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Ok, this is a simple problem, but I cant see how to get the answer. I calculated the activity coefficient, $\gamma_\pm$ = 0.389, and I need to find the activity $a_\pm$ for a 0.0120m solution of $Na_3PO_4$. I know the answer is 0.106 and that $$a_\pm = \frac{m_\pm}{m_0}\gamma_\pm$$ I must be pluggin in the wrong numbers. What is the top m? and isnt the bottom just 1? Thanks

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The top $m$ is the molality of the solution you have. You must know the concentration since you need it to compute the activity coefficient. The bottom $m_0$ is the standard molality which in general is 1 molal. It is there to make the activity dimensionless.

The paragraph above was my original answer and it contains a major omission. The quantity $m_\pm$ has to be calculated. The required formula is $$m_\pm^\nu = (m_+)^{\nu_+} (m_-)^{\nu_-}$$ where $m_-$ is the molality of the negative ions and $m_+$ is the molality of the positive ions. Also $\nu_-$ is the stoichiometric coefficient of the negative ions and $\nu_+$ is the same for the positive ions. Also $\nu_\pm = \nu_- + \nu_+$.

Thus for instance 0.1 molal $\ce{CaCl2}$ would have $\nu_- = 2$, $\nu_+ = 1$, and $\nu = 3$. Further $m_- = 0.2$ and $m_+ = 0.1$

Once again, sorry for the omission.

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