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.
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.
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