Question

In his acid-base studies, Arrhenius discovered an important fact involving reactions like the following:

$$\begin{gathered} {\mathbf{\text{KOH(aq) + HNO}}}_{\mathbf{\text{3}}}\mathbf{\text{(aq)n?}} \\ \mathbf{\text{NaOH(aq) + HCl(aq)n?}} \end{gathered}$$

(a) Complete the reactions and use the data for the individual ions in Appendix B to calculate each$\mathbf{∆}{\mathit{H}}_{\mathbf{r}\mathbf{\times }{\mathbf{n}}^{\mathbf{*}}}$

(b) Explain your results and use them to predict $\mathbf{∆}{\mathit{H}}_{\mathbf{r}\mathbf{\times }{\mathbf{n}}^{\mathbf{*}}}$for$\mathbf{\text{KOH(aq) + HCl(aq)}}\mathbf{\to }\mathbf{\text{?}}$

Short answer

a) For the reaction of NaOH and HCI,$∆H_{r\times n}^{\circ }=-55.9kJ/mol$

b)$∆H_{r\times n}^{\circ }=-55.9kJ/mol$

Step-by-step solution

Complete the reactions

(a)

First, write the ionic equation of the given reactants.

${\text{KOH and HNO}}_{\text{3}}$

${\text{K}}^{\text{+}}{\text{+ OH}}^{\text{-}}{\text{+H}}^{\text{+}}{\text{+ NO}}_{\text{3}}^{\text{-}}\to {\text{K}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{O + NO}}_{\text{3}}^{\text{-}}$

Cancel out the spectator ions and write the net equation.

${\text{OH}}^{\text{-}}{\text{+H}}^{\text{+}}\to {\text{H}}_{\text{2}}\text{O}$

Then, from Appendix B, calculate the$∆H_{r\times n}^{\circ }$.

role="math" localid="1663412574540" $$\begin{gathered} ∆H_{r\times n}^{\circ }=\sum {\text{n∆H}}_f^{\circ }\text{products -}\sum \text{m}∆H_f^{\circ }\text{reactants} \\ =∆H_f^{\circ },{\text{H}}_{\text{2}}\text{O(l) -[}∆H_f^{\circ },{\text{OH}}^{\text{-}}\text{(aq) +}∆H_f^{\circ },{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)]} \\ =-285.840kJ/mol-[\text{( - 229.94kJ/mol)+0]}∆H_{r\times n}^{\circ } \\ =\text{- 55.9kJ/mol} \end{gathered}$$

$\text{NaOH and HCl}$

${\text{Na}}^{\text{+}}{\text{+ OH}}^{\text{-}}{\text{+H}}^{\text{+}}{\text{+ Cl}}^{\text{-}}\to {\text{Na}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{O + Cl}}^{\text{-}}$

Cancel out the spectator ions and write the net equation.

${\text{OH}}^{\text{-}}{\text{+H}}^{\text{+}}\to {\text{H}}_{\text{2}}\text{O}$

Since it has the same net equation as the previous reaction, the$∆H_{r\times n}^{\circ }$will be the same. Therefore, for the reaction of NaOH and HCI,role="math" localid="1663412715397" $∆H_{r\times n}^{\circ }=-55.9kJ/mol$

Explain the results

(b)

The reactants in the reactions in the previous problem are all strong acids and bases. The reaction between a strong acid and strong base is a neutratization reaction and they will all have the same net equation.

$$\begin{gathered} {\text{K}}^{\text{+}}{\text{+ OH}}^{\text{-}}{\text{+H}}^{\text{+}}{\text{+ Cl}}^{\text{-}}\to {\text{K}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{O + Cl}}^{\text{-}} \\ {\text{OH}}^{\text{-}}{\text{+H}}^{\text{+}}\to {\text{H}}_{\text{2}}\text{O} \end{gathered}$$

Hence the $∆H_{r\times n}^{\circ }=-55.9kJ/mol$