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

Answer:

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 -> SrCl_2 + 2H_2O#

  • 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" xx "Volume"#

is equal to

#= "0.8 moles"/"1000 ml" * "525ml" = "420 millimoles"#

Here

#"milli" = 1/1000#

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 millimoles"/2 = "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"#