Question

An aqueous solution of calcium hydroxide is standardized by titration with a 0.164 M solution of hydroiodic acid. If 17.2 mL of base are required to neutralize 14.7 mL of the acid, what is the molarity of the calcium hydroxide solution?

Answer 1

0.07 M

Step-by-step explanation:

We have to start with the reaction between the Calcium hydroxide and the Hydroiodic acid:

`Ca(OH)_2~+~HF~->~CaF_2~+~H_2O`

Then we have to balance the reaction:

`Ca(OH)_2~+~2HF~->~CaF_2~+~2H_2O`

We have to keep in mind that we have to do use the volume in "L" units (14.7 mL= 0.0147L). When we plug the values into the equation we will get:

`M=(#mol)/(L)`

`0.164~M=(#mol)/(0.0147)`

`#mol=~0.0024`

Now, we have to use the information from the balanced reaction to finding the moles of Calcium hydroxide. When we check the reaction we found a molar ratio of 1:2 ( 1 mol of
`Ca(OH)_2`: 2 mol of
`HF`). With this molar ratio, we can find the moles of
`HF`.

`0.0024~mol~HF(1~mol~Ca(OH)_2)/(2~mol~HF)`

`0.0012~mol~Ca(OH)_2`

Finally, to find the molarity we have to divide the moles of
`Ca(OH)_2` and the volume of
`Ca(OH)_2` in liters (17.2 mL=0.0172L) so:

`M=(0.012)/(0.0172)=0.07`

The molarity of
`Ca(OH)_2` is 0.07M.