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In the laboratory, a student combines 35.8 mL of a 0.254 M calcium bromide solution with 12.4 mL of a 0.536 M iron(II) bromide solution. What is the final concentration of bromide anion?

Answer :

The final concentration of bromide anion should be 0.6530 M.

Concentration of mixture (C) = ( n₁ × C₁ × V₁ + n₂ × C₂ × V₂ ) / (V₁+V₂)

where C₁ = Concentration of 1 component = 0.254 M

V₁ = volume of 1 component = 35.8 ml

C₂ = Concentration of other component = 0.536 M

V₂ = volume of other component = 12.4 ml

n₁ = number of particle from 1 molecule of 1st component

= 2 as 1 molecule of CaBr₂ has 2 atom of Br = 2

n₂ = number of particle from 1 molecule of 2nd component

= 2 as 1 molecule of FeBr₂ has 2 atom of Br = 2

Now, put the values in above equation;

C = ( n1 × C1 × V1 + n2 × C2 × V2 ) / (V1+V2)

C = ( 2 × 0.254 × 35.8 + 2 × 0.536 × 12.4 ) / ( 35.8 + 12.4 )

C = ( 18.1864 + 13.2928 ) / ( 48.2 )

C = ( 31.4792 ) / ( 48.2 )

C = 0.6530 M

So, The final concentration of Bromide anion is 0.6530 M.

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