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If 525 mL of a 0.80 M HCl solution is neutralized with 315 mL of a [tex]$Sr(OH)_2$[/tex] solution, what is the molarity of the [tex]$Sr(OH)_2$[/tex]?

[tex]
\[ \text{Reaction: } \text{2 } HCl + Sr(OH)_2 \rightarrow SrCl_2 + 2 H_2O \]
[/tex]

Answer :

To solve the problem of finding the molarity of the [tex]\( Sr(OH)_2 \)[/tex] solution, let's go through the solution step-by-step:

1. Identify the Reaction:
The balanced chemical equation for the neutralization reaction is:
[tex]\[
2 \, \text{HCl} + \text{Sr(OH)}_2 \rightarrow \text{SrCl}_2 + 2 \, \text{H}_2\text{O}
\][/tex]
From the equation, we see that 2 moles of HCl react with 1 mole of [tex]\( \text{Sr(OH)}_2 \)[/tex].

2. Calculate Moles of HCl:
You are given 525 mL of a 0.80 M HCl solution. First, convert the volume from milliliters to liters:
[tex]\[
\text{Volume of HCl in liters} = \frac{525}{1000} = 0.525 \, \text{L}
\][/tex]
Use the molarity formula to find the moles of HCl:
[tex]\[
\text{Moles of HCl} = \text{Molarity} \times \text{Volume in liters} = 0.80 \times 0.525 = 0.42 \, \text{moles}
\][/tex]

3. Use Stoichiometry to Find Moles of [tex]\( Sr(OH)_2 \)[/tex]:
According to the stoichiometry of the reaction (2:1 ratio), for every 2 moles of HCl, 1 mole of [tex]\( \text{Sr(OH)}_2 \)[/tex] is needed. Therefore, calculate the moles of [tex]\( \text{Sr(OH)}_2 \)[/tex]:
[tex]\[
\text{Moles of } \text{Sr(OH)}_2 = \frac{\text{Moles of HCl}}{2} = \frac{0.42}{2} = 0.21 \, \text{moles}
\][/tex]

4. Calculate the Molarity of [tex]\( Sr(OH)_2 \)[/tex]:
You are given that 315 mL of [tex]\( Sr(OH)_2 \)[/tex] solution was used, so convert this volume to liters:
[tex]\[
\text{Volume of } \text{Sr(OH)}_2 \text{ in liters} = \frac{315}{1000} = 0.315 \, \text{L}
\][/tex]
Now find the molarity using the formula [tex]\( \text{Molarity} = \frac{\text{Moles}}{\text{Volume in liters}} \)[/tex]:
[tex]\[
\text{Molarity of } \text{Sr(OH)}_2 = \frac{0.21}{0.315} \approx 0.67 \, \text{M}
\][/tex]

So, the molarity of the [tex]\( Sr(OH)_2 \)[/tex] solution is approximately 0.67 M.

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