Question

Question 9 Suppose 15.3g of sodium bromide is dissolved in 250.mL of a 0.40 M aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the sodium bromide is dissolved in it.

Answer 1

Answer:

The answer is 0.59 M Br⁻

Explanation:

The molecular formula for sodium bromide is NaBr, and for silver nitrate is AgNO₃. The aqueous solution of AgNO₃ contains the silver and nitrate ions (Ag⁺ and NO₃)  according to its dissociation equilibrium:

AgNO₃ ⇆ Ag⁺ +NO₃

When a mass of NaBr is added to this solution, it dissociates as follows:

NaBr ⇄ Na⁺ + Br ⁻

Each mol of NaBr dissociates in one mol of Br⁻, so the final molarity of bromide ion (Br⁻) can be obtained directly from the molarity of NaBr if we know the molecular weight of NaBr, as follows:

Molecular weight of NaBr= 102.89 g/mol

250 ml = 0.25 L

$\frac{15.3 g NaBr }{0.25 L}$ × $\frac{1 mol NaBr}{102.89 g}$ = $\frac{0.59 mol}{L}$ = 0.59 M

According to that, the solution contains 0.59 mol of Br⁻ per liter.