# Structural Biochemistry/pH

Example of titration to the end-point.

The pH of a solution is defined as the negative logarithm of its hydrogen ion (H+) concentration. For example, If the concentration of H+ ion is 10^(-7),

            then the pH of the solution = -log (10)^(-7) = 7.


Therefore, as the Hydrogen Ion concentration increases, pH value decreases

         & as the Hydrogen Ion concentration decreases, pH value increases


The pH of a solution is a measure of the hydronium ion (H3O+) concentration on a logarithmic scale. The pH scale range is 0-14 from acidic to basic, respectively. The pH of a neutral compound, such as pure water at room temperature, is 7. The concentration of hydronium ions is related to the concentration of hydroxide ions by the dissociation of water:

H2O ${\displaystyle \rightleftharpoons }$ H+ + OH-

There is some confusion as to the meaning of the "p" in pH. Although many believe it to stand for "power" or "potential," (i.e. power of hydrogen), that may not actually be the case. According to Jens G Norby's research (Dept of Biophysics, University of Aarhus, DK), "power of hydrogen" is actually a false association made by W.M. Clark. In actuality, Sorensen, the originator of the term, "pH" did not explicitly state a meaning for the "p," and it appears to be an arbitrary selection of letters "in his initial explanation of the electrometric method." For more information about this misconception, see, "The origin and the meaning of the little p in pH."

pH is biologically important because it affects the structure and activity of macromolecules. pH is important in homeostatic processes. For example, most animals breathe not because they lack oxygen, but because CO2 buildup in the blood increases the blood acidity beyond normal levels. Enzyme function is especially sensitive to pH: each has an optimum pH at which they have maximum catalytic ability. Extreme pH levels can denature enzymes, completely disrupting their function. Other proteins are also destabilized by extreme pH levels.

The pH and pOH of a solution are related such that: pH+pOH=14. For example, if the pH of a solution is 5, the pOH of the same solution will be 14-5=9.

The pH of distilled water is 7, this is neutral. Any solution with a pH below 7 (i.e. pH 0 to pH 6.9) is acidic and any solution with a pH above 7 (i.e. pH 7.1 to pH 14) is basic.

Acidic solutions have a pH between 0 and 6.9 (the stomach contains HCl, which has a pH ~2). Alkaline solutions have a pH between 7.1-14 (the small intestine is pH 9). Neutral solutions are neither acidic nor alkaline so their pH is 7.

Hydrogen Chloride dissociates into hydrogen and chloride ions:

HCl(aq)= H+ + Cl-

Water dissociates to produce Hydrogen and hydroxide ions:

H2O(l) = H+ + OH-

Sodium Hydroxide dissociates to produce Sodium and hydroxide ions:

NaOH(aq) = Na+ + OH-

In each case the concentration of Hydrogen ions can be measured or calculated.[H+] represents hydrogen ions and the square brackets represent concentration.

• pH = -log[H+]

• pH + pOH = 14

• DECREASING pH, INCREASING pKa

• pH = pKa + log ([A-]/[HA])

• pH = pKa

## Mathematics

The pH of a solution is calculated by

pH = -log [H+]

By convention H+ is used to represent hydronium ions (H3O+). A log system simplifies the notation of the H+ concentration in a medium. A pH increase of 1 is indicative of a ten-fold decrease of H+ concentration. This negative logarithmic relationship between the hydrogen ion concentration and pH means that a lower pH indicates a higher concentration. Conversely a higher pH indicates a lower concentration of hydrogen ions.

The ionization of water is expressed by equilibrium constant: Keq = [H+][OH-]/[H2O], which is 1.0 * 10−14 at 25°C.

The tendency of an acid, HA, to lose a proton and form its conjugate base, A-, is defined by the acid dissociation constant, Ka. Another quantitative measurement of acidity is the pKa, which is calculated from the Ka (pKa = -log Ka). A smaller Ka means a higher value of pKa, thus a higher value of pKa is equivalent to a weaker acid due to less dissociation of the acid into H+ and its conjugate base. Similar to pH, a single integer difference in pKa represents a tenfold difference.

The pH of a solution is determined by relative concentrations of acids and bases.

Using the reaction of the dissociation of a weak acid, HA:

HA ${\displaystyle \rightleftharpoons }$ H+ + A-

The Ka can be written as:

Ka = [H+][A-] / [HA]

and rearranged as:

[H+] = K*[HA] / [A-]

Taking the negative log of each term:

-log[H+] = -log(Ka) + log([A-]/[HA])

Letting pH = -log[H+] results in:

pH = -logKa + log([A-]/[HA])

Letting pKa = -logKa gives:

pH = pKa + log([A-]/[HA])

This relationship of pH to pKa is known as the Henderson-Hasselbach equation.

When the concentration of an acid, HA, and its conjugate base, A-, are equal:

log([A-]/[HA]) = log 1 = 0

Yielding:

pH=pKa

meaning that the pH of the solution is equivalent numerically to the pKa of the acid. This point is also known as the half equivalence point.

This equation is useful for calculating the pH of a buffer solution, or a solution containing a known quantity of weak acid (or base) and its conjugate base (or conjugate acid). From a biochemistry perspective, Hederson-Hasselbach can be applied to amino acids. However, this equation does not account for the ionization of water in a solution and thus is not useful for calculating the pH of solutions strong acids or bases.

See Buffer for more information on weak acid or weak base calculations.

## pH Calculation for Polyprotic Acid

Polyprotic acid such as sulfuric acid, H2SO4 or phosphoric acid, H3PO4, is capable of being deprotonated more than once because of the presence of two or more H atoms within the molecule. However, the pKa of each form of a polyprotic acid determines the contributions of hydrogen atoms, and consequently a change in pH, to a solution.

For example, consider the dissociation phosphoric acid:

    H3PO4   ${\displaystyle \rightleftharpoons }$    H2PO4- + H+           Ka1 = 7.1 x 10−3    (1)
Ka1  =  ${\displaystyle {\tfrac {[H_{2}PO_{4}^{-}][H^{+}]}{[H_{3}PO_{4}]}}}$
H2PO4-  ${\displaystyle \rightleftharpoons }$    HPO42- + H+           Ka2 = 6.3 x 10−8    (2)
Ka2  =  ${\displaystyle {\tfrac {[HPO_{4}^{2-}][H^{+}]}{[H_{2}PO_{4}^{-}]}}}$
HPO42-  ${\displaystyle \rightleftharpoons }$    PO43- + H+            Ka3 = 4.5 x 10−13   (3)
Ka3  =  ${\displaystyle {\tfrac {[PO_{4}^{3-}][H^{+}]}{[HPO_{4}^{2-}]}}}$


Supposing we have 0.01 M H3PO4 we could calculate the pH of the solution based on the Ka values. Comparing the three pKa values of H3PO4, it is clear that the first deprotonation (1) contributes the most H+ because it is the highest value. At equilibrium, x can represent the M [H+] and this value is the same for [H2PO4-] as well. While [H3PO4] can be expressed as 0.01-x M. Therefore,

Ka1 = ${\displaystyle {\tfrac {(x)(x)}{0.01-x}}}$ = 7.1 x 10-3

After which the quadratic equation can be used: x2 + 0.0071x - 0.000071 = 0

To solve this equation, we use ${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$, and then obtain
x = 5.59 x 10-3 = [H+] = [H2PO4-]
pH = -log[H+] = -log 5.59 x 10-3 = 2.25

A less time-consuming (but also less accurate) approach can be used to avoid using the quadratic equation. If an acid dissociates less than 5% or if the acid concentration to Ka ratio is greater than 103, the following equation can be used:

      [H+] = ${\displaystyle {\sqrt {K_{a}\times C_{a}}}}$ ; where Ca is the concentration of acid.      (4)


However, in this case we see that 0.01M H3PO4 dissociates more than 5% in the first deprotonation.

${\displaystyle {\frac {0.00559*100}{0.01}}=56\%}$

Therefore, using quadratic equation to solve this problem is the right choice.

Since phosphoric acid has three acidic protons, the second and third deprotonations may also contribute to pH of the solution.

When phosphoric acid was first deprotonated, the [HPO42-] concentration was approximately 0.00559 (calculated above). The [HPO42-] to Ka2 ratio is greater than 103. In this case equation (4) can be used to approximate [H+] from the second deprotonation. From calculation, [H+] = ${\displaystyle {\sqrt {6.3\times 10^{-8}\times 5.59\times 10^{-3}}}=1.87\times 10^{-5}}$

As shown from calculation, [H+] contribution from the second deprotonation is much less than that from the first, and makes an insignificant change in pH. Similarly, calculations from the third deprotonation reveal that even less [H+] (approximately 2.9 x 10−9 M) is contributed. As shown by the differing pKa values, the majority of [H+] comes from the first deprotonation.

Now consider the multiple deprotonations of sulfuric acid:

    H2SO4   ${\displaystyle \rightleftharpoons }$    HSO4- + H+           Ka1 = large         (5)
Ka1  =  ${\displaystyle {\tfrac {[HSO_{4}^{-}][H^{+}]}{[H_{2}SO_{4}]}}}$
HSO4-   ${\displaystyle \rightleftharpoons }$    SO42- + H+           Ka2 = 1.2 x 10−2    (6)
Ka2  =  ${\displaystyle {\tfrac {[SO_{4}^{2-}][H^{+}]}{[HSO_{4}^{-}]}}}$


Equation (5) shows a large value for the first pKa, implying that sulfuric acid dissociates completely in the first deprotonation. (See acidic substances) Thus, if we have 0.01 M H2SO4, [H+] from the first dissociation would also be 0.01 M. In addition, there would be 0.01 M HSO4- in the solution, and is a source of protons by the second deprotonation (equation 6). Before using equation (4), check to see if the concentration of acid to Ka ratio is greater than 103 or not. Since this simplification cannot be used, the quadratic equation is used to determine [H+] from the second dissociation. The equation is as follows:

x2 + 0.012x - 0.00012 = 0
Again, ${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$ is used to obtain
x = 6.5 x 10-3 = [H+]

The second deprotonation gives a [H+] more than half of which was obtained from the first deprotonation. In contrast to the example with phosphoric acid, the second dissociation of sulfuric acid is contributes greatly to the concentration of H+. In total, [H+]tot = 0.01 + 0.065 = 0.0165 M

The pH = -log 0.0165 = 1.78

Sulfuric acid is a much stronger acid than phosphoric acid as seen by its H+ contribution in each deprotonation step.

## An Alternative Method to the Quadratic Equation

The quadratic equation can become messy and time consuming in certain situations. Lets consider the following hypothetical situation:

HA ${\displaystyle \rightleftharpoons }$ A- + H+

KHA = 6.3 x 10−3

[HA] = 2

[A-] = 0

[H+] = 1x10−1

Our equilibrium equation would therefore be:

${\displaystyle {\tfrac {[A^{-}][H^{+}]}{[HA]}}}$ = 6.3x10−3

Now we take into account the overall shift in equilibrium from the dissociation of HA and include the given concentrations.

${\displaystyle {\tfrac {[x][0.1+x]}{[2-x]}}}$ = 6.3x10−3

Because there is already a large concentration of protons in the solution the overall dissociation of HA, given by the variable x, will be very small. Therefore we are going to ignore the change in x in the denominator to yield an equation that is not quadratic.

${\displaystyle {\tfrac {[x][0.1]}{[2+x]}}}$ = 6.3x10−3

Solving for x yields approximately 0.11853 which will call x1 which we will use to successively obtain a more accurate number by plugging this number back into our original equation where we ignored x previously:

${\displaystyle {\tfrac {[x][0.1+x_{n}]}{[2-x]}}}$ = 6.3x10−3

Solving using x1 yield x2 to be 0.05604

In order to simplify the process the following equation is used to solve for xn+1 derived from our original rate expression.

xn+1 = ${\displaystyle {\tfrac {0.0126}{x_{n}+0.1+6.3E-3}}}$

The following successive approximations are calculated to be:

x3 = 0.07761

x4 = 0.06851

x5 = 0.07207

x6 = 0.07063

x7 = 0.07098

x8 = 0.07107

x9 = 0.07103

x10 = 0.07105

When xn starts repeating itself to a given amount of significant figures, it is safe to assume that the given answer is an accurate approximation and can be used for the final value of x. The overall number of successive approximations that must be calculated varies according to the situation. In some situations it is simpler to use the quadratic equation. An equation that required many successive approximations was used for the purposes of this example. Ingenuity is require to obtain an accurate in a time efficient manner.

## Contribution of Water in Dilute Acidic or Basic Solution

One more thing we need to consider in case of dilute acid or base is the contribution of H+ or OH- from water. As we have already known, at 25 oC water molecule could dissociate and H+ as well as OH- would form with the amount of 10−7 M each. For example, if we have 0.0000001 M acetic acid, we can determine the pH of the solution as follows:

Acetic acid is a monoprotic acid which has pKa 1.8 x 10−5. To determine the H+ of the solution, let's consider

    CH3COOH   ${\displaystyle \rightleftharpoons }$    CH3COO- + H+           Ka = 1.8 x 10−5
Ka  =  ${\displaystyle {\tfrac {[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}}}$


Regarding the previous section, we need quadratic equation to solve this problem, Thus

x2 + 0.000018x - 0.0000000000018 = 0

Then, we get x = 1 x 10−7 = [H+] the concentration of [H+] from 0.0000001 M acetic acid is nearly close to the ones obtained from water. In this case, we have to take into account the [H+] from water. Therefore,

[H+]tot = 2 x 10-7 M

Finally, the pH = -log 2 x 10−7 = 6.7

Remark For dilute basic solution, we can evaluate in the similar way.

## pH Scale Diagram

The scale is broken up into two regions: acidic (pH 0-7) and basic (pH 7-14) with a pH of 7 being said to be "neutral." Some acidic substances are gastric acid (pH ~ 2), vinegar (pH ~ 3) and coffee (pH ~ 5). Some basic substances are hand soap (pH ~ 10), ammonia (pH ~ 13), and lye (pH ~ 14). An example of a "neutral" substance would be pure water (pH ~ 7). pH values below 0 require measurements using specialized equipment. The reason is that typical pH measurements are done in an aqueous solution, which means that the measured pH will only be as strong as the weakest acid in solution (in this case H3O+ which has a pKa of 0). Therefore, measurements are typically made by calibrating with a strong acid like sulfuric acid. Similar arguments also apply for extremely strong bases.

## Acidic Substances

There are three definitions of acids based on Arrhenius, Bronsted-Lowry, and Lewis. By Arrhenius theory, acids are molecules that produce hydronium ions (H3O+) in aqueous solution. The Bronsted-Lowry definition of an acid is a molecule that can donate protons (H+). A Lewis acid is defined as a molecule that is able to accept electron pairs.

The most common strong acids include hydrochloric acid (HCl), hydrobromic acid (HBr), hydroiodic acid (HI), sulfuric acid (H2SO4), perchloric acid (HClO4), and nitric acid (HNO3). Where perchloric acid is the strongest with a pKa = -10 In calculations, strong acids are assumed to dissociate completely. For example: By convention 2.5 M HCl produces 2.5 mol of hydronium ions per liter.

## Basic Substances

Titration of a weak acid and strong acid with a strong base

Just as there are three definitions of acids, there are three corresponding definitions of bases. An Arrhenius base is defined as a molecule that produces hydroxide ions (-OH) in water. A Bronsted-Lowry base is a molecule able to accept hydrogen ions. Finally, a Lewis base is able to donate electron pairs.

Bases are usually characterized by having a negative charge (surplus of electron pairs), which according the to Lewis theory makes it capable of donating to another molecule. Bases can also have a neutral charge, but contain a lone pair of electrons (such as ammonia) which can be donated to other molecules.

Common strong bases include: Lithium hydroxide (LiOH), Sodium hydroxide (NaOH), Rubidium hydroxide (RbOH), Cesium hydroxide (CsOH), Potassium hydroxide (KOH), Magnesium hydroxide (Mg(OH)2) Calcium hydroxide (Ca(OH)2) Strontium hydroxide (Sr(OH)2) Barium hydroxide (Ba(OH)2). A strong base, like a strong acid, is defined as a compound that dissociates completely in aqueous solution.

## Application in Biochemistry

pH range for amino acids have Zwitterionic structure is 2-9. Below pH 2 the carboxyl group is not dissociated. Above pH 9, but amino group is not dissociated.

### Zwitterion

This is a isoleucine's zwitterion ion form

Since amino acids have both an acidic carboxylic acid group and a basic amine group, they can be either positively charged, negatively charged, or both at the same time depending on the pH. A zwitterion is a molecule that has both a positively charged group and a negatively charged group at the same time, thus carrying a total net charge of 0. The zwitterion is the most commonly found form of amino acids when the pH is in the neutral range, since in the neutral pH, the basic amine group is protonated (carries a positive charge) and the carboxyl group is deprotonated (carries a negative charge).

### Isoelectric Focusing

Since animo acids, and therefore proteins, have multiple sites that can be either positive or negative, the total overall charge of a protein depends on the pH. Isoelectic Focusing is a technique where proteins are separated based on the pH at which they become neutrally charged overall. This point is called the pI, or isoelectric point. Isoelectric Focusing is done by loading the sample proteins onto a gel with a pH gradient and then applying an electic current. The proteins will then move across the gel because of the electric current and will stop moving once they reach their pI. Isoelectric focusing can be combined with SDS-PAGE to increase the separation of proteins in a sample, a technique called Two-Dimensional Electrophoresis.

### The Bohr Effect

Christian Bohr, a Danish physiologist, defined the Bohr effect as the change in binding affinity of oxygen to hemoglobin based on hydrogen ion and carbon dioxide concentrations. Oxygen has higher affinity when the pH of the blood is high. CO2 and H+ exist in the following equilibrium in the blood:

CO2 + H2O ${\displaystyle \rightleftharpoons }$ H+ + HCO3

Thus, a change in either concentration will alter the pH. An increase in carbon dioxide will increase the concentration of hydrogen ions, and therefore lower both the pH and oxygen's affinity for hemoglobin. This benefits the body because areas that are high in carbon dioxide are likely deficient in oxygen. The low pH in those areas causes oxygen to diassociate from hemoglobin, releasing it into the body tissues that need it most.

As the pH increases, so does the ability for hemoglobin to bind oxygen. However, as the pH moves to a lower pH (as slight as a drop from 7.4 to 7.2), the ability to bind to oxygen decreases.

Carbon Dioxide

Carbon dioxide also affects the pH inside a red blood cell (which affects the ability of oxygen to bind to hemoglobin). As carbon dioxide flows into a red blood cell, carbonic anhydrase (an enzyme specific for this reaction) facilitates its interaction with water in forming carbonic acid, which then dissociates into bicarbonate and hydrogen ions.

CO2 + H2O ${\displaystyle \rightleftharpoons }$ H2CO3

H2CO3 ${\displaystyle \rightleftharpoons }$ H+ + HCO3-

The increase in H+ decreases the pH, and stabilizes the T-state of hemoglobin, also known as the deoxygenated state.

The transport of carbon dioxide into a red blood cell is carried through via membrane transport. It also directly interacts with hemoglobin, affected the binding and release of oxygen. The carbon dioxide decreases the oxygen binding affinity and stabilizes the T-state of hemoglobin even better than the effects of pH on oxygen binding.

Carbamate Group

Carbamate groups are negatively charged groups formed from the interaction of carbon dioxide and the last amine group in an amino acid. The carbamate groups stabilize the decrease in the binding of oxygen by increasing the salt bridge.