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

Answer:

Molarity of unknown $Ca(OH)_{2}$ solution is 0.304 M.

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 = $\frac{0.400}{1000}\times 38.04mol=0.0152mol$

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 $Ca(OH)_{2}$

= $\frac{C}{1000}\times 25.00mol$

Hence, $\frac{C}{1000}\times 25.00=0.00760$

or, C = 0.304

So, molarity of unknown $Ca(OH)_{2}$ solution is 0.304 M