Answer:
Molarity of unknown [tex]Ca(OH)_{2}[/tex] solution is 0.304 M.
Explanation:
Neutralization reaction: [tex]Ca(OH)_{2}+2HNO_{3}\rightarrow Ca(NO_{3})_{2}+2H_{2}O[/tex]
So, 2 moles of [tex]HNO_{3}[/tex] neutralize 1 mol of [tex]Ca(OH)_{2}[/tex]
Moles of [tex]HNO_{3}[/tex] added = [tex]\frac{0.400}{1000}\times 38.04mol=0.0152mol[/tex]
So, 0.0152 mol of [tex]HNO_{3}[/tex] neutralize 0.00760 mol of [tex]Ca(OH)_{2}[/tex]
Let's assume molarity of [tex]Ca(OH)_{2}[/tex] is C (M)
Then number of moles of [tex]Ca(OH)_{2}[/tex] in 25.00 mL of C (M) of [tex]Ca(OH)_{2}[/tex]
= [tex]\frac{C}{1000}\times 25.00mol[/tex]
Hence, [tex]\frac{C}{1000}\times 25.00=0.00760[/tex]
or, C = 0.304
So, molarity of unknown [tex]Ca(OH)_{2}[/tex] solution is 0.304 M
