Chemistry

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Solubility Product

We have come across many precipitation reactions in inorganic qualitative analysis. For example, dil HCl is used to precipitate Pb²⁺ ions as PbCl₂ which is sparingly soluble in water.

Kidney stones are developed over a period of time due to the precipitation of Ca²⁺ (as calcium oxalate etc.). To understand the precipitation, let us consider the solubility equilibria that exist between the undissociated sparingly soluble salt and its constituent ions in solution.

For a general salt X_mY_m

The equilibrium constant for the above is

In solubility equilibria, the equilibrium constant is referred as solubility product constant (or) Solubility product. In such heterogeneous equilibria, the concentration of the solid is a constant and is omitted in the above expression

K_{sp = [Xⁿ⁺]^m[Y^m]ⁿ}

The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co – efficient in a balanced equilibrium equation.

Solubility product finds useful to decide whether an ionic compound gets precipitated when solution that contains the constituent ions are mixed.

When the product of molar concentration of the constituent ions i.e., ionic product, exceeds the solubility product then the compound gets precipitated.

The expression for the solubility product and the ionic product appears to be the same but in the solubility product expression, the molar concentration represents the equilibrium concentration and in ionic product, the initial concentration (or) concentration at a given time ‘t’ is used.

In general we can summarise as,

Ionic product > K_sp, precipitation will occur and the solution is super saturated. Ionic product < K_sp, no precipitation and the solution is unsaturated.

Ionic product = K_sp, equilibrium exist and the solution is saturated.

Determination of Solubility Product from Molar Solubility

Solubility product can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution. For a solute X_mY_n,

X_mY_n(s) ⇄ mXⁿ⁺(aq) + nY^m-(aq)

From the above stoichiometrically balanced equation we have come to know that 1 mole of X_mY_n(s) dissociated to furnish ‘m’ moles of Xⁿ⁺ and ‘n’ moles of Y^m- is ‘s’ is molar solubility of X_mY_n, then

[Xⁿ⁺]=ms and [Y^m+]=ns
∴ K_sp = [Xⁿ⁺]^m[Y^m+]ⁿ

K_sp=(ms)^m(ns)ⁿ
K_sp=(m)^m(n)ⁿ(s)^m+n

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Salt Hydrolysis

When an acid reacts with a base, a salt and water are formed and the reaction is called neutralization. Salts completely dissociate in aqueous solutions to give their constituent ions. The ions so produced are hydrated in water. In certain cases, the cation, anion or both react with water and the reaction is called salt hydrolysis. Hence, salt hydrolysis is the reverse of neutralization reaction.

Salts of Strong Acid and a Strong Base

Let us consider the reaction between NaOH and nitric acid to give sodium nitrate and water.
NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

The salt NaNO₃ completely dissociates in water to produce Na⁺ and NO₃^– ions.
NaNO₃(aq) → Na⁺(aq) + NO₃^–(aq)

Water dissociates to a small extent as
H₂O(l) ⇄ H⁺(aq) + OH^–(aq)

Since [H⁺] = [OH^–], water is neutral
NO₃^– ion is the conjugate base of the strong acid HNO₃ and hence it has no tendency to react

with H⁺. Similarly, Na⁺ is the conjugate acid of the strong base NaOH and it has no tendency to react with OH^–.

It means that there is no hydrolysis. In such cases [H⁺] = [OH^–] pH is maintained and, therefore, the solution is neutral.

Hydrolysis of Salt of Strong Base and Weak Acid (Anionic Hydrolysis)

Let us consider the reactions between sodium hydroxide and acetic acid to give sodium acetate and water.
NaOH (aq) + CH₃COOH(aq) ⇄ CH₃COONa(aq) + H₂O(l)

In aqueous solution, CH₃COONa is completely dissociated as below

CH₃COONa (aq) → CH₃COO^–(aq) + Na⁺(aq)

CH₃COO^– is a conjugate base of the weak acid CH₃COOH and it has a tendency to react with H⁺
from water to produce unionised acid.

There is no such tendency for Na⁺ to react with OH^–.

CH₃COO^–(aq) + H₂O(l) ⇄ CH₃COOH(aq) + OH^–(aq) and therefore [OH^–]>[H⁺], in such cases, the solution is basic due to hydrolysis and the pH is greater than 7. Let us find a relation between the equilibrium constant for the hydrolysis reaction (hydrolysis constant) and the dissociation constant of the acid.

K_h value in terms of degree of hydrolysis (h) and the concentration of salt (C) for the equilibrium can be obtained as in the case of ostwald’s dilution law. K_h = h²C and i.e [OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}}\) and i.e [OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}}\).

pH of salt solution in terms of K_a and the concentration of the electrolyte

pH + pOH = 14
pH = 14 – p OH = 14 – {- log [OH^–]}
= 14 + log[OH^–]

Hydrolysis of Salt of Strong Acid and Weak Base (Cationic Hydrolysis)

Let us consider the reactions between a strong acid, HCl, and a weak base, NH₄OH, to produce a salt, NH₄Cl, and water.

HCl (aq) + NH₄OH(aq) ⇄ NH₄Cl(aq) + H₂O(l)
NH₄Cl(aq) → NH₄⁺ + Cl^–(aq)

NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with with OH^– from water to produce unionised NH₄OH shown below.

NH₄⁺(aq) + H₂O(l) ⇄ NH₄OH(aq) + H⁺(aq)

There is no such tendency shown by Cl^– and therefore [H⁺]>[OH^–]; the solution is acidic and the pH
is less than 7.

As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the K_h and k_b as K_h.K_b = K_w

Let us calculate the K_h value in terms of degree of hydrolysis (h) and the concentration of salt

Hydrolysis of Salt of Weak Acid and Weak Base (Anionic & Cationic Hydrolysis)

Let us consider the hydrolysis of ammonium acetate.
CH₃COONH₄(aq) → CH₃COO^–(aq) + NH₄⁺(aq)

In this case, both the cation (NH₄⁺) and anion (CH₃COO^–) have the tendency to
react with water

CH₃COO^– + H₂O ⇄ CH₃COOH + OH^–
NH₄⁺ + H₂O ⇄ NH₄OH + H⁺

The nature of the solution depends on the strength of acid (or) base i.e, if K_a > K_b; then the solution is acidic and pH < 7, if K_a < K_b; then the solution is acidic and pH < 7, if K_a < K_b; then the solution is basic and pH > 7, if K_a = K_b; then the solution is basic and pH > 7, if K_a = K_b; then the solution is neutral.

The relation between the dissociation constant (K_a, K_b) and the hydrolysis constant is given by the following
expression.

K_a.K_b.K_h = K_w

pH of the Solution

pH of the solution can be calculated using the following expression,
pH = 7 + 1/2 pK_a – 1/2 pK_b.

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Buffer Solution

Do you know that our blood maintains a constant pH, irrespective of a number of cellular acid – base reactions. Is it possible to maintain a constant hydronium ion concentration in such reactions? Yes, it is possible due to buffer action.

Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases, and this ability is called buffer action. The buffer containing carbonic acid (H₂CO₃) and its conjugate base HCO^–₃ is present in our blood. There are two types of buffer solutions.

1. Acidic Buffer Solution:

A solution containing a weak acid and its salt.
Example: Solution containing acetic acid and sodium acetate

2. Basic Buffer Solution:

A solution containing a weak base and its salt.
Example: Solution containing NH₄OH and NH₄Cl

Buffer Action

To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should contain both acidic as well as basic components so as to neutralize the effect of added acid (or) base and at the same time, these components should not consume each other. Let us explain the buffer action in a solution containing CH₃COOH and CH₃COONa. The dissociation of the buffer components occurs as below.

If an acid is added to this mixture, it will be consumed by the conjugate base CH₃COO^– to form the undissociated weak acid i.e, the increase in the concentration of H⁺ does not reduce the pH significantly.

CH₃COO^–(aq) + H⁺(aq) → CH₃COOH(aq)

If a base is added, it will be neutralized by H₃O⁺, and the acetic acid is dissociated to maintain the equlibrium. Hence the pH is not significantly altered.

These neutralization reactions are identical to those reactions that we have already discussed in common ion effect.

Les us analyse the effect of the addition of 0.01 mol of solid sodium hydroxide to one litre of a buffer solution containing 0.8 M CH₃COOH and 0.8 M CH₃COONa. Assume that the volume change due to the addition of NaOH is negligible. (Given: K₄ for CH₃COOH is 1.8 × 10^-5).

The dissociation constant for CH₃COOH is given by

The above expression shows that the concentration of H⁺ is directly proportional to

Let the degree of dissociation of CH₃COOH be α then,
[CH₃OOH] = 0.8 – α and [CH₃COO^–]
= α + 0.8

Given that

K_afor CH₃COOH is 1.8 × 10^-5
∴ [H⁺] = 1.8 × 10^-5; pH = – log (1.8 × 10^-5)
= 5 – log 1.8
= 5 – 0.26
pH = 4.74

Calculation of pH after adding 0.01 mol NaOH to 1 litre of buffer.

Given that the volume change due to the addition of NaOH is negligible
∴[OH^–] = 1.8 × 10^-5; pH = – log (1.8 × 10^-5)
= 5 – log 1.8
= 5 – 0.26
pH = 4.74

Caluculation of pH after adding 0.01 mol NaOH to 1 litre of buffer.

Given that the volume change due to the addition of NaOH is negligible
∴ [OH^–] = 0.01 M.

The consumption of OH^– are expressed by the following equations.

The addition of a strong base (0.01 M NaOH) increased the pH only slightly ie., from 4.74 to 4.75 . So, the buffer action is verified.

Buffer Capacity and Buffer Index

The buffering ability of a solution can be measured in terms of buffer capacity. Vanslyke introduced a quantity called buffer index, β , as a quantitative measure of the buffer capacity. It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pH by unity.

Here, β = \(\frac{dB}{d(pH)}\) …………. (8.19)
dB = number of gram equivalents of acid / base added to one litre of buffer solution.
d(pH) = The change in the pH after the addition of acid / base.

Henderson – Hassel Balch Equation

We have already learnt that the concentration of hydronium ion in an acidic buffer solution depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,

………….. (8.20)

The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

…………… (8.21)

Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution. Taking logarithm on both sides of the equation

……………. (8.22)

Reverse the sign on both sides

…………….. (8.23)

We know that

pH = – log [H₃O⁺] and pK_a = – logK_a

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Common ion Effect

When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. For example, the addition of sodium acetate to acetic acid solution leads to the suppression in the dissociation of acetic acid which is already weakly dissociated. In this case, CH₃COOH and CH₃COONa have the common ion, CH₃COO^–.

Let us analyse why this happens. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists. CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO^–(aq).

However, the added salt, sodium acetate, completely dissociates to produce Na⁺ and CH₃COO^– ion.

CH₃COONa(aq) → Na⁺(aq) + CH₃COO^–(aq)

Hence, the overall concentration of CH₃COO^– is increased, and the acid dissociation equilibrium is disturbed. We know from Le chatelier’s principle that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress.

So, in order to maintain the equilibrium, the excess CH₃COO^– ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of a weak acid (CH₃COOH) is suppressed. Thus, the dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

The common-ion effect refers to the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate. This behaviour is a consequence of Le Chatelier’s principle for the equilibrium reaction of the ionic association/dissociation.

The common ion effect is the phenomenon in which the addition of an ion common to two solutes causes precipitation or reduces ionization. An example of the common ion effect is when sodium chloride (NaCl) is added to a solution of HCl and water.

The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium. The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. The reaction is put out of balance, or equilibrium.

Aquifers which contain chalk and limestone, a common ion effect is used in these to obtain drinking water. Calcium carbonate is sparingly soluble in water, and it can be precipitated out by adding sodium chloride in the solution. In this way, the common ion effect is used in treatment of water.

According to Le Chatelier’s principle, addition of more ions alters the equilibrium and shifts the reaction to favor the solid or deionized form. In the case of an an acidic buffer, the hydrogen ion concentration decreases, and the resulting solution is less acidic than a solution containing the pure weak acid.

Le Chatelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

At a given temperature, the product of water concentrations and ions is known as the ionic product of water. With the rise in temperature, the value increases i.e. the concentration of H⁺ and OH^– ions increases with temperature increases.

If you have a solution and solute in equilibrium, adding a common ion (an ion that is common with the dissolving solid) decreases the solubility of the solute. This is because Le Chatelier’s principle states the reaction will shift toward the left (toward the reactants) to relieve the stress of the excess product.

pH gives us the measure of acid/base strength of any solution. pH scale ranges from 0-14, 7 is neutral, below 7 it represents acidic nature i.e. the compound is acidic and above 7 it represents basic nature i.e. the compound is basic i.e. pH is the negative logarithm of hydrogen ion.

The solubility of slightly soluble substances can be decreased by the presence of a common ion. This effect is known as common ion effect. Addition of a common ion to a slightly soluble salt solution will add up to the concentration of the common ion.

Le Chatelier’s principle (also known as “Chatelier’s principle” or “The Equilibrium Law”) states that when a system experiences a disturbance (such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.

Explanation:

This phenomenon is called the common-ion effect. When a compound dissolves in water it dissociates into ions. Increasing the concentration of one of these ions will shift the equilibrium towards the compound, thereby making it hard for the compound to dissolve in water (decreases solubility of compound).

When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas. When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.

The classic example of the practical use of the Le Chatelier principle is the Haber-Bosch process for the synthesis of ammonia, in which a balance between low temperature and high pressure must be found.

Le Chatelier’s Principle helps to predict what effect a change in temperature, concentration or pressure will have on the position of the equilibrium in a chemical reaction. This is very important, particularly in industrial applications, where yields must be accurately predicted and maximised.

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Ionisation of Weak Acids

We have already learnt that weak acids are partially dissociated in water and there is an equilibrium between the undissociated acid and its dissociated ions. Consider the ionisation of a weak monobasic acid HA in water.

Applying law of chemical equilibrium, the equilibrium constant K_c is given by the expression

…………. (8.9)

The square brackets, as usual, represent the concentrations of the respective species in moles per litre. In dilute solutions, water is present in large excess and hence, its concentration may be taken as constant say K. Further H₃O⁺ indicates that hydrogen ion is hydrated, for simplicity it may be replaced by H⁺. The above equation may then be written as,

………….. (8.10)

The product of the two constants K_C and K gives another constant. Let it be K_a

…………. (8.11)

The constant K_a is called dissociation constant of the acid. Like other equilibrium constants, K_a also varies only with temperature. Similarly, for a weak base, the dissociation constant can be written as below.

………….. (8.12)

Ostwald’s Dilution Law

Ostwald’s dilution law relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c). Degree of dissociation (α) is the fraction of the total number of moles of a substance that dissociates at equilibrium.

We shall derive an expression for ostwald’s law by considering a weak acid, i.e. acetic acid (CH₃COOH). The dissociation of acetic acid can be represented as

CH₃COOH ⇄ H⁺ + CH₃COO^–

The dissociation constant of acetic acid is,

Substituting the equilibrium concentration in equation (8.13)

…………. (8.14)

We know that weak acid dissociates only to a very small extent. Compared to one, α is so small and hence in the denominator (1 – α) ~ 1. The above expression (8.14) now becomes,

………….. (8.15)

Let us consider an acid with K_a value 4 × 10^-4 and calculate the degree of dissociation of that acid at two different concentration 1 × 10^-2M and 1 × 10^-4M using the above expression (8.15) For 1 × 10^-2M,

= 2 × 10^-1
= 0.2

For 1 × 10^-4M

= 2

When the dilution increases by 100 times, (Concentration decreases from 1 × 10^-2M to 1 × 10^-2M), the dissociation increases by 10 times. Thus, we can conclude that, when dilution increases, the degree of dissociation of weak electrolyte also increases. This statement is known as Ostwald’s dilution Law.

The concentration of H⁺ (H₃O⁺) can be caluculated using the K_a value as below.

[H⁺] = αC (Refer table) ………….. (8.16)

Equilibrium molar concentration of [H⁺] is equal to αC

Similarly, for a weak base

K_b = α² and α = \(\sqrt{\frac{K_{b}}{C}}\)
[OH^–] = αC
(or)
[OH^–] = \(\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}\) ………….. (8.18)