A 50.00-mL solution containing NaBr was treated with excess AgNO3 to precipitate 0.214 6 g of AgBr (FM 187.772). What was the molarity of NaBr in the solution?
The correct answer and explanation is :
To determine the molarity of NaBr in the solution, we can follow a step-by-step approach. The key here is to use the mass of AgBr precipitate formed to find the moles of AgBr, and from there, calculate the moles of NaBr, which is directly related to the molarity of the NaBr solution.
Step 1: Calculate the moles of AgBr formed
First, we use the given mass of AgBr to calculate the moles of AgBr. We know the molar mass (FM) of AgBr is 187.772 g/mol. The mass of AgBr is given as 0.2146 g.
Using the formula: Moles of AgBr=Mass of AgBrMolar mass of AgBr\text{Moles of AgBr} = \frac{\text{Mass of AgBr}}{\text{Molar mass of AgBr}}
Substitute the values: Moles of AgBr=0.2146 g187.772 g/mol=0.001142 mol\text{Moles of AgBr} = \frac{0.2146 \, \text{g}}{187.772 \, \text{g/mol}} = 0.001142 \, \text{mol}
Step 2: Use the stoichiometry of the reaction
The reaction between NaBr and AgNO3 is: NaBr+AgNO3→AgBr+NaNO3\text{NaBr} + \text{AgNO}_3 \rightarrow \text{AgBr} + \text{NaNO}_3
From the balanced equation, we see that the mole ratio between NaBr and AgBr is 1:1. This means that the moles of NaBr will be equal to the moles of AgBr that reacted, which is 0.001142 mol.
Step 3: Calculate the molarity of NaBr
Molarity (M) is defined as moles of solute divided by the volume of solution in liters. The volume of the NaBr solution is given as 50.00 mL, which is equivalent to 0.05000 L.
Using the formula: Molarity of NaBr=Moles of NaBrVolume of solution (L)\text{Molarity of NaBr} = \frac{\text{Moles of NaBr}}{\text{Volume of solution (L)}}
Substitute the values: Molarity of NaBr=0.001142 mol0.05000 L=0.02284 M\text{Molarity of NaBr} = \frac{0.001142 \, \text{mol}}{0.05000 \, \text{L}} = 0.02284 \, \text{M}
Final Answer:
The molarity of NaBr in the solution is 0.02284 M.
Explanation:
To summarize, the moles of AgBr formed were determined from its mass using its molar mass. The reaction between NaBr and AgNO3 proceeds in a 1:1 molar ratio, so the moles of NaBr are equal to the moles of AgBr. By knowing the volume of the NaBr solution and using the definition of molarity, we calculated the molarity of NaBr as 0.02284 M.