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**Download or View Detailed Notes for Chemistry Class 11**^{th}**Download or View Detailed Notes for Chemistry Class 12**^{th}**Ionic Equilibrium**Theory**Fundamentals of Acids, Bases & Ionic Equilibrium****Acids & Bases**When dissolved in water, acids release $\mathrm{H}^{+}$ ions, base release $\mathrm{OH}^{-}$ ions.**Arrhenius Theory**When dissolved in water, the substances which release (i) $\mathrm{H}^{+}$ ions are called acids (ii) $\mathrm{OH}^{-}$ ions are called bases**Bronsted & Lowry Concept**Acids are proton donors, bases are proton acceptors Note that as per this definition, water is not necessarily the solvent.*When a substance is dissolved in water, it is said to react with water e.g.*$\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}$*; HCl donates H+ to water, hence acid.*$\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \quad \rightarrow \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}$*; NH3 takes H+ from water, hence base.**For the backward reaction, $N H_{4}^{+}$ donate $H^{+}$ , hence it is an acid; $O H^{-}$ accepts H+, hence it is base. $N H_{3}$ (base) & $N H_{4}^{+}$ (acid) from conjugate acid base pair.***To get conjugate acid of a given species add $\mathrm{H}^{+}$ to it. e.g. conjugate acid of N2H4 is N2H5+. To get conjugate base of any species subtract H+ from it. e.g. Conjugate base of $\mathrm{NH}_{3}$ is $\mathrm{NH}_{2}^{-}$.***Conjugate acid and bases***Note:***Although $C I^{-}$ is conjugate base of HCl, it is not a base as an independent species. In fact,anions of all strong acid like $C I, N O_{3}^{-}, C l O_{4}^{-}$ etc. are neutral anions. Same is true for cations of strong bases like $K^{+}, N a^{+}, B a^{++}$ etc. When they are dissolved in water, they do not react with water (i.e. they do not undergo hydrolysis) and*these ions*do not cause any change in pH of water (others like $\left.C N^{-} d o\right)$.***Basic Anions : $\mathrm{CH}_{3} \mathrm{COO}^{-}, \mathrm{OH}^{-}, \mathrm{CN}^{-}$ (Conjugate bases of weak acids)***Some examples of :**Acid Anions: $H S O_{3}^{-}, H S^{-}$ etc. Note that these ions are*amphoteric*, i.e. they can behave both as an acid and as a base. e.g. for $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}:$**Lewis Concept :**Acids are substances which accept a pair of electrons to form a coordinate bond and bases are the substances which donate a pair of electrons to form a coordinate bond.***Important :**$\mathrm{Ca}+\mathrm{S} \rightarrow \mathrm{Ca}^{2+}+\mathrm{S}^{2-}$ is not a Lewis acidbase reaction since dative bond is not formed.*Lewis Acids :**As per Lewis concept, following species can acts as Lewis Acids :**(i) Molecules in which central atom has*incomplete octet. \text { (e.g. }\left.\mathrm{BF}_{3}, \mathrm{AlCl}_{3} \text { etc. }\right)*(ii) Molecules which have a central atom with empty d orbitals \text { (e.g. }\left.\operatorname{six}_{4}, \mathrm{GeX}_{4}, \mathrm{PX}_{3}, \mathrm{TiCl}_{4} \text { etc. }\right)**(iii)*Simple Cations:*Though all cations can be expected to be Lewis acids,**\mathrm{Na}^{+}, \mathrm{Ca}^{++}, \mathrm{K}^{+} \text {etc. } show no tendency to accept electrons. However \mathrm{H}^{+}, \mathrm{Ag}^{+} etc. act as Lewis acids.**(iv) Molecules having multiple bond between atoms of dissimilar electronegativity.***Lewis bases**are typically : (i) Neutral species having at least one lone pair of electrons.**\text { Autoprotolysis (or self-ionization) constant }\left(\mathrm{K}_{\mathrm{w}}\right)=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \text { Hence, } \mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{w}} \text { at all temperatures }***Autoprotolysis of water (or any solvent)**Condition of neutrality**\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right] \text {(for water as solvent) }*\text { At } 25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{W}}=10^{-14} \cdot \mathrm{K}_{\mathrm{W}} increases with increase in temperature. Accordingly, the neutral point of water \left(\mathrm{pH}=7 \text { at } 25^{\circ} \mathrm{C}\right) a*lso shifts to a value lower than 7 with increase in temperature.***Important***: \mathrm{K}_{\mathrm{W}}=10^{-14}**s a value at \text { (i) } 25^{\circ} \mathrm{C} \text { (ii) }for water only. If the temperature changes or if some other solvent is used, autoprotolysis constant will not be same.**** \quad \text { For dissociation of weak acids (eg. HCN), HCN + H_{ } 2 } \mathrm{O} \text { 1 } \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CN}^{-} \text {the equilibrium }***Ionisation Constant**constant expression is written as Ka =\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]}*** For the Polyprotic acids \left(\mathrm{e} \cdot \mathrm{g} \cdot \mathrm{H}_{3} \mathrm{PO}_{4}\right) sucessive ionisation constants are denoted \text { by } \mathrm{K}_{1}, \mathrm{K}_{2}, \mathrm{K}_{3} \text { etc. } For \mathrm{H}_{3} \mathrm{PO}_{4},*\text { Similarly, } \mathrm{K}_{\mathrm{b}} \text { denotes basic dissociation constant for a base. } \text { Also, } \mathrm{pK}_{\mathrm{a}}=-\log _{10} \mathrm{K}_{\mathrm{a}}, \mathrm{pK}_{\mathrm{b}}=-\log _{10} \mathrm{K}_{\mathrm{b}}**\text { Some Important Results: }\**left[\mathrm{H}^{+}\right] \text {concentration of }*Case (i) A weak acid in water**Similarly for a weak base, substitute \left[\mathrm{OH}^{-}\right] \text {and } \mathrm{K}_{\mathrm{b}} instead of \left[\mathrm{H}^{+}\right] \text {and } \mathrm{K}_{\mathrm{a}}*respectively in these expressions.*Case (ii)**(a)**A weak acid and a strong acid \left[\mathrm{H}^{+}\right]**is entirely due to dissociation of strong acid*(b)**A weak base and a strong base \left[\mathrm{H}^{+}\right]**is entirely due to dissociation of strong base*Neglect the contribution of weak acid/base usually.***Condition for neglecting : \text { If } \mathrm{c}_{0}**concentration of strong acid, c_{1} = concentration of*weak acid then neglect the contribution of weak acid if \mathrm{K}_{\mathrm{a}} \leq 0.01 \mathrm{c}_{0}^{2 /} \mathrm{c}_{1}***Proceed by the general method of applying two conditions (i) of electroneutrality (ii) of equilibria.***Case (iii) Two (or more) weak acids**The accurate treatement yields a cubic equation. Assuming that acids dissociate to**\text { a negligible extent }\left[\text { i.e. } c_{0}-x \approx c_{0}\right] \quad\left[\mathrm{H}^{+}\right]=\left(\mathrm{K}_{1} \mathrm{c}_{1}+\mathrm{K}_{2} \mathrm{c}_{2}+\ldots+\mathrm{K}_{\mathrm{w}}\right)^{1 / 2}**Case (iv) When dissociation of water becomes significant:**\text { Dissociation of water contributes significantly to }\left[\mathrm{H}^{+}\right] \text {or }\left[\mathrm{OH}^{-}\right] \text {only when for }*(i)**strong acids (or bases) : 10^{-8} \mathrm{M}<\mathrm{c}_{0}<10^{-6} \mathrm{M}**.*Neglecting ionisation of water at**10^{-6} \mathrm{M} \text { causes } 1 \% \text { error (approvable). Below } 10^{-8} \mathrm{M}, Neglecting ionisation of water at 10^{-6} \mathrm{M} \text { causes } 1 \% \text { error (approvable). } Below 10^{-8} \mathrm{M}, contribution of acid (or base) can be neglected and pH can be taken to be practically 7.***Weak acids (or bases) : \text { When } \mathrm{K}_{\mathrm{a}} \mathrm{c}_{0}<10^{-12}**then consider dissociation of water as well.**HYDROLYSIS*****Salts of strong acids and strong bases**do not undergo hydrolysis. ***Salts of a strong acids and weak bases**give an acidic solution. e.g. \mathrm{NH}_{4} \mathrm{Cl} when*dissolved, it dissociates to give \mathrm{NH}_{4}^{+} ions and \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \quad 1 \quad \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+}**\mathrm{K}_{\mathrm{h}}=\left[\mathrm{NH}_{3}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] /\left[\mathrm{NH}_{4}^{+}\right]=\mathrm{K}_{\mathrm{W}} / \mathrm{K}_{\mathrm{b}} of conjugate base of \mathrm{NH}_{4}^{+}**Important!**In general : \mathrm{K}_{\mathrm{a}}(\text { of an acid }) \mathrm{xK}_{\mathrm{b}} \text { (of its conjugate base) }=\mathrm{K}_{\mathrm{w}}**If the degree of hydrolysis(h) is small (<<1), \quad \mathrm{h}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}}**Otherwise h = \frac{-\mathrm{K}_{\mathrm{h}}+\sqrt{\mathrm{K}_{\mathrm{h}}^{2}+4 \mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}}}{2 \mathrm{c}_{0}}, \quad\left[\mathrm{H}^{+}\right]=\mathrm{c}_{0} \mathrm{h}******Salts of strong base and weak acid***give a basic solution (\mathrm{pH}>7) when dissolved in water, e.g. NaCN,**\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \quad 1 \quad \mathrm{HCN}+\mathrm{OH}^{-} \quad\left[\mathrm{OH}^{-}\right]=\mathrm{c}_{0} \mathrm{h}, \mathrm{h}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}}******Salts of weak base and weak acid***Assuming degree of hydrolysis to be same for the both the ions,**\mathrm{K}_{\mathrm{h}}=\mathrm{K}_{\mathrm{w}} /\left(\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}}\right),\left[\mathrm{H}^{+}\right]=\left[\mathrm{K}_{\mathrm{a}} \mathrm{K}_{\mathrm{w}} / \mathrm{K}_{\mathrm{b}}\right]^{1 / 2}***Note:**Exact treatment of this case is difficult to solve. So use this assumption in general cases. Also, degree of anion or cation will be much higher in the case of a salt of weak acid and weak base. This is because each of them gets hydrolysed, producing \mathrm{H}^{+} \text {and } \mathrm{OH}^{-} ions. These ions combine to form water and the hydrolysis equilibrium is shifted in the forward direaction.*Buffer Solutions**are the solutions whose pH does not change significantly on adding a small quantity of strong base or on little dilution.**These are typically made by mixing a weak acid (or base) with its conjugate base (or acid). e.g.**\mathrm{CH}_{3} \mathrm{COOH} with \mathrm{CH}_{3} \mathrm{COONa}, \mathrm{NH}_{3}(\mathrm{aq}) \text { with } \mathrm{NH}_{4} \mathrm{Cl} \text { etc. }**\text { If }\left.\mathrm{K}_{\mathrm{a}} \text { for acid (or } \mathrm{K}_{\mathrm{b}} \text { for base }\right) (or Kb for base) is not too high, we may write :**Henderson's Equation**\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \{[\mathrm{salt}] /[\mathrm{acid}]\} for weak acid with its conjugate base.**\text { or } \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \{[\mathrm{salt}] /[\text { base }]\} for weak base with its conjugate acid.**Important**: For good buffer capacity, [salt] : [acid ratios should be as close to one as possible. In such a case, \mathrm{pH}=\mathrm{pK}_{\mathrm{a}} (This also is the case at midpoint of titration)**Buffer capacity = (no. of moles of acid (or base) added to 1L) / (change in pH)***Indicators.**Indicator is a substance which indicates the point of equivalence in a titration by undergoing a change in its colour. They are weak acids or weak bases.*Theory of Indicators.**The ionized and unionized forms of indicators have different colours. If 90 % or more of a particular form (ionised or unionised) is present, then its colour can be distinclty seen.In general, for an indicator which is weak acid, HIn l H+ + In–, the ratio of ionized to unionized form can be determined from*This roughly gives the range of indicators. Ranges for some popular indicators are*Table 1 : Indicators***Equivalence point.**The point at which exactly equivalent amounts of acid and base have been mixed.*Acid Base Titration.**For choosing a suitable indicator titration curves are of great help. In a titration curve, change in pH is plotted against the volume of alkali to a given acid. Four cases arise.*(a)**Strong acid vs strong base.**The curve is almost vertical over the pH range 3.5-10. This abrupt change corresponds to equivalence point. Any indicator suitable. (b)**Weak acid vs strong base.**Final solution is basic 9 at equivalence point. Vertical region (not so sharp) lies in pH range 6.5-10. So, phenolphathlene is suitable. (c)**Strong acid vs weak base.**Final solution acidic. Vertical point in pH range 3.8-7.2. Methyl red or methyl orange suitable.*(d)**Weak acid vs weak base.**No sharp change in pH. No suitable indicator.***Note***: at midpoint of titration,**\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}, thus by pH measurements, Ka for weak acids (or K_{b} for weak bases) can be determined.**Polyprotic acids and bases.**\text { Usually } \mathrm{K}_{2}, \mathrm{K}_{3} \text { etc. can be safely neglected and only } \mathrm{K}_{1} \text { plays a significant }**Solubility product**\left(\mathbf{K}_{\mathbf{s p}}\right). For sparingly soluble salts (eg.**\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}) an equilibrium which exists is**Precipitation.**Whenever the product of concentrations (raised to appropriate power) exceeds the solubility product, precipitation occurs.**Common ion effects.**Suppression of dissociation by adding an ion common with dissociation products.**\text { e.g. } \mathrm{Ag}^{+} \text {or } \mathrm{C}_{2} \mathrm{O}_{4}^{2-} in the above example.***Simultaneous solubility.**While solving these problems, go as per general method i.e. (i) First apply condition of electroneutrality and*(ii) Apply the equilibria conditions.*