Questions: In the laboratory you dissolve 23.9 g of cobalt(II) bromide in a volumetric flask and add water to a total volume of 500 mL. What is the molarity of the solution? M. What is the concentration of the cobalt(II) cation? M. What is the concentration of the bromide anion? M.

Solution Steps

Cobalt(II) bromide, \(\text{CoBr}_2\), consists of one cobalt (Co) atom and two bromine (Br) atoms. The atomic masses are approximately:

• Cobalt (Co): 58.93 g/mol
• Bromine (Br): 79.90 g/mol

The molar mass of \(\text{CoBr}_2\) is calculated as follows: \[ \text{Molar mass of CoBr}_2 = 58.93 + 2 \times 79.90 = 218.73 \, \text{g/mol} \]

Using the mass of cobalt(II) bromide given (23.9 g), we calculate the number of moles: \[ \text{Moles of CoBr}_2 = \frac{23.9 \, \text{g}}{218.73 \, \text{g/mol}} = 0.1093 \, \text{mol} \]

Molarity (M) is defined as moles of solute per liter of solution. The total volume of the solution is 500 mL, which is 0.500 L. Therefore, the molarity of the \(\text{CoBr}_2\) solution is: \[ \text{Molarity of CoBr}_2 = \frac{0.1093 \, \text{mol}}{0.500 \, \text{L}} = 0.2186 \, \text{M} \]

Cobalt(II) bromide dissociates in water as follows: \[ \text{CoBr}_2 \rightarrow \text{Co}^{2+} + 2\text{Br}^- \] The concentration of \(\text{Co}^{2+}\) is equal to the molarity of the \(\text{CoBr}_2\) solution: \[ [\text{Co}^{2+}] = 0.2186 \, \text{M} \]

For every mole of \(\text{CoBr}_2\) that dissociates, two moles of \(\text{Br}^-\) are produced. Therefore, the concentration of \(\text{Br}^-\) is: \[ [\text{Br}^-] = 2 \times 0.2186 \, \text{M} = 0.4372 \, \text{M} \]

Final Answer