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    JEE Main Study Notes for Equilibrium: Important Tips, Previous Year Solved Questions and 45 Day Study Plan

    Anam Shams Anam Shams
    Content Curator

    Equilibrium is a relatively less important topic in JEE Main Chemistry. However, it forms an easy chunk of the Physical Chemistry syllabus. A total of 2 questions are usually asked carrying 8 marks from this topic. Therefore, the weightage of Equilibrium in the JEE Main is around 2-3%.

    Although a low weightage compared to other topics like Atomic Structure and Electrochemistry, the questions are quite scoring. Some of the topics tested under Equilibrium are Le Chatelier's Principle, Relation between Kp and Kc, Solubility Product, and Salt Analysis. More than one question was asked from Equilibrium in the JEE Main Session on January 8th, 2020. Check JEE Main Chemistry Syllabus

    Before you proceed towards studying the notes for Equilibrium, take a look at the following question -

    What is the order of the basic character?

    https://lh5.googleusercontent.com/bRytT6qmnwCEn8XhZKhqv-b1lknHRxylFnusgOr3r9yLM1uCOz1KoBJAGXpzIo7AxJIGq0JjjKzXQQFYGvsQ_6G-fu3lm25nYhNZ4EWcISTaWdGDUdslJmU9EZGr3jUsmovydMK3

    1) I > II > III > IV

    2) IV > III > I > II

    3) II > I > III > IV

    4) IV > I > II > III

    To ace Equilibrium in JEE Main Chemistry, refer to the study notes given below. 

    What is Equilibrium?

    Also Check 

    What is Equilibrium?

    Equilibrium is the state of a process in which the system's properties such as temperature, pressure, and concentration display no change over time. Two opposing processes are involved in all processes which attain equilibrium. It reaches equilibrium when the concentrations of the two competing processes are equal. The equilibrium is considered Physical Equilibrium because the competing mechanisms require only physical shifts. If chemical reactions are the opposite processes, the equilibrium is called Chemical Equilibrium

    Physical Equilibrium 

    Physical equilibrium can be described as equilibrium in physical processes. 

    ProcessConclusion
    Solid ⇔ Liquid H2O(s) ⇔ H2O(l)At a constant pressure, melting point is fixed.
    Liquid ⇔ Gas H2O(l) ⇔ H2O(g)PH2) constant at a specific temperature.
    Solute (s) ⇔ Solute (solution) Salt(Solid) ⇔ Salt(in solution)At a specific temperature, the conc. of solute in a solution is constant.
    Gas (g) ⇔ GAs (aq.) CO2(g) ⇔ CO2(in solution)At a specific temperature, gas (g) or gas(aq.) is constant. 

    Must Read

    a. Solid - liquid equilibrium

    An example of solid - liquid equilibrium is the relationship of ice and water. In a close network, ice and water achieve equilibrium at 0oC. Around this point the rate of ice melting is equal to that of water freezing. Equilibrium is seen here as -

    H2O(s) ⇔ H2O(l)

    b. Liquid - gas equilibrium

    Water evaporation in a closed tank is an example of liquid-gas equilibrium, where evaporation rate equals condensation rate. Equilibrium is seen here as -

    H2O(l) ⇔ H2O(g)

    c. Solid – Solution Equilibrium

    As you add more and more salt in water that is taken in a glass jar and mixed with a glass rod, after you dissolve it a bit, you will figure out that there is no more salt going to the water and it sets down at the bottom. The solution is now reported to be saturated and in a state of equilibrium. At this point, several salt molecules from the undissolved salt enter the solution (dissolution) and the same quantity of dissolved salt is deposited back (precipitation). Therefore, the rate of dissolution at equilibrium is equal to that of precipitation.

    Salt(Solid) ⇔ Salt(in solution)

    d. Gas –Solution equilibrium

    Within a closed vessel, the dissolution of a gas in a liquid under pressure forms a gas-liquid equilibrium. Cold drink bottles are the perfect example of this type of equilibrium. The equilibrium within the bottle is-

    CO2(g) ⇔ CO2(in solution)

    Download JEE Main Practice Paper


    Equilibrium in Chemical Process

    Equilibrium in Chemical Process (Reversible and Irreversible Reactions)

    A reaction is considered a reversible reaction when not only the reactants react to form the products under specific conditions but also the products react to form reactants under the same conditions. 

    For example, 

    1. 3Fe(s) + 4H2O(g) ⇔ Fe3O4(s) + 4H2(g)
    2. CaCO3(s) ⇔ CaO(s) + CO2(g)
    3. N2(g) + 3H2(g) ⇔ 2NH3(g)

    If a reaction can not take place in the reverse direction , i.e. the products formed do not react under the same condition to give back the reactants, it is called an irreversible reaction.

    For example, 

    1. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(g)
    2. 2M(g) + O2(g) → 2MgO(s)
    Note: If any of the products get eliminated from the system, a reversible reaction will turn into an irreversible one.

    Typically, a chemical equilibrium is represented as - 

    \fn_phv a\ A+b\ B \rightleftharpoons c\ C+b\ D where A and B are reactants, and C and D are products.

    The double arrow between the left part and the right part indicates that shifts exist in both directions. 

    Based on the degree of reaction, chemical reactions can be divided into three groups before equilibrium is reached. 

    A) Reactions which are almost complete. 

    B) Certain reactions which only continue to a very limited degree. 

    C) Those reactions which continue to such an extent that the equilibrium concentrations of reactants and products are comparable. 

    In fact, the state of equilibrium is fluid and not static. When the rate of forward reaction is equal to that of backward reaction, a reaction is assumed to have reached equilibrium.

    https://lh3.googleusercontent.com/4kaKga2CKVrAqVpt0W9aBTjVoU6ut9lGR1fE_epHNZleQTMjpzrQ0ULVLDqAZz_gRg3gl0HfoKnQH9gfv_HkmnI3k2NCz57KrZzNVpl56C18HSnozIFePwnNRXlv8AcnPuJoF1u1

    If all the reactants and products of any equilibrium reaction are in the same physical state, this is known as homogeneous equilibrium. 

    For example,

     N2(g) +  3H2(g) \rightleftharpoons2NH3 (g) 

    Here, all the reactants and products are in the same physical state.

    Heterogeneous equilibrium is the type of equilibrium in which the state of one or more reacting species varies, i.e. both reactants and products are not in the same physical condition.

    For example,

    2NaHCO3(s) \rightleftharpoonsNa2CO3(s) + CO2(g) + H2O(l)

    Here, the reactant is in the solid state while the product is in the solid, liquid, and gaseous matter states. 


    Ionic Equilibrium

    Ionic Equilibrium

    Chemical reactions occur mostly in solutions. Solution chemistry plays a very critical role in the process. All chemical substances consist of either polar or nonpolar units (called ions). The operation of these substances is more apparent and noticeable in solutions. The action of these substances is based on the nature and conditions of the medium in which they are applied. Consequently , it is important to understand the principles which regulate their actions in solution. 

    This kind of equilibrium is found in quickly ionizing substances, or in polar compounds where ionization can be caused. Because of the ease of ionization in solvent medium, ionic and polar substances are more easily soluble in polar solvents. These solutions become rich in mobile charge carriers (ions) with the dissolution of ionic and polar substances in the solvent, and thus can conduct electricity. Substances capable of conducting electricity are called electrolytes, while non-conducting substances are called non-electrolytes.
    Check Gujarat Board Released Question Banks for JEE Main


    Characteristics of Equilibrium State

    Characteristics of Equilibrium State

    i) When the reversible reaction is conducted in a closed system, equilibrium can be achieved. 

    ii) This can be done from either side of the reaction. 

    iii) A catalyst can speed up the equilibrium approach but does not change the equilibrium state.

    iv) It is of a dynamic nature, i.e. the reaction does not end but reactions both forward and backward take place at the same time. 

    v) Changes in pressure, concentration, or temperature favor one of the reactions (forward or backward) resulting in a shift in one direction of equilibrium point.

    Read How to Effectively Prepare for the Coming JEE Main Session from Home?

    Law of Mass Action and Equilibrium Constant 

    Guldberg and Waage developed a relationship between the rate of chemical reaction and the concentration of the reactants or, in the context of the law of mass action, their partial pressure. According to this law, “the rate at which a compound reacts is directly proportional to its active mass, and the rate of a chemical reaction is directly proportional to the sum of the active masses of reactants each elevated to a value equal to the related stoichiometric coefficient occurring in the balanced chemical equation”.

    https://lh3.googleusercontent.com/FB5L8q77Fk-CO_1yU_ta2wANGyidwLJdgocjGtglz_Lb5Ev9pNe6Uz_HcCOpD5BPEaXFEL9JKQkULDsWHWe12DPynUMBd7QnM4H-Bg2NBLGEmz-NvGc6j2Z1BVK3h4BEosFUbCPm

    rate of reaction ∝ [A]a.[B]b

    rate of reaction = K[A]a[B]b

    where K is the rate constant or velocity constant of the reaction at that temperature.

    Unit of rate constant (K) = [moles/lit]1–n time–1 (where n is order of reaction).

    Note: For unit reactant concentration, the reaction rate is equal to that of constant or similar rate of reaction. 

    The rate of forward and backward reactions is equal in equilibrium and the concentration of reactants and products remains constant as shown in the figure after achieving equilibrium.

    https://lh5.googleusercontent.com/dNAkR-VS_ZzGer6nBcwKJELHYSKQfLtpwIJDfNswTgUBt9F8_PHiO3NGzRJYxHzxA2dKbDs10lprxQ-y4ewIHCNzaglYb9q3WAZydl44sPPNY6wg3FG5N20VUEHxPbQppvQpRT-A

    Note: Active mass is the molar concentration of the reacting substances that are actually partaking in the reaction.

    Hence,

    Active mass = number of moles/volume in litres

    Active mass of solid is taken to be unity.

    Also, Active mass of reactant (a) = Conc. × activity coefficient

    i.e. a = Molarity × f for dilute solution f = 1

    Applying Law of mass action for general reversible reaction,

    aA + bB \rightleftharpoonscC + dD

    Rate of forward reaction \alpha[A]a[B]b

    or Rf = Kf [A]b [B]b

    Similarly for backward reaction

    Rb = Kb[C]c [D]d

    At equilibrium Kf[A]a[B]b = Kb[C]c[D]d

    \Rightarrow \frac{K_f}{K_b} =\frac{[C]^c[D]^d}{[A]^a[B]^b} =K_c

    The above equation is called equilibrium equation and Kc is known as equilibrium constant.

    https://lh4.googleusercontent.com/ZYRIYdWTD07N3o2GFZ996s9aWEW9ef78KiRzjGK5lpOoV5G5s_H0UvUI7FsBtAZsoyB1LZ-GJJsX0pjZ8ZZmy0iohZj_sehWoLvoakWGuMICTVOCrSzlkeH4vuwlBtsLgfCM6-84

    The subscript ‘c’ shows that Kc is displayed in concentrations of mol L–1. 

    At a specific temperature, the product of concentrations of the reaction products raised in the balanced chemical equation to the respective stoichiometric coefficient divided by the product of concentrations of the reactants raised to their corresponding stoichiometric coefficients has a constant value. This is known as the Law of Chemical Equilibrium or Equilibrium Law. 

    In the reverse reaction, the equilibrium constant is the inverse of the equilibrium constant for the forward path reaction.

    \Rightarrow K_f =\frac{1}{K_b}

    Equilibrium Constant in terms of Partial Pressures (Kp)

    For reactions that involve gases,

    pV = \textup{nRT} \Rightarrow p = \frac{\textup{n}}{v}\ \textup{RT}

    \Rightarrow p = \textup{cRT}

    it is more feasible to display the equilibrium constant in terms of partial pressure.

    The ideal gas equation is represented as,

    where c or n/v is the concentration expressed in mol/m3 (mol/L3) & R= 0.0831 bar litre/mol K.

    This is an indication that at constant temperature, the pressure of any gas is directly proportional to its concentration.

    For reaction in equilibrium,

    \textup{aA(g)+bB(g)}\rightleftharpoons \textup{cC(g)+dD(g)}

    \textup{K}= \frac{[C]^c[D]^d}{[A]^a[B]^b} = \textup{K}_\textup{c}

    \textup{or}\ \textup{K} = \frac{[p_C]^c[p_D]^d}{[p_A]^a[p_B]^b} = \textup{K}_\textup{p}

    We can write, With the usage of Ideal Gas Equation we get,

    PA = CART = [A] RT

    PB = CBRT= [B] RT

    Pc = CCRT = [C] RT

    Pd = CDRT= [D] RT

    Substituting the above values, 
    \textup{K}_\textup{p} =\frac{[C]^c[D]^d(RT)^(^c^+^d^)}{[A]^a[B]^b(RT)^(^a^+^b^)}

    \Rightarrow \textup{K}_\textup{p} =\frac{[C]^c[D]^d}{[A]^a[B]^b} (RT)^(^c^+^d^-^a^-^b^)

    \Rightarrow \textup{K}_\textup{p} =\textup{K}_\textup{c} (RT)^\Delta ^n

    where n = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation. The equation given above represents the relationship between kc and kp.

    Equilibrium Constant of a Reaction that involves Condensed Phase

    The representation for the equilibrium constant of a reaction between a condensed substance (solid or liquid) and a gaseous state is determined by considering only the gaseous species concentrations or partial pressures. Steady integration of condensed phases is combined with constant equilibrium.


    Facts about Equilibrium Constant

    Facts about Equilibrium Constant

    a) A huge value of KC or KP signals that the forward reaction reaches completion or very nearly so.

    b) A low numerical value of KC or KP signals that the forward reaction does not reach to any significant extent.

    c) A reaction will most probably reach a state of equilibrium in which both, reactants and products are present if the numerical value of KC or Kp is not very huge or low. 

    d) The equilibrium constant of a forward reaction and its backward reaction are reciprocal to each other.

    https://lh6.googleusercontent.com/34RBi-b_oaUoh4l2hW-Jk2FmRoBg91gJx_Z9izmiNwxnKI5W1N8kVTURy7maS9BZKfu8c2JTc-voJvHgmqMLrQ2VCySZlgym4XBLOw8wahSdvrGaja71SAtyFSeB4_j-ZoDPczYO

    e) If a chemical equation is multiplied by a specific factor, its equilibrium constant should be increased to a power equal to that factor so as to avail the equilibrium constant for the new reaction.

    https://lh6.googleusercontent.com/sGNKC_fLJ8H3oHanP8PqVQtVU6rhD5xTQGvhnecJQ0Z91db0btnNnaZwCfhrRmhDGbulyHei_VnnwqntTEowotGTemZ2OdsK4DCdEmHX47gdLYFJtv0xXnJg6y4xoAIQh-MFXx4v

    The Reaction Quotient “Q”

    Look at the following equilibrium.

    PCl5 (g) 39_equal.JPGPCl3(g) + Cl2 (g)

    At equilibrium,

    \frac{[Cl_2].[PCl_3] }{ [PCl_5]} = K_c

    When the reaction is not in equilibrium, this ratio is known as ‘QC’. QC is the general term used for the ratio given above at any point of time. At equilibrium, QC becomes KC.

    In a similar manner, PCl2 PPCl2 / PPCl2 is known as QP and at equilibrium, it turns into KP.

    Suppose the reaction is at equilibrium, Q = Kc

    A net reaction progresses from left to right (forward direction) if Q < KC.

    A net reaction progresses from right to left (the reverse direction) if Q >Kc

    A net reaction proceeds from left to right (forward direction) if Q < KC.  A net reaction proceeds from right to left (the reverse direction) if Q >Kc


    Le Chatelier’s Principle

    Le Chatelier’s Principle 

    "When an equilibrium is exposed to a change of concentration, temperature or external pressure, the equilibrium will move in that direction where the effects of these changes are nullified." 

    It can be understood by example given below. Overall, we can also foresee the direction of equilibrium with the theoretical assumption in mind.

    PCl5 ——> PCl3 + Cl2

    Assume that this reaction is at equilibrium and the moles of Cl2, PCl3, and PCl5 at equilibrium are a, b, and c respectively, and the total pressure is PT.

    2363_equation.JPG

     Since,

    PT  =  (a+b+c) RT / V  

    KP = abRT / cV  

    Suppose d moles of PCl3 is included in the system, the value of Q would be, a(b+d)RT / cV. This is evdiently more than KP. Therefore, the system will move reverse to reach equilibrium.

    If the volume of the system is increased, the Q becomes abRT / cV' where V' > V. Q becomes lesser, and the system will move forward to obtain equilibrium.

    If a noble gas is added at constant pressure, it leads to an increase of the system’s volume and hence, the reaction moves forward.

    If the noble gas is added at constant volume, the expression of Q remains as Q = abRT / cV and the system remains in equilibrium.

    Thus, for using the Le Chatlier’s principle, change the expression of KP and KC into basic terms and later notice the effect of different changes.

    Solubility Product 

    A solution is said to be saturated when it stays in contact with undissolved solute.

    In a saturated solution of an electrolyte, two equilibria exist. This can be represented as - 

    AB  \rightleftharpoons  AB    \rightleftharpoons        solublity-product                  

    Solid unionized              ions

    (dissolved)

    On application of the law of action to the ionic equilibrium,

      K=\frac{[A^+][B^-]}{[AB]}

    As the solution is saturated, the concentration of unionised molecules of the electrolyte is stable at a specific temperature, i.e., [AB] = K'= constant.

    Hence,

    [A+] [B-] = K[AB] = KK’ = Ks (constant)

    Ks is called the solubility product. It is described as the product of the concentration of ions in a saturated solution of an electrolyte at a particular temperature.

    Note: Solubility product is the ionic product only when the solution is saturated and not in all conditions.

    Salt Hydrolysis 

    SaltNatureDegreeHydrolysis ConstantpH
    NaCl (Strong acid + Strong Base) 2. Ch3COONa (Weak acid + Strong base) 3. NH4Cl (Strong acid + Weak base) 4. CH3COONH4 (Weak acid + Weak base)Neutral Base Acidic *No Hydrolysis h = √kw/Cka h = √kw/Ckb h = √kw/(ka + kb)- Kh = kw/ka Kh = kw/Ckb Kh = kw/(ka + kb)- pH=1/2[pkw + pka + logC] pH=1/2[pkw- pkb - logC] pH=1/2[pkw + pka - pkb]

    In the scenario of salt of weak acid and weak base, the nature of medium after hydrolysis is determined in the following manner:

    (i) If Ka = Kb, the medium should be neutral.

    (ii) If Ka > Kb, the medium should be acidic.

    (iii) If Ka < Kb, the medium should be basic.

    The degree of hydrolysis of salts of weak acids and weak bases is not affected by dilution as there is no concentration term in the expression of degree of hydrolysis.

    Note: The degree of hydrolysis always rises with an increase in temperature because at a high temperature, increase in Kw is more when compared to Ka and Kb.

    Points to Remember in Equilibrium

    Points to Remember in Equilibrium

    1. Remember,

      • CH3COOH has Ka = 1.8*10^(–5) and pKa = 4.74;
      • For NH3, pKb = 4.77
      • Sq.rt. of 1.8 and 18 are 1.34 and 4.2 respectively.
    2. Acid-base theories- Bronsted and Lewis theory

      • CAB pairs. (add H+ to get CA and remove H+ to get CB)
      • Na+, K+, NH4+ are not electrophiles or lewis acids.
      • [SbF6]-, [SiF6]2- are non-nucleophilic anions. (why?)
    3. Acid-base equilibrium

    1. Acid-base reactions progress towards the formation of weaker acids and weaker bases.

    Previous Year Solved Questions

    Previous Year Solved Questions on Equilibrium

    Q1. Find out the pH of the solution when 0.1 M CH3 COOH (50 ml) and 01. M NaOH (50 ml) are mixed, [Ka (CH3COOH)=10-5]

    Solution:

    CH3 COOH \rightleftharpoons CH3 COO_ + H+ …(I)

    NaOH → Na+ + OH-

    H+ + OH_ \rightleftharpoonsH2O …(II)

    (I) + (II)

    CH3COOH + OH- CH3COO- + H2O . (III)

    0.05-X 0.05-x x

    Keq of eq. (III) = Ka/Kw

    Conc. of H2O remains constant

    109 = x/(0.05-x)2

    because the value of eq. Const. is very high

    Here for x» 0.05

    Let 0.05-x=a

    109=0.05/a2

    a = 7.07\times10-6

    pOH= 6-log 7.07

    pOH= 6 – 0.85

    pH= 14-6+0.85 = 8.85

    Q2. The solubility product of Pb3 (PO4)2 is 1.5 x 10-32. Calculate the solubility in gms/litre.

    Solution:

    Solubility product of Pb3 (PO4)2 = 1.5 \times10–32

    Pb3 (PO4)2 \rightleftharpoons3Pb2+ + 2PO43-

    Suppose x is the solubility of Pb3 (PO4)2

    Then, Ksp = (3x)3 (2x)2 = 108 x5

    https://lh5.googleusercontent.com/NH2eNqcNAbfalxAGg8u23thsAp35FEg_FHYfplMcs0lVYJjr3WFnxdSA6L_MfeUpBiNaTnNeIeimjhOZw-jjFhzL8CcrJ4ab2NH-Y8anpZgTG_O_kI5PTVcWbi41dsaLvLs9qq4v

    x = 1.692 \times10–7 moles/lit

    Molecular mass of Pb3(PO4)2 = 811

     x = 1.692 ´ 10–7 ´ 811 g/lit = 1.37 \times10–4 g/lit

     Solubility product is

     Ksp(SrC2O4) = [Sr2+] [C2O42–] = (5.4 \times10–4)2- = 2.92 \times10–7

    Q3. Calculate the concentration of H+, HCO3- and CO32-, in a0.01M solution of carbonic acid if the pH is 4.18.

    Ka1(H2CO3) = 4.45 \times10–7 and Ka2 = 4.69 \times10–11

    Solution: 

    pH = – log[H+]

    4.18 = – log [H+]

    [H+] = 6.61 \times10–5

    H2CO3 \rightleftharpoonsH+ + HCO3-

    https://lh6.googleusercontent.com/Q84vlmdga7IXdXMbW0rHUgB_WZEGMeqIheAKVPkB_hw3QLkojHf3wdKMXgjl7lv9lX44ZEURE7K2BfcD6iJJTWro-eE8RyNVMbNCwkeGj47RyRbq3Lz0fVGG8_az2ngD0Bcx_Z8P

    HCO3- \rightleftharpoonsH+ + CO32-

    https://lh3.googleusercontent.com/ND7xLD0R7_wWO-vzYxgYnWbI98cNcvgTpRH5y_a8fFZH8MInUvcgsahziKF3BB0YoYaSvoVPsSs3ugiZB9kxATl6gFQhPk17FMXCr_8_TV0E1oEIWY16P6t5CDRdVWyZbrM1KwV5

     [CO32-] = 4.8 \times10–11

    Q4. Find out the molar solubility of Mg(OH)2 in 1MNH4Cl

    KspMg(OH)2 = 1.8 \times10–11

    Kb(NH3) = 1.8 \times10–5

    Solution:

    Mg(OH)2(s) \leftrightharpoonsMg++ + 2OH– K1 = Ksp

    2NH4+ + 2OH- \leftrightharpoons2NH4OH K2 = 1/K2b

    https://lh5.googleusercontent.com/jIX9NNruN3c9mypDH9ekoKidt-K2VqBADIVs3AVaM8LADnCCrpFuYp95Ot8BRwA4AJNpShSMvssbtTv_uifPzoYqySRzWG74Yfa8HUq5-V1KMwDMVywOdrhJyKmgzEyWKhfyQdYC

    Q5. How much of AgBr can dissolve in 1.0 L of 0.4 M NH3? Assume that [Ag (NH3)2]+ is the sole complex formed if Kf [Ag(NH3)2+]=1.0\times108, Ksp (AgBr)= 5.0\times10-13

    Solution: 

    AgBr \rightleftharpoonsAg+ + Br- 

    Ag+ + 2NH3 \rightleftharpoonsAg (NH2)2+

    Let x= solubility ,

    Then, x= [Br-]=[Ag+]+[Ag(NH3)2+]

    https://lh5.googleusercontent.com/mB82MPLX4V5F-MGSmXMU3Ea-e_fJs_ciXnEhZxf_x2vQ2S-tC5JMxxeR0tqCgNRBXKWEQnFPIn0xwWBmpFF0TwS8bcmfy74CkZQegAnrViQ9Q2PJTtg6QCgoXn8z8G-XXmdcXARY

    x2=8.0´10-6 

    \Rightarrowx=2.8´10-3 M 

    Revision Tips for Equilibrium

    1. For the auto-protolysis of water-
    • Calculate the degree of dissociation of water and its dissociation constant at 298K.
    • Ionic product of water (Kw) and its reliance on temperature.
    1. Use common-ion effect in Salt-analysis
    2. Remember that for theory of indicators, the pH range of the buffer or indicator is pKa+1, pKa-1 i.e. two units.

    Tricks to Solve Questions from Equilibrium 

    Q1. Trick to solve the question: Consider the solubility to be x. Balance the equation by splitting the compounds - Ag2Cr04 and AgNO3. Enter the values found in the final equation for Ksp and find the answer to x. 

    Solution: (b)

    Q2. Trick to solve the question: Split HA into its elements. Correlate the rates of HA andHX to frame an equation. Equate HX to 1M. Enter the values found and derive the answer. 

    Solution: (a)

    Q3. Trick to solve this question: Write down the formula for pH. Substitute the values of 1 and 2 given in the formula. Use M1V1 = M2V2 to find the value of V2. 

    Solution: (b)

    Tips to Study Chemistry for JEE Mains 

    1. Concentrate on all three sections – organic, physical and inorganic chemistry. 
    2. A must-know is periodic table and chemical bonding. Grasp complete reaction and its mechanism. 
    3. You ought to have a clear knowledge of basic Chemistry principles. 
    4. Thoroughly learn all the key formula. 
    5. Give special attention to topics such as GOC, chemical bonding, thermodynamics, electro-chemistry, and coordination compounds. 
    6. Make personal and short notes for yourself.
    7. Take JEE mock exams and go through the analysis to figure out what your shortcomings are. Work on your shortcomings, and increase your scores. 
    8. Take chapter-wise JEE Practice tests to boost the knowledge fundamentals as and when you keep preparing for every chapter.

    45 Day Chemistry Study Plan for JEE Main 

    45 Day Chemistry Study Plan for JEE Main 

    WeeksActivity
    Week 1 (June 7 to 14)Physical Chemistry has the most weightage in JEE Main Chemistry. Study important chapters from this section such as Atomic Structure, Electrochemistry, and Equilibrium first. Then, move on to study the chapters with lesser weightage like Surface Chemistry and Solid State. Choose two topics everyday and go through them. Revise the basics well. Use NCERT books.
    Week 2 (June 14 to 20, 2020)Study Organic Chemistry. Pay more attention to important chapters such as Aromatic Compounds, Alkyl Halides, Alcohol, and Ether. Then, move on to study less important chapters such as Biomolecules and Carbonyl Compounds. Use NCERT books.
    Week 3 (June 21 to 27, 2020)Study Inorganic Chemistry. Start by studying the most important chapters like Chemical Bonding, p Block, and s Block. Then study less important chapters like Metallurgy and Qualitative Analysis. Use NCERT books.
    Week 4 (June 28 to -July 4, 2020)Revise all the formulas and important concepts. Solve as many previous year’s papers as you can. Pick up questions from the Chemistry section and attempt them as fast as you can (within 45 minutes-1 hour).
    Week 5 (July 5 to 11, 2020)Attempt mock tests everyday. 
    Week 6 (July 12 to 17, 2020)Do one final revision. 
    JEE Main 2020 (July 18 to 23, 2020)Research the analysis of the ongoing papers.

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