A 15.0 mL solution of Sr(OH)2 is neutralized with 38.5 mL of 0.350 M HCI. What is the concentration of the original Sr(OH)2 solution?

The required molarity of 15.0 mL solution is 0.449 M and the further calculation can be defined as follows:

Given that:

Volume of solution = 15.0 mL

n-factor of solution = 2

Molarity of HCl solution = 0.350 M

Volume of HCl solution = 38.5 mL

n-factor of HCl solution = 1

To find:

Molarity of solution required.

Formula used to calculate the molarity of solution is:

where, are the n-factor, molarity and volume of acidic solution while are the n-factor, molarity and volume of basic solution.

Put the values in the above formula:

So, the required molarity of solution is 0.449 M.

Learn more about neutralization reaction: link

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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it means there are 3 atoms of oxygen