Question

25.00 mL of unknown calcium hydroxide solution is titrated with 0.400 M standard nitric acid solution. If 38.04 mL of the standard acid solution is required to reach a phenolphthalein endpoint, what is the molarity of the unknown calcium hydroxide solution

Answer 1

Molarity of unknown
`Ca(OH)_(2)` solution is 0.304 M.

Step-by-step explanation:

Neutralization reaction:
`Ca(OH)_(2)+2HNO_(3)\rightarrow Ca(NO_(3))_(2)+2H_(2)O`

So, 2 moles of
`HNO_(3)` neutralize 1 mol of
`Ca(OH)_(2)`

Moles of
`HNO_(3)` added =
`(0.400)/(1000)* 38.04mol=0.0152mol`

So, 0.0152 mol of
`HNO_(3)` neutralize 0.00760 mol of
`Ca(OH)_(2)`

Let's assume molarity of
`Ca(OH)_(2)` is C (M)

Then number of moles of
`Ca(OH)_(2)` in 25.00 mL of C (M) of
`Ca(OH)_(2)`

=
`(C)/(1000)* 25.00mol`

Hence,
`(C)/(1000)* 25.00=0.00760`

or, C = 0.304

So, molarity of unknown
`Ca(OH)_(2)` solution is 0.304 M