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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

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