If 525 mL of 0.80 M HCl solution is neutralized with 315 mL of Sr(OH)2 solution what is the molarity of the Sr(OH)2? __ HCl + __ Sr(OH)2  __ SrCl2 + __ H2O

1 Answer
Vijay Raghavendran · Stefan V.
Jun 7, 2014

The molarity of Sr(OH)_2Sr(OH)2 is "0.67 M"0.67 M

Explanation:

Acids and bases neutralize each other. Strontium hydroxide will neutralize hydrochloric acid to produce a salt and water. The reaction will be

Sr(OH)_2 + 2HCl -> SrCl_2 + 2H_2OSr(OH)2+2HCl→SrCl2+2H2O

  • the volume of strontium hydroxide is "315 mL"315 mL
  • the volume of hydrochloric acid is "525 mL"525 mL
  • the molarity of hydrochloric acid is "0.80 M"0.80 M

Steps:

The number of moles of HClHCl consumed in the process

"Moles" = "Concentration" xx "Volume"Moles=Concentration×Volume

is equal to

= "0.8 moles"/"1000 ml" * "525ml" = "420 millimoles"=0.8 moles1000 ml⋅525ml=420 millimoles

Here

"milli" = 1/1000milli=11000

For every mole of HClHCl, half a mole of strontium hydroxide is neutralized as observed from the stoichiometry (or balanced equation).

Therefore, the number of moles of strontium hydroxide is equal to

"420 millimoles"/2 = "210 millimoles"420 millimoles2=210 millimoles

Therefore, the concentration of strontium hydroxide will be

"210 millimoles" / "315 ml" = "0.66667 millimoles/mL" = "0.67 moles/L" = "0.67 M"210 millimoles315 ml=0.66667 millimoles/mL=0.67 moles/L=0.67 M

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