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abruzzese [7]
2 years ago
11

A 25.0 mL solution of Sr(OH)₂ is neutralized with 31.6 mL of 0.150 M HBr. What is the concentration of the original Sr(OH)₂ solu

tion?
Chemistry
1 answer:
Savatey [412]2 years ago
4 0

A 25.0 mL solution of Sr(OH)₂ is neutralized with 31.6 mL of 0.150 M HBr, then the concentration of the original Sr(OH)₂ solution is 0.189M.

<h3>How do we calculate the concentration?</h3>

Concentration of the solution will be calculated by using the below chemical reaction as:

M₁V₁ = M₂V₂, where

M₁ & V₁ is the molarity and volume of HBr solution and M₂ & V₂ is the molarity and volume of original Sr(OH)₂ solution.

On putting values on above equation by taking from question, we get

M₂ = (0.15)(31.6) / (25) = 0.189 M

Hence required molarity of Sr(OH)₂ solution is 0.189M.

To know more about molarity, visit the below link:

brainly.com/question/24305514

#SPJ1

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<h3>What is the net ionic equation?</h3>

The term net ionic equation refers to the equation that shows the ions that underwent a change in the reaction. We have to note that the reaction species here must be ionic species which are able to dissociate into ions in solutions.

Now the first step is to put down the molecular equation. The molecular equation shows the reaction of the compounds as follows;

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