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2 years ago

A 15.0 mL solution of Sr(OH)2 is neutralized with 38.5 mL of 0.350 M

Chemistry

1 answer:

Darya [45]2 years ago

7 0

Answer:

Molarity of Sr(OH)₂ = 0.47 M

Explanation:

Given data:

Volume of Sr(OH)₂ = 15.0 mL

Volume of HCl = 38.5 mL (0.0385 L)

Molarity of HCl = 0.350 M

Concentration/Molarity of Sr(OH)₂ = ?

Solution:

Chemical equation:

Sr(OH)₂ + 2HCl → SrCl₂ +2H₂O

Number of moles of HCl:

Molarity = number of moles/ volume in L

0.350 M = number of moles/0.0385 L

Number of moles = 0.350 mol/L× 0.0385 L

Number of moles = 0.0135 mol

Now we will compare the moles of HCl with Sr(OH)₂.

HCl : Sr(OH)₂

2 : 1

0.0135 : 1/2×0.0135 = 0.007 mol

Molarity/concentration of Sr(OH)₂:

Molarity = number of moles / volume in L

Molarity = 0.007 mol /0.015 L

Molarity = 0.47 M

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