x 0.023 /1000 = 0.0115 Moles

7.5mg of barium chloride  = 7.5 / 208.233 x 1000 =   3.601x 10-5 Moles

According to the above equation 3.601x 10-5 Moles  of  barium iodate will produce

3.601x 10-5 Moles  of  barium iodate =  3.601x 10-5 X 487.132 =  0.017545 gm or 17.55 mg

3.601x 10-5 Moles of  barium iodate forms with double the amount of sodium iodate

                                           2 x  3.601x 10-5 = 7.203 x 10-5M

hence 7.203 x 10-5 M Moles of sodium iodate will used to react with barium chloride. Rest of the ions will be in equilibrium .

Let us calculate  = 0.0115 -  7.203 x 10-5   =  0.01142797 Moles

Molarity  =  0.01142797 x 1000 /500 =  0.02285594 Moles

0.02285594 Moles  of iodate ion will be in equilibrium

No barium ion willl be left . hence it is not in equilibrium

17.55 mg of Ba(IO3)2 will be formed.

Learn more about equilibrium reaction

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