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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 _x0001_ __ SrCl2 + __ H2O
The of ##Sr(OH)_2## is ##"0.67 M"##
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 (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"##