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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"##