Final answer:
The total concentration of bromide ions (Br−) in the solution after dissolving 0.15 mol of CaBr2 and 0.25 mol of LiBr and diluting to 400 ml is 1.375 M.
Step-by-step explanation:
The concentration of Br− ions in the final solution can be determined by first calculating the total moles of Br− ions provided by both CaBr2 and LiBr. Since both compounds dissociate completely in water, CaBr2 provides 2 moles of Br− for every mole of CaBr2 and LiBr provides 1 mole of Br− for every mole of LiBr.
Therefore, for CaBr2: 0.15 mol × 2 = 0.30 mol Br−
And for LiBr: 0.25 mol × 1 = 0.25 mol Br−
Total moles of Br− = 0.30 mol + 0.25 mol = 0.55 mol
The final concentration of Br− is then calculated by dividing the total moles of Br− by the total volume of solution in liters. The solution volume is 400 ml which is equivalent to 0.400 L.
Concentration of Br− = 0.55 mol / 0.400 L = 1.375 M