Examples for activities

2.2.1 Examples for activities

With \(m_+ = \nu_+ m_B\), and \(m_-=\nu_-m_B\) we get from Eq. (2.13)

\begin{equation*} a_B = \left(\gamma_\pm\left(\frac{m_\pm}{m^0}\right)\right)^\nu = \gamma_\pm^\nu \frac{(\nu_+m_B)^{\nu+}(\nu_-m_B)^{\nu-}}{(m^0)^\nu} \quad . \label{act_2} \end{equation*}

(2.14)

\(\text{NaCl}\):

\begin{equation*} a_{\text{NaCl}} = a_{\text{Na}+}a_{\text{Cl}-} = \gamma_\pm^2 \frac{(m_{\text{NaCl}})^2}{(m^0)^2} \quad . \label{act_NaCl} \end{equation*} (2.15)

\(\text{Fe}(\text{ClO}_4)_3\):

\begin{equation*} \begin{split} a_{\text{Fe}(\text{ClO}_4)_3} & = a_{\text{Fe}3+}\left(a_{\text{ClO4}-}\right)^3 \\ & = \gamma_\pm^4 \frac{m_{\text{Fe}(\text{ClO}_4)_3} (3\,m_{\text{Fe}(\text{ClO}_4)_3})^3}{(m^0)^4} = \gamma_\pm^4 \frac{27 (m_{\text{Fe}(\text{ClO}_4)_3})^4}{(m^0)^4} \quad ; \label{act_FeClO43} \end{split} \end{equation*}

(2.16)

here \(m_{\text{Fe}3+} = m_{\text{Fe}(\text{ClO}_4)_3}\) and \(m_{\text{ClO}4-} = 3\,m_{\text{Fe}(\text{ClO}_4)_3}\) has been used.

Thus the key-point is always how to calculate (or measure) \(\gamma_\pm\).

With frame

Chemical potential and activity of ionic solutions/salts