3. (1 pt) The sum of the neutralization of HCI (aq) by NaOH (aq) (Part B) and th

Question

3. (1 pt) The sum of the neutralization of HCI (aq) by NaOH (aq) (Part B) and the ionization of HCaH302 equals the neutralization of HC2H302 (aq) by NaOH (aq) (Part C). Write the equation for Parts B and C to demonstrate how these two add together to yield the ionization of HC2H302 i.e., HC2H302 (aq) H (aq) + C2H302 (aq) Using your data and Hess's law, determine the enthalpy of ionization of HC2H302. Hint: one or more of the equations may need to be reversed. (L } HG H3O2 (aq) H + (aq) + C2H3O2-(aq) Conclusion (0.25 pt):

Explanation / Answer

HCl + NaOH ==== NaCl + H2O (B)

Neutralization of HC2H3O2 will be C

HC2H3O2 + NaOH ==== NaC2H3O2 + H2O

lets add the equations, we will need to reverse equation B since we want to remove the NaOH to do that we need 1 mole of NaOH on the reactants side and 1 mole on the products side, water is also removed

HC2H3O2 + NaOH ==== NaC2H3O2 + H2O H = -64.76 KJ / mol

NaCl + H2O ===== HCl + NaOH H = + 64.14 KJ / mol

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HC2H3O2 + NaCl====NaC2H3O2 + HCl

Now get the ionic equation

HC2H3O2 + Na + Cl====Na + C2H3O2 + H + Cl (The ions are separated, the acid does not dissociate since it is weak that is why is not splitted into ions on the left side)

Remove the ions that appear on both sides

HC2H3O2 ====C2H3O2 + H

The enthalpy according to your values will be

-64.76 + 64.14 = -0.62 KJ / mole

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