A student carried out a titration using HC2H3O2(aq)HC2H3O2(aq) and NaOH(aq)NaOH(aq). The net ionic equation for the neutralization reaction

Question

A student carried out a titration using HC2H3O2(aq)HC2H3O2(aq) and NaOH(aq)NaOH(aq). The net ionic equation for the neutralization reaction that occurs during the titration is represented above. The NaOH(aq)NaOH(aq) was added from a buret to the HC2H3O2(aq)HC2H3O2(aq) in a flask. The equivalence point was reached when a total of 20.0mL20.0mL of NaOH(aq)NaOH(aq) had been added to the flask. How does the amount of HC2H3O2(aq)HC2H3O2(aq) in the flask after the addition of 5.0mL5.0mL of NaOH(aq)NaOH(aq) compare to the amount of HC2H3O2(aq)HC2H3O2(aq) in the flask after the addition of 1.0mL1.0mL of NaOH(aq)NaOH(aq), and what is the reason for this result

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Thiên Di 4 years 2021-08-16T16:55:43+00:00 1 Answers 166 views 0

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  1. Gerda
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    2021-08-16T16:57:18+00:00
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    Answer:

    The amount of HC₂H3₃2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.

    Explanation:

    Equation of the reaction between acetic acid, HC₂H₃O₂(aq) and sodium hydroxide, NaOH(aq) is given below:

    CH₃COOH (aq) + NaOH (aq) —-> CH₃COONa (aq) + H₂O

    The equation of the reaction shows that acetic acid andsodium hydroxide will react in a 1:1 ratio

    Since the concentration of NaOH was not given, we can assume that the concentration is 0.01 M

    Moles of NaOH in 5.0 mL of 0.01 M NaOH = 0.01 × 5/1000 = 0.00005 moles

    Moles of NaOH in 1.0 mL of 0.01 M NaOH = 0.01 ×1/1000 = 0.0001 moles

    Ratio of moles of NaOH in 5.0 mL to 1.0 mL = 0.00005/0.00001 = 5

    There are five times more moles of NaOH in 5.0 mL than in 1.0 mL and this means that 5 times more the quantity of HC₂H₃O2(aq) required to react with 1.0 mL NaoH is needed to react with 5.0 mL NaOH.

    Therefore, the amount of HC₂H₃O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.

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