Answer: The concentration of [tex]Cu(OH)_{2}[/tex] is 0.369 M.
Explanation:
Given: [tex]V_{1}[/tex] = 72.4 mL, [tex]M_{1}[/tex] = ?
[tex]V_{2}[/tex] = 47.8 mL, [tex]M_{2}[/tex] = 0.56 M
Formula used to calculate the concentration of [tex]Cu(OH)_{2}[/tex] is as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
Substitute the values into above formula as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}\\M_{1} \times 72.4 mL = 0.56 M \times 47.8 mL\\M_{1} = \frac{0.56 M \times 47.8 mL}{72.4 mL}\\= 0.369 M[/tex]
Thus, we can conclude that the concentration of [tex]Cu(OH)_{2}[/tex] is 0.369 M.
