The final molarity of bromide anion in the solution is 0.200 M.
The x moles of sodium bromide (NaBr) are dissolved in y liters of an aqueous solution of potassium carbonate (K₂CO₃). Calculate the final molarity of bromide anion (Br⁻) in the solution.
When sodium bromide is dissolved in water, it dissociates into its constituent ions:
NaBr(aq) → Na⁺(aq) + Br⁻(aq)
Therefore, the moles of bromide ions in the solution are equal to the moles of sodium bromide that were dissolved. The molarity of bromide ions can then be calculated using the following equation:
Molarity = moles of solute / volume of solution (in liters)
In this case, the moles of solute are x moles of sodium bromide and the volume of solution is y liters. Therefore, the molarity of bromide ions can be calculated as follows:
Molarity of Br⁻ = x moles / y liters
Example:
Suppose 0.100 moles of sodium bromide are dissolved in 0.500 liters of an aqueous solution of potassium carbonate. Calculate the final molarity of bromide anion in the solution.
Using the Equation above, we can calculate the molarity of bromide ions as follows:
Molarity of Br⁻ = 0.100 moles / 0.500 liters = 0.200 M .