Question

A 72.4 mL solution of Cu(OH) is neutralized by 47.8 mL of a 0.56 M H2(C204) solution. What is the concentration of the Cu(OH)?​

Answer 1

Answer: The concentration of
`Cu(OH)_(2)` is 0.369 M.

Step-by-step explanation:

Given:
`V_(1)` = 72.4 mL,
`M_(1)` = ?

`V_(2)` = 47.8 mL,
`M_(2)` = 0.56 M

Formula used to calculate the concentration of
`Cu(OH)_(2)` is as follows.

`M_(1)V_(1) = M_(2)V_(2)`

Substitute the values into above formula as follows.

`M_(1)V_(1) = M_(2)V_(2)\\M_(1) * 72.4 mL = 0.56 M * 47.8 mL\\M_(1) = (0.56 M * 47.8 mL)/(72.4 mL)\\= 0.369 M`

Thus, we can conclude that the concentration of
`Cu(OH)_(2)` is 0.369 M.