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During a titration the following data were collected. A 20.0 mL portion of solution of an unknown acid HX was titrated with 2.0 M KOH. It required 60.0 mL of the base to neutralize the sample. What is the molarity of the acid HX?

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Answer:

The molarity of the acid HX is 6.0 M.

Explanation:

We determine the amount of moles of KOH used to neutralize the acid:

[tex]\frac{2.0moles_{KOH}}{1000ml} *60ml[/tex]=0.12 moles KOH

Then, we calculate the amount of moles of acid:

0.12 moles KOH×[tex]\frac{1 mole HX}{1 moles KOH}[/tex]=0.12 moles HX

The molarity of HX is:

[tex]\frac{0.12 moles HX}{20ml} *\frac{1000ml}{1l}[/tex]=6.0 M

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