# 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

Jun 7, 2014

The molarity of $S r {\left(O H\right)}_{2}$ is $\text{0.67 M}$

#### Explanation:

Acids and bases neutralize each other. Strontium hydroxide will neutralize hydrochloric acid to produce a salt and water. The reaction will be

$S r {\left(O H\right)}_{2} + 2 H C l \to S r C {l}_{2} + 2 {H}_{2} O$

• the volume of strontium hydroxide is $\text{315 mL}$
• the volume of hydrochloric acid is $\text{525 mL}$
• the molarity of hydrochloric acid is $\text{0.80 M}$

Steps:

The number of moles of $H C l$ consumed in the process

$\text{Moles" = "Concentration" xx "Volume}$

is equal to

$= \text{0.8 moles"/"1000 ml" * "525ml" = "420 millimoles}$

Here

$\text{milli} = \frac{1}{1000}$

For every mole of $H C l$, 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

$\text{420 millimoles"/2 = "210 millimoles}$

Therefore, the concentration of strontium hydroxide will be

$\text{210 millimoles" / "315 ml" = "0.66667 millimoles/mL" = "0.67 moles/L" = "0.67 M}$