Question

50.00 mL of unknown calcium hydroxide solution is titrated with 0.250 M standard nitric acid solution. If 43.43 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

The molarity of the unknown calcium hydroxide solution is determined by the stoichiometry of its reaction with nitric acid in a titration. After calculating the number of moles of nitric acid used, this number is halved to find the moles of calcium hydroxide due to the 1:2 mole ratio. Dividing by the original volume of calcium hydroxide solution gives the molarity, which is 0.108575 M.

Step-by-step explanation:

The molarity of the unknown calcium hydroxide solution can be calculated using the volume and molarity of the titrant (nitric acid), and the stoichiometry of the neutralization reaction between calcium hydroxide and nitric acid, which is:

Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(l)

Since 2 moles of nitric acid reacts with 1 mole of calcium hydroxide, the moles of calcium hydroxide that reacted are half the moles of nitric acid. First, calculate the moles of HNO3 used:

# moles HNO3 = volume of HNO3 (L) × molarity of HNO3 = (0.04343 L) × (0.250 M) = 0.0108575 mol

Now, calculate the moles of Ca(OH)2:

# moles Ca(OH)2 = 0.0108575 mol HNO3 / 2 = 0.00542875 mol

Finally, calculate the molarity of the calcium hydroxide solution:

Molarity of Ca(OH)2 = moles of Ca(OH)2 / volume of solution in liters = 0.00542875 mol / 0.05000 L = 0.108575 M

Answer 2

Answer: Molarity of the unknown calcium hydroxide solution is 0.217 M.

Step-by-step explanation:

Given:
`V_(1)` = 50.00 mL,
`M_(1)` = ?

`V_(2)` = 43.43 mL,
`M_(2)` = 0.250 M

Formula used 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) * 50.00 mL = 0.250 M * 43.43 mL\\M_(1) = (0.250 M * 43.43 mL)/(50.00 mL)\\= 0.217 M`

Thus, we can conclude that molarity of the unknown calcium hydroxide solution is 0.217 M.