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Hootenanny

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**An Introduction to pH and Buffer Solutions Calculations [work in progress]**

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`This is a work in progress`

__1. pH and Strong Acids__pH is defined as the negative logarithm of the hydrogen ion concentration in a solution; pH = - log

_{10}[H

^{+}(aq)]. (From this point forward 'log' will represent log to the base ten). The square brackets above indicate that it is the concentration we are concerned with. Now, since a strong acid effectively dissociates completely into its respective ions when reacted with water; the amount of moles of protons in solution is equal to the number of moles of H

^{+}in the acid. This is not actually the case since some water auto-ionises; however, this effect is insignificant when dealing with acids with a concentration greater than around 5x10

^{-7}mol.dm

^{-3}. The auto-ionisation of water will be discussed later. One must be careful here to note whether a monoprotic or diprotic acid is used. A monoprotic acid has a single ionizable hydrogen atom per acid molecule; a diprotic has two, polyprotic many.

__Worked Examples__

*Calculate the pH of a solution of Hydrochloric acid with a concentration of 0.01 mol.dm*^{-3}First we must write the equation for the dissociation of the acid in water;

[tex]HCl(aq) + H_{2}O(l) \rightarrow H_{3}O^{+}(aq) + Cl^{-}(aq)[/tex]

We can further simplify this to

[tex]HCl(aq) \rightarrow H^{+}(aq) + Cl^{-}(aq)[/tex]

The water and hence the oxonium ion can be ignored since the water is present in excess. From the latter equation we can see that Hydrochloric acid is a monoprotic strong acid. Therefore, the concentration of protons

^{*}in solution is equal to the concentration of the acid, hence the pH is;

[tex]pH = - \log\left[ H^{+}(aq)\right] = - \log\left[ 0.01\right] = 2[/tex]

**Calculate the pH of a solution of Sulphuric Acid with a concentration of 0.01 mol.dm**^{-3}Again we should start by writing an equation for the dissociation;

[tex]H_{2}SO_{4}(aq) \rightarrow 2H^{+}(aq) + SO_{4}^{2-}(aq)[/tex]

Note that sulphuric acid is a diprotic acid, therefore to obtain the concentration of protons on solution we must double the concentration of the acid;

[tex]pH = - \log\left[ H^{+}(aq)\right] = -\log\left[ 2\times 0.01\right] \approx 1.70[/tex]

So although the sulphuric acid and the hydrochloric acid are the same concentration, sulphuric acid has a lower pH since it is diprotic.

We can also use the pH to calculate the concentration of a particular acid.

*A solution of Nitric Acid has a pH of 1.2, determine the concentration of the acid used*The first step as with all these calculations is to write the dissociation equation for the acid;

[tex]HNO_{3}(aq) \rightarrow H^{+}(aq) + NO_{3}^{-}[/tex]

Rearranging the standard equation for pH we obtain;

[tex]pH = -\log\left[ H^{+}(aq)\right] \Leftrightarrow \left[ H^{+}(aq)\right] = 10^{-pH}[/tex]

[tex]\left[ H^{+}(aq)\right] = 10^{-1.2} \approx 0.0631 mol.dm^{-3}[/tex]

Since nitric acid is monoprotic the concentration of hydrogen ions in solution is equal to the concentration of the acid; therfore, the nitric acid has a concentration of

__0.0631 mol.dm__

^{-3}(3sf)

_________________________________________

__2.Weak Acids__Weak acids do not completely dissociate when in solution, instead an equilibrium is set up;

[tex]HA(aq)\rightleftharpoons H^{+}(aq) + A^{-}(aq)[/tex]

We can write an equilibrium expression for this reaction. In this case the constant is known as the acidity constant or acid dissociation constant and is given the symbol K

_{a};

[tex]K_{a} = \frac{\left[ H^{+}(aq)\right] \left[ A^{-}(aq) \right]}{\left[ HA(aq) \right]}[/tex]

Here, we can now make two assumptions which will grately simplify the calculations;

**Assumption One: [H**

^{+}(aq)] = [A^{-}(aq)]This means that we are ignoring the effect of the ionisation of water, this is a valid assumption for most 'normal' concentrations.

**Assumption Two:**As weak acids dissociate very little, we can assume that the concentration of the acid at equilibrium is the same as if the acid did not dissociate

[tex]K_{a} = \frac{\left[ H^{+}(aq)\right]^2}{\left[ HA(aq) \right]}[/tex]

The K

_{a}values are spread of a large range, therefore it is often more convenient to define the term pK

_{a}= - log

_{10}K

_{a}

__Worked Example__

*Knowing that K*_{a}for ethanoic acid is 1.7x10^{-5}mol.dm^{-3}at 298K, calculate the pH of a solution of enthanoic acid with a concentration of 1 mol.dm^{-3}at 298K.The temperature is important here as it effects the position of the equilibrium by Le chatelier's principle, therefore K

_{a}will be affected.

Rearranging our equation we have:

[tex]K_{a} = \frac{\left[ H^{+}(aq)\right]^2}{\left[ HA(aq) \right]} \Leftrightarrow \left[ H^{+}(aq)\right] = \sqrt{K_{a} \cdot \left[ HA(aq) \right]}[/tex]

Thus:

[tex]pH = -\log \left( \sqrt{K_{a} \cdot \left[ HA(aq) \right]} \right) =

-\log \left( \sqrt{(1.7\times10^{-5}) \cdot [ 1 ]} \right)[/tex]

[tex]pH \approx 2.38 (\text{at 289K})[/tex]

_________________________________________

__3. Auto-Ionisation of Water__Although water is not considered ionic, it does dissociate to some extent. Water is amphoretic (able to act both as a base and an acid) and will auto-dissociate into the oxonium ion H

_{3}O

^{+}and the hydroxide ion;

[tex]2H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + OH^{-}(aq)[/tex]

We can write now write an expression for the equilibrium constant, which in this case is known as the ionic product of water (K

_{w});

[tex]K_{w} = \frac{\left[ H_{3}O^{-}(aq) \right] \left[OH^{-}(aq) \right]}{[H_{2}O(l)]}[/tex]

In pH calculations we can treat the oxonium ion as a proton, and since the water is present in excess the [H

_{2}O

_{(l)}] term is effectively constant and therefore can be omitted; thus the K

_{w}expession becomes;

[tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right][/tex]

*This expression will be important when dealing with bases*

Water has a K

_{w}value of 1 x 10

^{-14}mol

^{2}.dm

^{-6 at 298K (note the units here, since we are multiplying two concentrations together we must square the units), hence the pH of water at 298K is; [tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right][/tex] In pure water [H+(aq)] = [OH-(aq)], therefore; [tex] \left[ H^{+}(aq) \right] = \sqrt{K_{w}}[/tex] [tex]pH = - \log \left| \sqrt{k_{w}} \right| = -\log \left| \sqrt{1\times 10^{-14}} \right|[/tex] [tex]pH = 7[/tex] An interesting point to note, is that the auto-dissociation of water is an endothermic process, therefore, by Le Chatelier's principle if we increase the temperature the position of the equilibrium will shift to oppose that change, the value of Kw will decrease and therefore the pH of water falls as we increase the temperature. _________________________________________* Note that a H+ is a proton. Both terms will be used interchangeably here Code: Sections to Follow; [LIST] [*]pH of Strong Bases [*]Weak Bases [*]Buffer Solutions, applications, mechanism and Calculations [*]Advance treatments taking into account the autoionisation of water [/LIST] Please feel free to PM with any suggestions, comments or corrections. }

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