Suppose 26.0g of sodium bromide is dissolved in 150mL of a 0.60M 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.
NaBr + AgNO3 ==> AgBr(s) + NaNO3
mols NaBr = grams/molar mass = estimated 0.25
mols AgNO3 = M x L = estd 0.09
So all of the AgNO3 will be used and you will have 0.25-0.09 = 0.163 mols NaBr unreacted.
(NaBr) = (Br^-) = mols/L = 0.163/0.150 =?
I have assumed that the amount of Br^- furnished by the slightly soluble AgBr is negligible. I'm sure that is so although I didn't calculate how much that would be.
1.087
To calculate the final molarity of bromide anion in the solution, we need to determine how many moles of sodium bromide and silver nitrate are present in the solution.
Step 1: Calculate the moles of sodium bromide.
Given:
Mass of sodium bromide = 26.0 g
Molar mass of sodium bromide = 22.99 g/mol (sodium) + 79.90 g/mol (bromide)
Molar mass of sodium bromide = 102.89 g/mol
Using the formula:
moles = mass / molar mass
moles of sodium bromide = 26.0 g / 102.89 g/mol
moles of sodium bromide = 0.2527 mol
Step 2: Calculate the moles of silver nitrate.
Given:
Volume of silver nitrate solution = 150 mL = 0.150 L
Initial molarity of silver nitrate = 0.60 M
Using the formula:
moles = volume (L) x molarity
moles of silver nitrate = 0.150 L x 0.60 mol/L
moles of silver nitrate = 0.090 mol
Step 3: Determine the limiting reagent.
The limiting reagent is the reactant that is completely consumed and limits the amount of product formed. In this case, sodium bromide and silver nitrate react in a 1:2 ratio according to the balanced equation:
2NaBr + AgNO₃ -> 2AgBr + NaNO₃
To find the limiting reagent, we compare the moles of sodium bromide and silver nitrate. Since we have 0.2527 mol of sodium bromide and 0.090 mol of silver nitrate, silver nitrate is the limiting reagent.
Step 4: Calculate the moles of bromide anion.
From the balanced equation, we can see that each mole of silver nitrate produces 2 moles of bromide anion.
moles of bromide anion = 0.090 mol silver nitrate x 2 mol bromide/mol silver nitrate
moles of bromide anion = 0.180 mol
Step 5: Calculate the final molarity of bromide anion.
Given:
Volume of solution = 150 mL = 0.150 L
Using the formula:
Molarity = moles / volume (L)
Molarity of bromide anion = 0.180 mol / 0.150 L
Molarity of bromide anion = 1.20 M
Therefore, the final molarity of bromide anion in the solution is 1.20 M.
To calculate the final molarity of bromide anion in the solution, we need to determine the number of moles of bromide anion and then divide it by the volume of the solution.
First, let's find the number of moles of sodium bromide:
Given mass of sodium bromide = 26.0 g
To determine the moles, we will use the molar mass of sodium bromide (NaBr), which is the sum of the atomic masses of its constituent elements (sodium and bromine).
Molar mass NaBr = atomic mass of Na + atomic mass of Br
= 22.99 g/mol + 79.90 g/mol
= 102.89 g/mol
Now, using the formula:
Number of moles = mass / molar mass
Number of moles of sodium bromide = 26.0 g / 102.89 g/mol
≈ 0.2526 mol
Next, let's calculate the amount of bromide anion in moles:
Since sodium bromide dissociates in water to give one mole of bromide anion per mole of sodium bromide, the number of moles of bromide anion is equal to the number of moles of sodium bromide.
Number of moles of bromide anion = 0.2526 mol
Finally, we need to determine the final molarity of bromide anion. Molarity is calculated by dividing the moles of solute by the volume of the solution in liters.
Given volume of solution = 150 mL = 150/1000 L = 0.15 L
Final molarity of bromide anion = Number of moles of bromide anion / Volume of solution
= 0.2526 mol / 0.15 L
≈ 1.68 M
Therefore, the final molarity of bromide anion in the solution is approximately 1.68 M.