Equilibrium

 of hydrogen ion. In dilute solutions (< . M), activity of hydrogen ion (H + ) is equal in magnitude to molarity represented by [H + ]. It should be noted that activity has no units and is defined as: = [H + ] / mol L – From the definition of pH, the following can be written, pH = – log a H+ = – log {[H + ] / mol L – } Thus, an acidic solution of HCl ( – M) will have a pH = .

Similarly, a basic solution of NaOH having [OH – ] = – M and [H O + ] = – M will have a pH = . At °C, pure water has a concentration of hydrogen ions, [H + ] = – M. Hence, the pH of pure water is given as: pH = –log( – ) = Acidic solutions possess a concentration of hydrogen ions, [H + ] > – M, while basic solutions possess a concentration of hydrogen ions, [H + ] < – M. thus, we can summarise that Acidic solution has pH < Basic solution has pH > Neutral solution has pH = Now again, consider the equation ( .

) at K K w = [H O + ] [OH – ] = – Taking negative logarithm on both sides of equation, we obtain –log K w = – log {[H O + ] [OH – ]} = – log [H O + ] – log [OH – ] = – log – p K w = pH + pOH = ( . ) Note that although K w may change with temperature the variations in pH with temperature are so small that we often ignore it. p K w is a very important quantity for aqueous solutions and controls the relative concentrations of hydrogen and hydroxyl ions as their product is a constant. It should be noted that as the pH scale is logarithmic, a change in pH by just one unit also means change in [H + ] by a factor of .

Similarly, when the hydrogen ion concentration, [H + ] changes by a factor of , the value of pH changes by units. Now you can realise why the change in pH with temperature is often ignored. Measurement of pH of a solution is very essential as its value should be known when dealing with biological and cosmetic applications. The pH of a solution can be found roughly with the help of pH paper that has different colour in solutions of different pH.

Now-a-days pH paper is available with four strips on it. The different strips have different colours (Fig. . ) at the same pH.

The pH in the range of - can be determined with an accuracy of ~ . using pH paper. Fig. .

pH -paper with four strips that may have different colours at the same pH For greater accuracy pH meters are used. pH meter is a device that measures the pH-dependent electrical potential of the test solution within . precision. pH meters of the size of a writing pen are now available in the market.

The pH of some very common substances are given in Table . (page ). Problem . The concentration of hydrogen ion in a sample of soft drink is .

× – M. what is its pH ? pH = – log[ . × – ] = – {log[ .

Therefore, the pH of the soft drink is . and it can be inferred that it is acidic. Problem . Calculate pH of a .

× – M solution of HCl. Table . The Ionization Constants of Some Selected Weak Acids (at 298K) Acid Ionization Constant, K a Hydrofluoric Acid (HF) . × – Nitrous Acid (HNO ) .

× – Formic Acid (HCOOH) . × – Niacin (C H NCOOH) . × – Acetic Acid (CH COOH) . × – Benzoic Acid (C H COOH) .

× – Hypochlorous Acid (HCIO) . × – Hydrocyanic Acid (HCN) . × – Phenol (C H OH) . × – 2H O (l) H O + (aq) + OH – (aq) K w = [OH – ][H O + ] = – Let, x = [OH – ] = [H O + ] from H O.

The H O + concentration is generated (i) from the ionization of HCl dissolved i.e., HCl(aq) + H O(l) H O + (aq) + Cl – (aq), and (ii) from ionization of H O. In these very dilute solutions, both sources of H O + must be considered: [H O + ] = – + x K w = ( – + x)(x) = – or x + – x – – = [OH – ] = x = . × – So, pOH = . and pH = .

. . Ionization Constants of Weak Acids Consider a weak acid HX that is partially ionized in the aqueous solution. The equilibrium can be expressed by: HX(aq) + H O(l) H O + (aq) + X – (aq) Initial concentration (M) c Let α be the extent of ionization Change (M) -c α +c α +c α Equilibrium concentration (M) c-c α c α c α Here, c = initial concentration of the undissociated acid, HX at time, t = .

α = extent up to which HX is ionized into ions. Using these notations, we can derive the equilibrium constant for the above discussed acid-dissociation equilibrium: K a = c α / c( - α ) = c α / - α K a is called the dissociation or ionization constant of acid HX. It can be represented alternatively in terms of molar concentration as follows, K a = [H + ][X – ] / [HX] ( . ) At a given temperature T , K a is a measure of the strength of the acid HX i.e., larger the value of K a , the stronger is the acid.

K a is a dimensionless quantity with the understanding that the standard state concentration of all species is 1M. The values of the ionization constants of some selected weak acids are given in Table . . Table .

The pH of Some Common Substances Name of the Fluid pH Name of the Fluid pH Saturated solution of NaOH ~ Black Coffee . . M NaOH solution Tomato juice ~ . Lime water .

Soft drinks and vinegar ~ . Milk of magnesia Lemon juice ~ . Egg white, sea water . Gastric juice ~ .

Human blood . 1M HCl solution ~ Milk . Concentrated HCl ~– . Human Saliva .

The pH scale for the hydrogen ion concentration has been so useful that besides p K w , it has been extended to other species and quantities. Thus, we have: p K a = –log ( K a ) ( . ) Knowing the ionization constant, K a of an acid and its initial concentration, c, it is possible to calculate the equilibrium concentration of all species and also the degree of ionization of the acid and the pH of the solution. A general step-wise approach can be adopted to evaluate the pH of the weak electrolyte as follows: Step .

The species present before dissociation are identified as Brönsted-Lowry acids/bases. Step . Balanced equations for all possible reactions i.e., with a species acting both as acid as well as base are written. Step .

The reaction with the higher K a is identified as the primary reaction whilst the other is a subsidiary reaction. Step . Enlist in a tabular form the following values for each of the species in the primary reaction (a) Initial concentration, c. (b) Change in concentration on proceeding to equilibrium in terms of α , degree of ionization.

(c) Equilibrium concentration. Step . Substitute equilibrium concentrations into equilibrium constant equation for principal reaction and solve for α . Step .

Calculate the concentration of species in principal reaction. Step . Calculate pH = – log[H O + ] The above mentioned methodology has been elucidated in the following examples. The following proton transfer reactions are possible: ) HF + H O H O + + F – K a = .

× – ) H O + H O H O + + OH – K w = . × – As K a >> K w , [ ] is the principle reaction. HF + H O H O + + F – Initial concentration (M) . Change (M) – .

α + . α + . α Equilibrium concentration (M) . – .

α . α . α Substituting equilibrium concentrations in the equilibrium reaction for principal reaction gives: K a = ( . α ) / ( .

– . α ) = . α / ( – α ) = . × – We obtain the following quadratic equation: α + .

× – α – . × – = The quadratic equation in α can be solved and the two values of the roots are: α = + . and – . The negative root is not acceptable and hence, α = .

This means that the degree of ionization, α = . , then equilibrium concentrations of other species viz., HF, F – and H O + are given by: [H O + ] = [F – ] = c α = . × . = .

× – ) = . Problem . The pH of .1M monobasic acid is . .

Calculate the concentration of species H + , A – Problem . The ionization constant of HF is . × – . Calculate the degree of dissociation of HF in its .

M solution. Calculate the concentration of all species present (H O + , F – and HF) in the solution and its pH. Table . The Values of the Ionization Constant of Some Weak Bases at K Base K b Dimethylamine, (CH ) NH .

× – Triethylamine, (C H ) N . × – Ammonia, NH or NH OH . × – Quinine, (A plant product) . × – Pyridine, C H N .

× – Aniline, C H NH . × – Urea, CO (NH ) . × – and HA at equilibrium. Also, determine the value of K a and p K a of the monobasic acid.

pH = – log [H + ] Therefore, [H + ] = –pH = – . = . × – [H + ] = [A – ] = . × – Thus, K a = [H + ][A - ] / [HA] [HA] eqlbm = .

= . × – p K a = – log( – ) = Alternatively, “Percent dissociation” is another useful method for measure of strength of a weak acid and is given as: Percent dissociation = [HA] dissociated /[HA] initial × % ( . ) Problem . Calculate the pH of .08M solution of hypochlorous acid, HOCl.

The ionization constant of the acid is . × – . Determine the percent dissociation of HOCl. HOCl(aq) + H O (l) H O + (aq) + ClO – (aq) Initial concentration (M) .

Change to reach equilibrium concentration (M) – x + x +x equilibrium concentartion (M) . – x x x K a = {[H O + ][ClO – ] / [HOCl]} = x / ( . –x) As x << . , therefore .

× – , thus, x = . × – [H + ] = . × – M. Therefore, Percent dissociation = {[HOCl] dissociated / [HOCl] initial }× = .

Ionization of Weak Bases The ionization of base MOH can be represented by equation: MOH(aq) M + (aq) + OH – (aq) In a weak base there is partial ionization of MOH into M + and OH – , the case is similar to that of acid-dissociation equilibrium. The equilibrium constant for base ionization is called base ionization constant and is represented by K b . It can be expressed in terms of concentration in molarity of various species in equilibrium by the following equation: K b = [M + ][OH – ] / [MOH] ( . ) Alternatively, if c = initial concentration of base and α = degree of ionization of base i.e.

the extent to which the base ionizes. When equilibrium is reached, the equilibrium constant can be written as: K b = (c α ) / c ( - α ) = c α / ( - α ) The values of the ionization constants of some selected weak bases, K b are given in Table . . Many organic compounds like amines are weak bases.

Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by another group. For example, methylamine, codeine, quinine and nicotine all behave as very weak bases due to their very small K b . Ammonia produces OH – in aqueous solution: NH (aq) + H O(l) NH + (aq) + OH – (aq) The pH scale for the hydrogen ion concentration has been extended to get: p K b = –log ( K b ) ( . ) Problem .

The pH of .004M hydrazine solution is . . Calculate its ionization constant K b and p K b . NH NH + H O NH NH + + OH – From the pH we can calculate the hydrogen ion concentration.

Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have: [H + ] = antilog (–pH) = antilog (– . ) = . × – [OH – ] = K w / [H + ] = × – / .

× – = . × – The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to .004M. Thus, K b = [NH NH + ][OH – ] / [NH NH ] = ( .

. Problem . Calculate the pH of the solution in which .2M NH Cl and .1M NH are present. The pK b of ammonia solution is .

. NH + H O NH + + OH – The ionization constant of NH , K b = antilog (–p K b ) i.e. K b = – . = .

× – M NH + H O NH + + OH – Initial concentration (M) . . Change to reach equilibrium (M) –x +x +x At equilibrium (M) . – x .

+ x x K b = [NH + ][OH – ] / [NH ] = ( . + x)(x) / ( . – x) = . × – As K b is small, we can neglect x in comparison to .1M and .2M.

Thus, [OH – ] = x = . × – Therefore, [H + ] = . × – pH = – log[H + ] = . .

. . Relation between K a and K b As seen earlier in this chapter, K a and K b represent the strength of an acid and a base, respectively. In case of a conjugate acid-base pair, they are related in a simple manner so that if one is known, the other can be deduced.

Considering the example of NH + and NH we see, NH + (aq) + H O(l) H O + (aq) + NH (aq) K a = [H O + ][ NH ] / [NH + ] = . × – NH (aq) + H O(l) NH + (aq) + OH – (aq) K b =[ NH + ][ OH – ] / NH = . × – Net: H O(l) H O + (aq) + OH – (aq) K w = [H O + ][ OH – ] = . × – M Where, K a represents the strength of NH + as an acid and K b represents the strength of NH as a base.

It can be seen from the net reaction that the equilibrium constant is equal to the product of equilibrium constants K a and K b for the reactions added. Thus, K a × K b = {[H O + ][ NH ] / [NH + ]} × {[NH + ] [OH – ] / [NH ]} = [H O + ][OH – ] = K w = ( . × – ) × ( . × – ) = .

× – M This can be extended to make a generalisation. The equilibrium constant for a net reaction obtained after adding two (or more) reactions equals the product of the equilibrium constants for individual reactions: K NET = K × K × …… ( . ) Similarly, in case of a conjugate acid-base pair, K a × K b = K w ( . ) Knowing one, the other can be obtained.

It should be noted that a strong acid will have a weak conjugate base and vice-versa. Alternatively, the above expression K w = K a × K b , can also be obtained by considering the base-dissociation equilibrium reaction: B(aq) + H O(l) BH + (aq) + OH – (aq) K b = [BH + ][OH – ] / [B] As the concentration of water remains constant it has been omitted from the denominator and incorporated within the dissociation constant. Then multiplying and dividing the above expression by [H + ], we get: K b = [BH + ][OH – ][H + ] / [B][H + ] ={[ OH – ][H + ]}{[BH + ] / [B][H + ]} = K w / K a or K a × K b = K w It may be noted that if we take negative logarithm of both sides of the equation, then p K values of the conjugate acid and base are related to each other by the equation: p K a + p K b = p K w = (at 298K) Problem . Determine the degree of ionization and pH of a .05M of ammonia solution.

The ionization constant of ammonia can be taken from Table . . Also, calculate the ionization constant of the conjugate acid of ammonia. The ionization of NH in water is represented by equation: NH + H O NH + + OH – We use equation ( .

) to calculate hydroxyl ion concentration, [OH – ] = c α = . α K b = . α / ( – α ) The value of α is small, therefore the quadratic equation can be simplified by neglecting α in comparison to in the denominator on right hand side of the equation, Thus, K b = c α or α = √ ( . × – / .

× – pH = –log( . × – ) = . . Now, using the relation for conjugate acid-base pair, K a × K b = K w using the value of K b of NH from Table .