Question

100.0 mL of 0.500MCaBr²and 50.0 mL of 1.00MNaBr are mixed. What is the concentration of bromide ion in the resulting solution? A) 0.333M

B) 0.500M
(C) 0.667M
D) 0.750M
(C) 1.00M
​(100 mL)(.500)+(50 mL)(1 m)=

Answer 1

The concentration of bromide ions in the mixed solution of 0.500M CaBr² and 1.00M NaBr is 0.667M, which corresponds to answer choice (C).

Step-by-step explanation:

To calculate the concentration of bromide ions in the resulting solution after mixing 0.500M CaBr² and 1.00M NaBr, we must account for both the initial moles of bromide from each source and the final volume of the solution.

From the calcium bromide solution (CaBr²):
Moles of bromide = 0.500 moles/L × 0.100 L = 0.050 moles of Br⁻ (remember each CaBr² provides 2 Br⁻)

From the sodium bromide solution (NaBr):
Moles of bromide = 1.00 moles/L × 0.050 L = 0.050 moles of Br⁻ (each NaBr provides 1 Br⁻)

Total moles of bromide = 0.050 + 0.050 = 0.100 moles of Br⁻

The final volume after mixing = 0.100 L + 0.050 L = 0.150 L

Concentration of bromide in the final solution is:

Molarity (M) = Moles of solute / Volume of solution
Molarity of Br⁻ = 0.100 moles / 0.150 L = 0.667M

Thus, the correct answer is (C) 0.667M.

Answer 2

To find the concentration of bromide ions in the resulting solution, calculate the moles of bromide ions from each solution and then determine the concentration by dividing the total moles by the total volume of the solution.

Step-by-step explanation:

To find the concentration of bromide ions in the resulting solution, we need to calculate the total moles of bromide ions and then determine the concentration.

First, calculate the moles of bromide ion from each solution using the formula n = C * V, where C is the concentration in moles per liter and V is the volume in liters. For CaBr2, n = 0.500 M * 0.100 L = 0.050 moles. For NaBr, n = 1.00 M * 0.050 L = 0.050 moles.

Next, add the moles of bromide ion from each solution: 0.050 moles + 0.050 moles = 0.100 moles.

Finally, calculate the concentration of bromide ion in the resulting solution by dividing the moles by the total volume of the solution: 0.100 moles / (0.100 L + 0.050 L) = 0.667 M.