Suppose 300 mL of 0.50 M lithium bromide solution and 300 mL of 0.70 M rubidium bromide solution are combined. What is the concentration of the bromide ion in the resulting solution?

The concentration of Br ions in the resulting solution of LiBr and RbBr has been 0.6 M.

The addition of LiBr and RbBr has been dissociated into the equal moles of Li, Rb, and Br.

Thus 1 mole of LiBr = 1 mole Br

1 mole RbBr = 1 mole Br.

The moles of LiBr in 0.5 M solution:

Molarity =

0.5 =

Moles of LiBr = 0.15 mol

The moles of Br from LiBr = 0.15 mol.

The moles of RbBr in 0.7 M solution:

Molarity =

0.7 =

Moles of RbBr = 0.21 mol

The moles of Br from RbBr = 0.21 mol.

The total moles of Br ions from LiBr and RbBr has been :

= 0.15 + 0.21

= 0.36 mol.

The total volume of the solution will be:

= 300 + 300 ml

= 600 ml.

The concentration of the Br ion has been:

Molarity =

Molarity of Br ions =  

Molarity of Br ions = 0.6 M.

The concentration of Br ions in the resulting solution of LiBr and RbBr has been 0.6 M.

For more information about the concentration of the sample, refer to the link:

link

0.60 mol·L⁻¹  

Explanation:

Data:  

LiBr: c = 0.50 mol/L; V =300 mL

RbBr: c = 0.70 mol/L; V =300 mL

1. Calculate the moles of Br⁻ in each solution

(a) LiBr

(b) RbBr

2. Calculate the molar concentration of Br⁻

(a) Moles of Br⁻

n = 0.150 mol  + 0.210 mol = 0.360 mol

(b) Volume of solution

V = 300 mL + 300 mL = 600 mL = 0.600 L

(c) Molar concentration

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c) should be the answer.