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Sav [38]
3 years ago
12

For the reaction H2PO4- HAsO4 2-HPO4 2-+H2AsO4- what species are a conjugate acid-base pair?

Chemistry
2 answers:
aleksklad [387]3 years ago
8 0

Ionic equation:

\text{H}_2\text{PO}_4^{-} + \text{HAsO}_4^{2-} \to \text{HPO}_4^{2-} + \text{H}_2\text{AsO}_4^{-}

The acid and base in a conjugate pair differ by only one proton \text{H}^{+}. The acid loses one proton to produce a conjugate base, whereas the base gains a proton to produce its conjugate acid.

\text{H}_2\text{PO}_4^{-} loses one proton to produce \text{HPO}_4^{2-} in this reaction.

\text{H}_2\text{PO}_4^{-} \to \text{H}^{+} + \text{HPO}_4^{2-}

Meanwhile, \text{HAsO}_4^{2-} gains one proton to form \text{H}_2\text{AsO}_4^{-}.

\text{HAsO}_4^{2-} + \text{H}^{+} \to \text{H}_2\text{AsO}_4^{-}

Therefore

  • \text{H}_2\text{PO}_4^{-} is the conjugate acid  \text{HPO}_4^{2-}, its conjugate base.
  • \text{HAsO}_4^{2-} is the conjugate base of \text{H}_2\text{AsO}_4^{-}, its conjugate acid.
kondaur [170]3 years ago
7 0

Ionic equation:




The acid and base in a conjugate pair differ by only one proton . The acid loses one proton to produce a conjugate base, whereas the base gains a proton to produce its conjugate acid.


loses one proton to produce  in this reaction.




Meanwhile,  gains one proton to form .




Therefore


  • is the conjugate acid  , its conjugate base.
  • is the conjugate base of , its conjugate acid.


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<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]

For the given chemical reaction:

C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

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