Question

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

Answer 1

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"`