Question

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?

Answer 1

Answer:

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

$\text{Moles} = \text{0.300 L} \times \dfrac{\text{0.50 mol}}{\text{1 L}} = \text{ 0.150 mol}$

(b) RbBr

$\text{Moles} = \text{0.300 L} \times \dfrac{\text{0.70 mol}}{\text{1 L}} = \text{ 0.210 mol}$

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

$c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.360 mol}}{\text{0.600 L}} = \textbf{0.60 mol/L}$

 

Answer 2

0.60 mole is the answer I did it in it was right