Question

Consider the formation of solid Ba(IO 3 ) 2 from the mixing of 7.5mg of barium chloride being dissolved in 500ml of 0.023M sodium iodate.

a) How much solid barium iodate form

b) What is the concentration of barium ions and iodate ions at equilibrium?

Answer 1

Balanced equation:
BaCl₂ + 2 NaIO₃ = Ba(IO₃)₂ + 2 NaCl
Reaction type: double replacement

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