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.
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
Suppose 0.112 g of zinc bromide is dissolved in 100. mL. of a 27.0 mM aqueous solution of potassium carbonate
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