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How many mL of 0.14 Sr(OH)2 are needed to neutralize 35 mL of 0.16 M HCl ?

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How many mL of 0.14 Sr(OH)2 are needed to neutralize 35 mL of 0.16 M HCl ?

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The reaction between strontium hydroxide and hydrochloric acid is given by the chemical equation: Sr(OH)2 + 2HCl --> SrCl2 + 2H2O

From the chemical equation it can be seen that to neutralize 1 mole of Sr(OH)2, 2 moles of HCl are required.

The concentration of the HCl solution available is 0.16 M. 35 mL of 0.16 M Sr(OH)2 has 0.16*35/1000 = 0.0056 moles of HCl. This can neutralize 0.0028 mole of Sr(OH)2.

The concentration of the Sr(OH)2 solution is 0.14 M. Let the volume of Sr(OH)2 solution required for the neutralization reaction be V.

V*0.14 = 0.0028

V = 0.0028/0.14 = 0.02 L

The volume of a 0.14 M solution of Sr(OH)2 required to neutralize 35 mL of a 0.16 M solution of HCl is 20 mL.

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