Question

If 225mL of a 0.200 mol/L calcium bromide solution is added to a 155mL of a 0.450 mol/L cobalt iii bromide solution what is the final concentration of bromide ions?

Answer 1

First, let's separate both solutions:

We assume solutions are aditive.

Solution 1)

225mL of a 0.200 mol/L calcium bromide

Volume 1 = 225 mL = 0.225 L

Concentration 1 = 0.200 mol/L

Compound = calcium bromide = CaBr2

CaBr => Ca+2 + 2Br- (Bromide ions)

Concentration of bromide ions = 2 x 0.200 mol / L = 0.400 mol/L

Then, the moles of bromide ions = 0.225 L x 0.400 mol/L = 0.09 moles

Bromide ions 1 = 0.0900 moles

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Solution 2)

We proceed the same as solution 1

Cobalt (III) bromide = CoBr3

CoBr3 => Co+3 + 3Br-

Concentration of Bromide ions = 3 x 0.450 mol/L = 1.35 mol/L

Moles of bromide ions = 1.35 mol/L x 0.155 L (155 mL) = 0.209 moles

Bromide ions 2 = 0.209 moles

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The mixture of these 2 compounds

Total moles of Bromide ions = 0.0900 moles + 0.209 moles = 0.299 moles

Total volume = 0.225 L + 0.155 L = 0.380 L

Final concentration = 0.299 moles / 0.380 L = 0.787 mol/L

Answer: 0.787 mol/L