Question

100.0 ml of 0.500 m cabr2 and 50.0 ml of 1.00 m nabr are mixed. what is the concentration of bromide ion in the resulting solution?

Answer 1

The concentration of bromide ion in the resulting solution is 1.00 M.

Step-by-step explanation:

To find the concentration of bromide ion in the resulting solution, we need to calculate the moles of bromide in each solution and then add them together.

For the first solution, 100.0 mL of 0.500 M CaBr2:

Moles of CaBr2 = (0.500 mol/L) * (0.100 L) = 0.050 mol

Moles of Br- ion = 0.050 mol * 2 = 0.100 mol

For the second solution, 50.0 mL of 1.00 M NaBr:

Moles of NaBr = (1.00 mol/L) * (0.050 L) = 0.050 mol

Moles of Br- ion = 0.050 mol * 1 = 0.050 mol

The total moles of Br- ion in the resulting solution is 0.100 mol + 0.050 mol = 0.150 mol.

Now, we need to find the final volume of the resulting solution:

Volume of the resulting solution = 100.0 mL + 50.0 mL = 150.0 mL = 0.150 L.

Finally, we can calculate the concentration of bromide ion in the resulting solution:

Concentration of bromide ion = moles of Br- ion / volume of solution = 0.150 mol / 0.150 L = 1.00 M.

Answer 2

To find the concentration of bromide ion in the resulting solution, we need to calculate the total moles of bromide ion in the final solution and divide by the total volume of the solution.

First, we can find the number of moles of bromide ion in the initial solution of cabr2 by multiplying the volume of the solution (100.0 mL) by the concentration (0.500 M):

moles = (100.0 mL) * (0.500 M) = 50.0 moles

And also in Nabr by multiplying the volume of the solution (50.0 mL) by the concentration (1.00 M)

moles = (50.0 mL) * (1.00 M) = 50.0 moles

Next, we add the total moles of bromide ion of the two solutions, which would be 50.0 moles + 50.0 moles = 100.0 moles

The final volume of the solution is 150.0 ml

Then, we divide the total number of moles by the total volume of the solution to find the concentration:

concentration = moles / volume = 100.0 moles / 150.0 mL = 0.667 M

So the concentration of bromide ion in the resulting solution is 0.667 M (in units of Molarity)