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Question

A solution is prepared by mixing 75.0 mL of 0.027 M BaCl2 and 115 mL of 0.040 M K2SO4. What are the concentrations of barium and sulfate ions in this solution? Assume only SO42- ions (no HSO4-) are present.

Answer 1

BaCl2(aq) + K2SO4(aq) --> Ba(SO4)s + 2KCl(aq)

So we will lose 75 ml*0.027moleBa*/1000ml soln
= 0.002025 moles BaSO4 produced

K2SO4 started with = 115 ml( 0.04mole/1000ml)
= 0.0046 moles K2SO4 started with

so K2SO4 left = 0.0046 - 0.002025 = 0.0026 moles SO4-2

There is NO Ba+2 left in solution

Answer 2

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