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)2 is 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→SrCl2+2H2O

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

Steps:

The number of moles of HCl consumed in the process

Moles=Concentration×Volume

is equal to

=0.8 moles1000 ml⋅525ml=420 millimoles

Here

milli=11000

For every mole of HCl, 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 millimoles2=210 millimoles

Therefore, the concentration of strontium hydroxide will be

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

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