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I am not from chemistry background. Could you tell me in simple words what the activity of hydrogen $a_{\ce{H+}}$ is? It arises in the context that the negative logarithm of the activity of hydrogen gives the pH: $\text{pH} = -\log_{10} a_{\ce{H+}}$.

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Building on @Philipp and @chipbuster's answers-

What is Activity?

When we have dilute solutions or small quantities of a gas or other substance, the substance behaves as if the particles are just points, rather than having a size. In these settings, we often use a simplified set of equations.

For example, with the hydrogen ion, we use $\mathrm{pH} = -\log_{10}[\ce{H+}]$ where $[\ce{H+}]$ is the concentration, i.e. the number of particles (in moles) divided by the volume (in litres).

Once you get a lot of these particles, they bang into each other and take up room (think of billiard balls on a pool table or a room full of tennis balls flying around).

In these circumstances, we talk about ‘activity’, which is a way to correct for this. Basically, it is what the concentration or amount appears to be if we were using the simplified laws (the effective concentration). For the hydrogen ion, it is also what the concentration appears to be to physiological systems. In other words, it uses a correction factor for the simplified laws.

Activity, Concentration and Activity Coefficient

The thermodynamic definition of activity is:

  • $a_{i} = \operatorname{e}^{\frac{\mu_i - \mu_i^⦵ }{RT}}$

Which in low partial pressures and dilute solutions is simplified to:

  • $a = \gamma\cdot\frac{c}{c^⦵}$ for a solution
  • $a = \gamma\cdot\frac{p}{p^⦵}$ for a gas

where:

  • $a$ is the activity
  • $\gamma$ is the activity coefficient (a constant number based on what the substance is, what the solvent is if there is one, the temperature and other factors)
  • $c$ is the concentration (for something dissolved); $p$ is the partial pressure (for a gas)
  • $c^⦵$ is the standard state concentration; $p^⦵$ is the standard state pressure

pH and Activity

The original version of pH as defined by Sørensen is:

  • $\mathrm{pH} = -\log_{10}[\ce{H+}]$ (hereafter referred to as $\mathrm{p}[\ce{H+}]$)

However, the modern definition that takes into account activity is:

  • $\mathrm{pH} = -\log_{10}a_{\ce{H+}}$

$\gamma$ is found experimentally. In the case of the hydrogen ion, it is found using electrochemical methods.

Activity of Hydrogen Ion in Dilute Solutions

Standard thermodynamic state for $\ce{H+}$ is considered to be 1 mol/L ($\mathrm{p}[\ce{H+}]$ 0); thus:

  • $a = \gamma\cdot\frac{c}{c^⦵}$ (as a dilute solution)
  • $a_{\ce{H+}} = \gamma\cdot\frac{c}{c^⦵} = \gamma\cdot\frac{[\ce{H+}]}{1\ \mathrm{mol/L}} = \gamma\cdot [\ce{H+}]$

Also, in dilute solutions, activity can be approximated/calculated using an extension of the Debye-Hückel equation, the Güntelberg equation, which applies to concentrations of < 0.1 mol/L:

  • $\log_{10} \gamma_i=\frac{-Az_{i}^{2}\sqrt{I}}{1+\sqrt{I}}$

where:

  • $A$ and $B$ are constants that depend on solvent and temperature (0.51 and $0.33\times10^8$ respectively for water at 25 °C)
  • $z$ is the charge of the ion (1 for $H^+$)
  • $I$ is concentration of the ion

For $\ce{H+}$ in a solution of water at 25 °C this would be:

  • $\log_{10}\gamma=\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

Substituting, we get:

  • $a_{\ce{H+}}=\gamma[\ce{H+}]$
  • $\log_{10}a_{\ce{H+}}=\log_{10}[\ce{H+}]+\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$
  • $-\log_{10}a_{\ce{H+}}=-\log_{10}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

and so for dilute solutions only:

  • $\mathrm{pH} = \mathrm{p}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$
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  • $\begingroup$ Seem to be some problems with this answer. First problem: the jump from I (ionic strength) to [H+] is not explained; seems the final equation would only be true for a solution where there is H+ and a univalent counter ion, and no other ions (like HCl solution). Second problem, says "A and B are constant", but there is no B in the equation. $\endgroup$ – DavePhD Apr 29 '15 at 12:59
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The activity is some sort of dimensionless measure for the concentration. In thermodynamics, a lot of important equations are derived for the so-called ideal gas, which is a model system where the particles are treated as if they had no own size (point particles) and don't interact with one another (apart from elastic collisions).

This model is a very crude simplification of reality but nevertheless gives very good results, when the temperature is not too high and the pressure is low. So, the equations one can derive from this model are rather simple and yet quite usable for real systems. But of course, the applicability of these equation has limits, since in real systems the particles interact a lot with each another (via electrostatic and quantum mechanical forces) and have a certain size. But, in order to retain the simple form of the equations derived for ideal gases, some physical quantities like the pressure or the concentration are scaled by empirical coefficients to reflect the deviation from the ideal behavior and get the right results for real systems. An example is the activity $a$, which is equal to the dimensionless concentration, the so-called mole fraction $x$, scaled by the activity coefficient $f$ \begin{equation} a = f x \ . \end{equation} So, for example, instead of the concentration (or more correctly the mole fraction) one uses the activity in some equations derived for ideal gases to describe real systems accurately.

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I'm not quite sure what your source means by "activity." Generally, the formula we use for pH is: $\mathrm{pH} = -\log[\ce{H}^{+}]$ where $[\ce{H}^{+}]$ is the concentration of the hydrogen ion, in moles per liter.

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    $\begingroup$ Your equation only holds for dilute solutions which show nearly ideal behavior. The formulation with acitivities is the correct one. But nevertheless, for convenience the difference is often overlooked. $\endgroup$ – Philipp Aug 7 '13 at 19:13
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    $\begingroup$ Darnit, and I learned about activities not more than a few months ago. I don't know why that slipped my mind. Thanks much! $\endgroup$ – chipbuster Aug 7 '13 at 21:50

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