You can never know too much chemistry: Part 3: pH

I remember asking students when I first came to the UTC: "what does the 'p' stand for in pH?" I was surprised at the answers: proportion? percentage? part? Not unreasonable guesses if you haven't been taught the origins of the scale used to measure the precise level of acidity or alkalinity of a solution. This question came hot on the heels of my first (in retrospect) more challenging question: can pH be greater than 14 or less than 1? It was clear that pH needed explaining properly. But before I move on, the p is thought by some to refer to the "power" as in force of the hydrogen ion in solution, but is formally defined as the decimal logarithm (log base 10) of the reciprocal hydrogen ion activity. Before I confuse you completely the term activity is often used synonymously with concentration, and at this stage in your studies I shall not burden you with the difference, but I will write about activity and concentration separately in respect of the buffers we use in Biological experiments. If you want to read more about p then there is a further comment here. Mathematically, pH  is written as follows:

The consequence of this logarithmic relationship, is that pH has no units (or it is said to be dimensionless). (I will not dwell on the use of logs here since we have covered this in lab classes last year, but again, it is a topic for the future and for now, you can find a basic explanation here). The use of the term pH was pioneered by the Danish Physical Chemist Soren Sorensen (pictured top left).

There are a few simple consequences of the above relationship. First as the pH increases, the concentration of H+ ions (sorry sometimes superscripting seems too many mouse clicks away) decreases, or alternatively, as the pH decreases, the concentration of H+ ions increases and consequently acidity increases. Secondly, the range of pH that we tend to work with in the laboratory is between a strong acid (pH3) and a strong base (or alkali) at pH11. However, in vivo, apart from localised exceptions, we maintain our net intracellular and body fluid pH at around 7. The problem with pHs beyond 3 (or 11) are related to the instrument used to measure pH (and more importantly the Nernst equation, relating pH and electrode potential) utilises an ion selective glass electrode, and the properties of glass electrodes become non-Nernstian at extremes of pH. It is worth noting that absolute measurements of Tris buffers require a dedicated Tris electrode (see here for an explanation). However, in principle pH can exceed the boundaries of 1-14, it just isn't commonly observed in most laboratories.

The ionization of water (and in fact we refer to pH more precisely in terms of the hydronium ion) is given as follows:

K_eq = [H+][OH-]/[H₂O], which is 1.0 * 10⁻¹⁴ at 25°C.

Therefore, a few simple things follow:

The pH of pure water is 7. Why? Can you work this out for yourselves?

The basic elements of acids (proton donors) and bases (as proton acceptors) is covered very concisely at this Chemguide site. 

Look up a definition for the term Ka and consider the form of the Henderson- Hasselbalch equation which is given below (and explained in full at the Chemwiki site). 

I will use the splendid Excel resources provided by Professor Scott Sinex, to illustrate the value of the H-H equation and I would encourage you to take a look yourselves, in advance (and after). 

Let me give you an indication of the value of this relationship in respect of amino acids. Take an amino acid such as glycine or histidine. IBoth have an acidic group (COO-) and a basic group (NH3+): these define the amino acid nature of these two molecules. In addition, it whilst Gly has two protons, one of which is referred to as the R group or the side chain, His has what is referred to as an imidazole side chain. For each of the ionizable protons, we can re-arrange the H-H equation, and produce a value for pKa, which is given by:

pKa = pH + log [HA] / [A^-]

(note the inversion of the acid (HA) and its conjugate base (A-). Hence the values of the pKas of the amino acid side chains gives an indication of their potential role in acid base catalysis. See here for more details and a list of values.

Why is this relationship useful?

• It tells us that when the pH = pKa then log [HA] / [A^-] = 0 therefore [HA] = [A^-]   i.e. equal amounts of the two forms, the acid and the conjugate base.

• If we make the solution more acidic, i.e. lower the pH, so pH < pKa, then log [HA] / [A^-] has to be > 0 so [HA] > [A^-]. This makes sense as it tells us that a stronger acid will cause the formation of HA, the protonated form.

• If instead we make the solution more basic, ie raise the pH, so pH > pKa and log [HA] / [A^-] has to be < 0 so [HA] < [A^-]. This makes sense as it tells us that a stronger base will cause the formation of A^- , the deprotonated form.

These features of amino acid side chains (shown in the figure left), are thought to be a key element of the mechanism of catalysis in a number of enzymes. In the example shown, two His residues of Ribonuclease A, perform a pincer movement on the RNA substrate, an example of amino acid side chains participating in acid base catalysis. You can see a detailed series of images here. The ability of the "ring nitrogens" to act as acid or base provides one of the pathways for removing the activation energy in this and a number of other enzymes.

You can never know too much chemistry Part 2

In the second Blog aimed at supporting students taking  Unit 16, Chemistry for Biology Technicians. There are three areas that I would like to address: Rates of Reaction, the principles of acids and bases in the form of pH measurements and finally a word or two about chemical equilibria. Having spent most of my PhD measuring the initial rates of glutamate dehydrogenase enzymes, I moved on to work with enzymes that are much more difficult to assay. It is important to choose the easiest and most robust enzymes for developing and testing kinetic ideas. Hence the work of Biochemists in the last century tended to feature focus on NAD(H)-linked enzymes or proteases, for which reliable and sensitive spectroscopic methods were widely available. However, our understanding of catalysis must be considered in the light of the "natural" tendency of a molecule to decompose or oxidise (for example) in the absence of any enzymes. Or, for two or more molecular species to unite to form new comopunds. In other words, before we can appreciate the catalytic power of enzymes, we need tobe able to understand the reactivity of molecules in the context of time.  This is the field of reaction kinetics, which is of critical importance not only in Biology, but in all applications of chemistry from drugs to the synthesis of tonne quantities of plastics.

Figure 1

Take for example a solution of glucose (RHS) at neutral pH and room temperature (I will use pH as a defining parameter here, and will discuss the details below), and a solution of hydrogen peroxide. The image of glucose suggests that it forms a cyclic structure in solution. In fact glucose can adopt several conformations, but this is the most stable. However, it shows no tendency to decompose or pick up any other atmospheric atoms (eg oxygen). Which in principle poses a problem for living organisms where combustion of glucose in oxygen, provides us with the energy that we ultimately store as ATP. On the other hand, the concentration of hydrogen peroxide in a beaker left on the bench at 9am on  Monday will have diminished overnight, owing to its tendency to release oxygen and form water. Therefore if we measure the release of oxygen at a series of specific times, from a solution of H2O2, we will obtain the rate of decomposition under a given set of conditions. These experiments can be carried out very easily and have given rise to the following graphical relationship for the decomposition of H2O2 (the data can be plotted either as H2O2 left in solution, or alternatively O2 released: the former will decline and the latter will rise). 

The first thing to note is that the graph is not a straight line, but it does follow a a simple graphical trajectory that we call "exponential".  If we start with 100 mmoles/litreof H2O2, and we have 90 mmoles/litre after 10 minutes, and if the rate of decomposition was linear, then how many molecules would we have after 40 minutes (sketch a simple graph)? If we look at the graph in Figure 2,we start with 22.5 mmoles/litre and after 10 minutes, we have 14 mmoles/litre. What would this value be if there was a linear decomposition (try drawing a straight line between the first two data points in Fig. 2). 

This tells us that the relationship between the rate of decomposition of H2O2 under these conditions follows a "first order" reaction pathway. This is a description of the mathematical relationship that provides us with the best fit to the observed rate of decomposition measured in our experiment. 

I want to stress here, that nobody made the maths of reaction kinetics deliberately difficult (I am not a conspiracy theorist; well at least not yet): we simply make the measurements and find the best mathematical equation to fit the data. Physical chemist then try to attach physical phenomena to the algebraic terms consistent with the mathematical relationship (or model). This is a really important feature of Science in general. It is standard practice to use the simplest mathematical model to fit the experimental data. This model is then tested in order to establish its "robustness" for predictive purposes. You will all be familiar with one of the most complex mathematical models in common use: that describing the behaviour of British weather over time. The predictive powers of the model do not allow the Met Office to accurately forecast on more than a daily basis: long term forecasts are notoriously unreliable. However, as the Bard from Duluth Minnesota once commented so appositely: "You don't need a weatherman to know which way the wind blows".

Getting back to reaction kinetics. We have made one observation for H2O2, which sets it apart from glucose. Namely that the former molecule has a greater tendency to decompose compared with a solution of glucose (I should have added that the glucose must be kept sterile in these experiments. Why?). We now need to look at the rate equation that fits the data.  The standard equation takes the form of 

V = k[A]^x

When there are two or more reactants, the equation is expanded as follows:

V = k[A]^x[B]^y[C]^z

V is the measured rate, k is the rate constant (determined experimentally from the slope) and the indices (or exponents) refer to the order of the reaction. The terms [A] etc refer to the molar concentration of the reactant(s) at the start of the reaction and for a first order reaction, x = 1, etc. One consequence of this mathematical model, is that the rate of the reaction is dependent on the concentration of the reactants. For a first order reaction like the decomposition of H₂O₂, the rate of reaction is proportional to the concentration of the peroxide. This is readily seen by varying the starting concentration and measuring the rate of oxygen production. This approach forms the core of reaction kinetics. It is standard practice to measure product formation, or substrate depletion, with time in order to determine the value of rate constants. The units of reaction rates are expressed as Moles per second. Therefore the rate constant will have units of 1/second (or reciprocal time).

In the case of Glucose combining with a phosphate group (provided by ATP) in the first committed step of glycolysis, the reaction rate will be a function of both reactants:

V = k.[Glucose].[ATP] 

it is sometimes necessary to fix one concentration (say [A]) and vary the second reactant ([B]). By systematically measuring the rates of reaction it becomes possible to construct "rate equations" which allow a reaction rate to be "forecast" under a given set of temperature, pressure, pH and reactant concentrations. This experimental method and the values obtained, forms the basis of determining whether a particular reaction pathway is likely to be operational in a Biological setting and in fact in any chemical process.


What makes one  molecule decompose, oxidise or combine with another molecule? Reactivity is synonymous with molecular stability. It is related to the relative energies of the reactants and the products. In the case of H₂O_2, it rapidly dissociates into water and oxygen, largely as a result of the instability of the O-O peroxide bond, there are also differences when the molecule is dissolved in water, or neat (an old word meaning undiluted). In fact H₂O₂ can be used in rocket propulsion! However, when diluted down to a 1-5% solution, it can be used to produce that aesthetically appealing hair colour we refer to as bleached blond! 

I will just add here that when a reaction is independent of the concentration of the reactant(s) it is said to be zero order. I will not discuss this further. When two or more molecules react, there is a requirement for collision; more precisely productive collision. I think it is pretty logical to appreciate that in order for two chemical groups on a molecule to combine and form a new bond (it could be the peptide bond in protein synthesis, for example), there will be an optimal geometry involved. Think lock and key.

Coming back to the issue of why do some reactants react readily (or we often say spontaneously) and others require heat (or in Life, an enzyme or two)? This  BBC web site is worth a look if you need a refresher.  In order for a reactant,or group of reactants to be transformed into products, there is finite amount of energy required to initiate the reaction, which then reaches a peak, beyond which the reaction accelerates to the formation of products. The "height"of this peak was termed "activation energy" by the famous physical chemist Arrhenius. In chemistry we generally drive the reactants over the activation energy barrier with heat: in Biology we deploy enzymes (just think of pushing a piano up the stairs and when you get to the top pushing it out of the bedroom window!). Acting as catalysts, enzymes (as you will be aware from the Biochemistry Unit) have evolved to overcome challenging activation energy barriers by breaking them down into several smaller steps; a little like walking around a mountain range to get to the other side, rather than climbing over the top.

Why then is a reaction dependent on the concentration of the reactants? It's simple: if two molecules need to collide productively in order to react, then the more there are in a given volume, the greater the frequency of productive collision. Equally, if the reactants are lazy, then by putting a little heat into the reaction vessel, makes them more agitated and therefore they make more collisions than at the lower temperature. Another factor that plays a role in determining the rate of a reaction is the size of the particles in the reaction vessel. This is not an issue that arises in Biology, but if you compare the rate at which a sugar cube dissolves compared with a spoonful of sugar (assuming the same mass), you'll get the idea straight away. It is that the greater the exposed "surface" of the reactant, the faster the reaction proceeds. So in summary, reactions proceed as a function of the intrinsic energy in the reactants and the products and this in turn means that some reactions are spontaneous, some require heat (or enzymes) and some are so slow, that they are effectively inert. Let's finish this section by putting all of this into the context of Equilibrium Constants.

If we write a reaction in the following form 

A + B = P + Q

Where  A and B represent reactants and P and Q are the products, the reaction reaches an equilibrium when the reactants and products reach the end of the reaction. This governed by two rate constants: one for the forward reaction called k with the subscript f, + or sometimes just 1and the reverse reaction by k with the subscript r, - or 2. The equilibrium position is a reflection of the probability of the reactants converting to products, but in reality it is usually a dynamic process in which at the point of equilibrium the concentrations of A, B, P and Q are defined but the forward and reverse reactions proceed in a balanced way. We refer to this as a dynamic equilibrium at the molecular level.

The equation that defines the equilibrium constant, Keq,  is conventionally written as 

Keq = [Products]/[Reactants] or [P][Q]/[A][B]

and this is equal to the ratio of the corresponding rate constants:

kf/kr

You can read more about the  concepts at the Chemwiki site here. I now want to look at the application of these concepts to understanding acids and bases and redox reactions. I am going to add this to the post over the next few days, so keep an eye out for updates!

Electron-half-equations

When magnesium reduces hot copper(II) oxide to copper, the ionic equation for the reaction is:




The equation identifies the positive charge on the Cu and Mg, but the electrons, whose departure from the metals, give rise to the charged state, are not shown. This is the convention, but it can also be written as 



These two equations are "half equations" or "half reactions". Any redox reaction is made up of two half reactions: one defining the gain and the other the loss of electrons 

Important to Remember:

Oxidation is a loss of electrons

Reduction is a gain of electrons

You can find a detailed account of the steps underlying the formalism for writing half reactions at this Chemguide site. It isnt necessary for me to reinvent this particular wheel when it is so clear!

Here is an example of how our understanding of Redox reactions allows us to reconcile the relationship between electron transfer reactions and the generation of energy in oxidative phosphorylation (this is for advanced level students).

Consider the overall reaction of the oxidation of NADH paired with the reduction of O₂: the potential can be calculated as shown below.

Reduction Potentials	ereduction
NAD+ + 2H+ + 2e- --> NADH + H+	-0.32 V
(1/2) O2 + 2H+ + 2e- --> H2O	+0.82 V

NADH + H+ --> NAD+ + 2H+ + 2e-	eoxidation = 0.32 V
(1/2) O2 + 2H+ + 2e- --> H2O	ereduction = 0.82 V
net: NADH + (1/2)O2 + H+ -->             H2O + NAD+	erxn = 1.14 V

ΔG= -nFerxn

The overall reaction is

The electrical potential (e_rxn) is related to the free energy (ΔG) by the following equation:

where n is the number of electrons transferred (in moles, from the balanced equation), and F is the Faraday constant (96,485 Coulombs/mole). (Using this equation, ΔG is given in Joules; one Joule =1 Volt x 1 Coulomb.)

Hence the overall reaction for the oxidation of NADH paired with the reduction of O₂ has a negative change in free energy (ΔG =-220 kJ); i.e., it is spontaneous.

Thus, the higher the electrical potential of a reduction half reaction, the greater the tendency for the species to accept an electron.

The importance of pH, I have decided warrants a separate Part, so look out for it in a few days. In the meantime, I came across a superb resource for real time visualisation of the manipulation of thermodynamic and kinetic equations at Scott Sinex's resource page in the USA.