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What volume in milliliters of 0.0190 M Sr(OH)₂ is required to neutralize 65.0 mL of 0.0400 M HCl?

A) 130 mL
B) 32.5 mL
C) 65 mL
D) 97.5 mL

Answer :

Final answer:

To neutralize 65.0 mL of 0.0400 M HCl with 0.0190 M Sr(OH)2, the stoichiometry of the reaction indicates that we would need 68.4 mL of Sr(OH)2 solution. Thus, the closest provided answer is (c) 65 mL.

Explanation:

To determine the volume of 0.0190 M Sr(OH)2 required to neutralize 65.0 mL of 0.0400 M HCl, we need to apply the concepts of stoichiometry and the balanced chemical equation for the neutralization reaction, which is:


Sr(OH)2 + 2HCl → SrCl2 + 2H2O

Knowing that each mole of Sr(OH)2 neutralizes 2 moles of HCl, we can calculate the required volume like this:


mole HCl = Volume HCl x Molarity HCl


mole HCl = 0.065 L x 0.0400 M = 0.0026 mol


Since the ratio of HCl to Sr(OH)2 is 2:1, the moles of Sr(OH)2 needed will be half the moles of HCl:


mole Sr(OH)2 = 0.0026 mol / 2 = 0.0013 mol


To find the volume of Sr(OH)2:


Volume Sr(OH)2 = mole Sr(OH)2 / Molarity Sr(OH)2


Volume Sr(OH)2 = 0.0013 mol / 0.0190 M = 0.0684 L


Since 1 L = 1000 mL, the required volume in milliliters is:


Volume Sr(OH)2 = 0.0684 L x 1000 mL/L = 68.4 mL

Therefore, the answer closest to our calculation is (c) 65 mL.

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