Question

Suppose 0.934 g of sodium bromide is dissolved in 50. mL of a 0.10 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. Round your answer to 2 significant digits x10

Answer 1

NaBr + AgNo_3 rightarrow AgBr downarrow + NaNO_3 moles of AgNo_3 = m times v/1000 = 0.10 times 50/1000 moles of NaBr = weight/molecular weight = 0.934/102.8 = 0.009 Therefore AgNo_3 is the limiting reagent Therefore 0.005 miles of AgBr is formed [AgBr] = 0.005 times v/1000 = 0.005 times 50/1000 = 0.00025 m and remaining moles of NaBr = 0.009 - 0.005 = 0.004 Therefore Remaining concentration of [NaBr] = 0.004 times 50/1000 total concentration of [Br] = [AgBr] + remaining condition [NaBr] = 0.00025 + 0.0002 = 0.0045 = 0.45 times 10^- 3 m