Question

Suppose 0.112 g of zinc bromide is dissolved in 100. mL. of a 27.0 mM aqueous solution of potassium carbonate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the zinc bromide is dissolved in it. 

Be sure your answer has the correct number of significant digits. 

Answer 1

ZnBr₂ + K₂CO₃ → ZnCO₃(s) + 2KBr

moles = (grams / molecular weight)

moles = liters *Molarity

Molecular Weight of ZnBr₂ = 65.38 + (2*79.90)

                                               = 225.18 g/mole

moles ZnBr₂ = 0.112 g ZnBr2 / 225.18 g/mole

                       = 4.97*10^-4 moles

moles K₂CO₃ = 0.1 l*0.027 M = 0.0027 moles

Molarity of bromide ion = moles/Volume

                                          = (2*4.97*10^-4)/0.1

                                          = 9.94*10^-3 M