A titration problem

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?

The molarity of Sr(OH)2 is 0.67 M

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.

Volume of Strontium Hydroxide is 315 mL
Volume of Hydrochloric acid is 525 mL
Molarity of HCl is 0.80 M

Steps:

Moles of HCl consumed in the process (Moles = Concentration * Volume)

= (0.8 moles/1000 ml)*(525ml)
= 420 millimoles

(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, moles of Strontium hydroxide is 240 millimoles.

Volume of Strontium hydroxide is 315 ml.

Therefore concentration = moles / volume

= 240 millimoles / 315 ml = 0.6666 millimoles / mL = 0.67 moles / L
=0.67 M

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