1. The neutralization of HC2H3O2(aq) by NaOH(aq) can be considered to be the sum
ID: 513258 • Letter: 1
Question
1. The neutralization of HC2H3O2(aq) by NaOH(aq) can be considered to be the sum of the neutralization of H+(aq) by OH- and ionization of HC2H3O2 (HC2H3O2<==> H+ + C2H3O2). Write the net ionic equation for each of these reactions and demonstrate how two of them add together to yield the third. using your data Hess's law, determine the enthalpy of ionization of HC2H3O2.
(25 ml of 1M HC2H3O2 and 25 mL of 1M of NaOH at room temperature)
2. The enthalpy of neutralization of HCN(aq) with NaOH(aq) is -12.0 KJ/mol. Using your value for the enthalpy of neutralization of H+(aq) and OH-(aq) and the technique used in question 1, determine the enthalpy of ionization of HCN. compare the relative amounts of ionization of HCN and HC2H3O2. Which is the stronger acid?(the stronger acid, the lower the H of ionization)
3. assume that 100 mL of 1M NaOH react at room temperature with 100 mL of 1M HCl. calculate the following: q(solution), q(rxn), t, H.
Explanation / Answer
Answer 1.)
NaOH -----> Na+ + OH-
CH3COOH -------> CH3COO- + OH-
When the two react, the ionic reaction can be written as:
Na+ + OH- + CH3COO- + H+ --------> Na+ + CH3COO- + H2O reaction [1]
To write the net ionic equation, cross out the spectator ions from reaction 1. After the removal of spectator ions, the ions that remain behind are:
H+ + OH- -------> H2O Net ionic equation
Heat of neutralization is given as:
H+ + OH- -------> H2O delta H = -57 kJ/mol [1]
Net ionic equation for neutralization of acetic acid is:
CH3COOH + OH- ---------> CH3COO- + H2O delta H = -55.2 kJ/mol [2]
We have to find out delta H for [3] below:
CH3COOH -------> CH3COO- + OH- [3]
[2] - [1] will give you delta H for [3]
-55.2 - (-57) = +1.8 kJ/mol
The enthalpy of ionization of acetic acid is +1.8 kJ/mol
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