Suppose 3.50g of zinc bromide is dissolved in 300.mL of a 65.0mM aqueous solution of potassium carbonate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the zinc bromide is dissolved in it.
ZnBr2 + K2CO3 ==> ZnCO3(s + 2KBr.
Note that all of the Br2 stays in solution so just assume there is no reaction.
Final ZnBr2 (if no ZnCO3 fomeed) is 3.50 g ZnBr2/molar mass ZnBr2 = mols ZnBr2.
Then (ZnBr2) = mols ZnBr2/L solution = mols ZnBar2/0.300 L = ?
Then final (Br^-) is twice that of ZnBr2.
To calculate the final molarity of bromide anion in the solution, we need to determine the number of moles of zinc bromide and the resulting moles of bromide anion.
1. Calculate the number of moles of zinc bromide:
- Given mass of zinc bromide = 3.50 g
- Molar mass of zinc bromide = atomic mass of Zn + 2 * atomic mass of Br
= (65.38 g/mol) + 2 * (79.90 g/mol) = 224.18 g/mol (rounded to 2 decimal places)
- Moles of zinc bromide = mass / molar mass
= 3.50 g / 224.18 g/mol = 0.0156 mol (rounded to 4 decimal places)
2. Calculate the volume of the potassium carbonate solution:
- Given volume of the solution = 300.0 mL = 0.300 L
3. Calculate the moles of bromide anion present in the solution after the reaction:
- The reaction between zinc bromide and potassium carbonate produces zinc carbonate and potassium bromide:
ZnBr2 + K2CO3 -> ZnCO3 + 2KBr
- The stoichiometric ratio is 1:2 for ZnBr2:KBr. Therefore, the moles of bromide produced is twice the moles of zinc bromide used.
- Moles of bromide anion = 2 * moles of zinc bromide = 2 * 0.0156 mol = 0.0312 mol (rounded to 4 decimal places)
4. Calculate the final molarity of bromide anion in the solution:
- Molarity (M) is defined as moles of solute divided by liters of solution:
Molarity = moles of solute / volume of solution
- Volume of solution = 0.300 L (given)
- Final molarity of bromide anion = moles of bromide anion / volume of solution
= 0.0312 mol / 0.300 L = 0.104 M (rounded to 3 decimal places)
Therefore, the final molarity of bromide anion in the solution is 0.104 M.
To calculate the final molarity of bromide anion in the solution, we need to consider the stoichiometry of the reaction between zinc bromide (ZnBr2) and potassium carbonate (K2CO3). The reaction can be written as:
ZnBr2 + K2CO3 -> ZnCO3 + 2KBr
From the balanced equation, we can see that for every 1 mole of ZnBr2, we get 2 moles of KBr. This means that the number of moles of bromide anion (Br-) produced is twice the number of moles of ZnBr2.
1. Calculate the number of moles of ZnBr2:
We are given the mass of ZnBr2, which is 3.50 g. To find the number of moles, we divide the mass by the molar mass of ZnBr2 (which can be obtained from the periodic table). The molar mass of ZnBr2 is approximately 225.20 g/mol.
moles of ZnBr2 = 3.50 g / 225.20 g/mol
2. Calculate the number of moles of bromide anion:
As mentioned earlier, the number of moles of bromide anion is twice the number of moles of ZnBr2.
moles of Br- = 2 * moles of ZnBr2
3. Calculate the new total volume of the solution:
We are given that the initial volume of the solution is 300 mL. Since the volume of the solution does not change when the zinc bromide is dissolved in it, the final volume of the solution remains the same.
4. Calculate the final molarity of bromide anion:
Molarity is defined as moles of solute divided by the volume of the solution in liters. Here, the solute is the bromide anion (Br-) and the volume of the solution is 300 mL or 0.300 L.
Molarity (M) of Br- = moles of Br- / volume of solution in L
Now that we have all the necessary information, we can plug in the values into the formula to calculate the final molarity of bromide anion.