Problem: Consider the following acid-neutralization reactions involving the strong base NaOH(aq):HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)NH4+(aq) + NaOH(aq) → NH3(aq) + Na+(aq) + H2O(l)In the third equation NH4+(aq) is acting as an acid. Based on the value of ΔH˚ for this reaction, do you think it is a strong or a weak acid?

FREE Expert Solution

$\boxed{\mathbf{\Delta H}{\mathbf{°}}_{\mathbf{rxn}}\mathbf{=}\mathbf{\Delta H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}\mathbf{ }\mathbf{prod}}\mathbf{-}\mathbf{\Delta H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}\mathbf{ }\mathbf{react}}}$

HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)

$$\begin{gathered} \mathbf{\Delta H}{\mathbf{°}}_{\mathbf{rxn}}\mathbf{=}\left[\mathbf{(}\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{NaNO}}_{\mathbf{3}}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{-}\mathbf{446}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{kJ}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{NaNO}}_{\mathbf{3}}}}\mathbf{\right)}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{)}\left(\frac{\mathbf{-}\mathbf{285}\mathbf{.}\mathbf{83}\mathbf{ }\mathbf{kJ}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\right)\right] \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\left[\mathbf{(}\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{HNO}}_{\mathbf{3}}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{-}\mathbf{206}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{kJ}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{HNO}}_{\mathbf{3}}}}\mathbf{\right)}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{NaOH}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{-}\mathbf{469}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{kJ}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{NaOH}}}\mathbf{\right)}\right] \end{gathered}$$

ΔH°_rxn = -55.83 kJ/mol

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)