Tiêu đề 1.IONIC EQUILIBRIUM FINAL SSR.pdf ✅

NISHITH Multimedia India (Pvt.) Ltd., 1 JEE MAINS - CW - VOL - I IONIC EQUILIBRIUM NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III SYNOPSIS * Relative Strength of Inorganic Acids * Hydracids of the elements of the same period: As we know, if the charge is spread over larger volume then, charge density is small and basic character of the ion is also small because of the ability of ion to attract a proton becomes less and hence acidic strength of conjugate acid increases. · Now, consider the hydracids of 2nd period CH4 , NH3 , H2O, HF. We find that, from left to right · Size of central atom over which the negative charge is present is decreasing · The volume of central atom overlapped by hydrogen is 3/4th, 2/3rd, 1⁄2 and 1 respectively. – C – N – O – F H H H H H H – CH3 – NH2 – OH – F Increasing volume available to electron Increasing electron delocalisation Decreasing electron density Decreasing bascity · Volume available for negative charge is increasing in the conjugate bases 3 3 (CH NH OH F )        , therefore, charge density of the conjugate bases is decreasing · Basicity of the conjugate bases is decreasing and acidity of the acids is increasing · Stability of conjugate bases is increasing 3 3 (CH NH OH F )        · Acid dissociation constant is increasing CH4 (10-58) < NH3 (10-35) < H2O (10-14)< HF (10-4) · Basicity constant is decreasing therefore, the acidic character is increasing 4 3 2 (CH NH H O HF)    . * Hydracids of the elements of the same group: In each Group : from top to bottom Atomic size of the central atom is increasing. Volume available for the negative charge is increasing Charge density is decreasing Basicity of the ions is decreasing Acidity of conjugate acids is increasing Acidity constant is increasing Basicity constant is decreasing Therefore; order of acidic character can be also explained as following along with the above reasons a) VII A group (Halogens) - HF < HCl < HBr < HI (Due to decreasing bond energy of H - X bond). b) VI A group - H2O < H2 S < H2 Se < H2 Te (Due to decreasing trend in electron donor ability of OH- , HS- , HSe- , HTe- ions). c) V A group - NH3 < PH3 < AsH3 < BiH3 (Due to decreasing order of electron density). d) IV A group - CH4 < SiH4 < GeH4 < SnH4 < PbH4 (Due to decreasing order of electronegativity.) e) III A group - BH3 < AlH3 < GaH3 < InH3 < TlH3 (Due to decreasing order of electronegativity.) * Among hydrides of elements with same electronegativity, acidic strength increases with size of the central atom e.g. CH H S HI 4 2   . * The ability of Borontrihalides to act as Lewis acid increases in the order of BF3 < BCl3 < BBr3 < BI3 due to decreasing order of strength of  bond. IONIC EQUILIBRIUM
IONIC EQUILIBRIUM 2 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III NISHITH Multimedia India (Pvt.) Ltd., B X X X B X X X * Acids that require only one electron pair to complete an outer shell are stronger than those requiring two. Thus GaCl3 > ZnCl2 . * Oxyacids * Oxyacids with same central atom at different oxidation state: Among oxyacids of the same central atom having general formula. (HO)m ZOn Where Z = central atom m = no of oxygen atoms attached to z and attached to H. n = No of oxygen atoms attached to z but not attached to H. Acidic strength increases with increasing oxidation no e.g.- a) 1 3 5 7 2 3 very weak weak strong very strong HO Cl HO ClO HO ClO HO ClO        They give OCl, OClO, OClO , OClO 2 3 anions after removal of a proton. The species which have more than one oxygen atom show resonance. We see that from left to right, acidic strength increases Due to : · More the no. of oxygen atoms, more the resonance. · Volume available for the negative charge is increasing · Charge density is decreasing · Basicity of the anion is decreasing · Acidity of the conjugate acid is increasing. b) 4 6 H SO H SO 2 3 2 4    c) 3 5 H NO H NO 2 3    * Oxyacids with different central atom of same oxidation state and same configuration. a) HOI < HOBr < HOCl b) HIO4 < HBrO4 < HClO4 c) HPO3 < HNO3 d) H3AsO4 < H3 PO4 . Among above; acidity increases from left to right Due to: · Decreasing size of the central atom · Increasing order of the electro negativity · Increasing order of pull of electron of O-X bond towards X. · Resulting into increasing positive character on oxygen. · Ultimately, increasing order of release of H+ . * The strength of oxyacids increases from left to right across a period e.g. - H2 SiO4 < H3 PO4 < H2 SO4 < HClO4 , Which is attributable to increase in electronegativity of the central atom, * Hydrated metal ions: Under favourable conditions, one or more protons may dissociate to form the coordinated aqua groups. n 2 6 2 (Hydrated metal ion) [M(H O) ] H O      n 1 M(H O) OH H O 2 5 3    For hydrated metal ion or aqua acids, acidity increases with the increase of positive charge of the central metal ion and with decreasing ionic radius. [Fe(H2O)6 ] 3+ > [Fe(H2O)6 ] 2+ . Exceptions are some times seen due to the effect of covalent bonding. 2 3 2 6 2 6 [Fe(OH ) ] [Fe(OH ) ]    3 2 2 6 2 6 [Al(OH ) ] [Hg(OH ) ]     * Oxyacids of Phosphorous H PO H PO H PO 3 2 3 3 3 4   Relative Strength of Inorganic Bases * In Period: The basicity of the compound decreases from left to right along a period e.g.: NH H O HF 3 2   Due to: a) Increase in the electronegativity of the atom holding the electron pair(s)
NISHITH Multimedia India (Pvt.) Ltd., 3 JEE MAINS - CW - VOL - I IONIC EQUILIBRIUM NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III b) Decrease in the availability of electron pair(s) for sharing with the proton. * In group: The basicity of a compound decreases from top to bottom in a group e.g. : a) F  > Cl > Br  > I b) O2  > S2  c) NH3 > PH3 > AsH3 > SbH3 > BiH3 . Due to: a) Increase in the size of atom holding the unshared electron pair(s). b) Decrease in the availability of electrons. * Presence of negative charge on the atom holding the electron pair(s) increases the basicity while presence of positive charge decreases basicity e.g. H O H — OH H O3    * Alkali and Alkaline Hydroxides: Basic nature increases on going down the group e.g. a) LiOH < NaOH < KOH < RbOH < CsOH b) Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 . pH OF THE MIXTURES : (A) pH for the mixture of Weak Acid and Strong Acid Let strong acid be HB whose conc. is C1 HB  H+ + B- 0 C1 C1 and weak acid whose concentration is C2 and degree of dissociation of  HA = H+ + A C2 (1- ) C2 C2 Total (H+ ) conc. = C1 + C2 pH = - log [C1 + C2 ] (B) pH calculation of solution of a mixture of two weak Monobasic Acids in water Let two weak acids be HA and HB and their conc. are C1 and C2 , 1 is the degree of dissociation of HA in presence of HB (due to common ion effect) and 2 be degree of dissociation of HB in presence of H A . In aqueous solution of HA and HB following equilibrium exists. HA + H2O(l)   H3O+ + A- C1 (1 -  1 ) C1  1 + C2 2 C1 1 HB + H2O(l)   H3O+ + B- C2 (1 -  2 ) (C1 1 + C2 2 ) C2 2 3 1 1 2 2 1 1 a[HA] 1 1 [H O ][A ] [C C ][C ] K [HA] C (1 )           3 1 1 2 2 2 2 a[HB] 2 2 [H O ][B ] [C C ][C ] K [HB] [C (1 )]           1 1 2 2 pH – log[H ] – log[C C ]       = - log a 1 a 2 1 2 K C K C  where a1 K = Ka(HA) and a2 K = Ka(HA) (C) pH of a dibasic Acids and Polyprotic Acid Let's take the e.g. of a dibasic acid H2A. Assuming both dissociation is weak. Let the initial conc. of H2A is C and  1 and  2 be degree of dissociation for first and second dissociation. H2A   HA- + H+ C(1 -  1 ) C 1 (1 -  2 ) C 1 + C 1  2 HA-   H+ + A2  C 1 (1 -  2 ) C 1 + C 1  2 C 1  2 . 1 2 [HA ][H ] Ka [H A]    1 2 1 1 2 1 1 [C (1 )][C C ] Ka C(1 )           1 1 2 1 2 2 1 2 [H ][A ] [C C ][C ] Ka [HA ] [C (1 )]                After solving for 1 and 2. We can calculate the H+ conc. [H+ ] = C 1 + C 1  2 pH = - log [C 1 + C 1 2 ]
IONIC EQUILIBRIUM 4 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - III NISHITH Multimedia India (Pvt.) Ltd., (D) pH of mixture of acids Let one litre of an acidic solution of pH = 2 be mixed with two litre of other acidic solution of pH = 3. The resultant pH of the mixture can be evaluated in the following way. Sample 1 Sample 2 pH = 2 pH = 3 [H+ ] = 10-2 M [H+ ] = 10-3 M V = 1 litre V = 2 litre M1V1 + M2V2 = MR (V1 + V2 ) 10-2  1 + 10-3  2 = MR (1 + 2) 3 R 12 10 M 3    3 R 4 10 M    (Here, MR = Resultant molarity) –3 R pH – log[4 10 ] 2.3980    pKa and pKb for a conjugate acid-base pair For an acid HX – HX H X      – a [H ][X ] K [HX]   ............... (A) For conjugate base – X of acid HX – – X H O HX OH   2   – b – [HX][OH ] K [X ]  ..................... (B) By eqs. (A) and (B), – K K [H ][OH ] K a b w     or a b w pK pK pK 14    Note : 1. Stronger is acid, weaker is its conjugate base. 2. Higher is the value of a pK of an acid, lower is acid strength and higher is basic strength of its conjugate base. * Hydrolysis of Salts Salts are strong electrolytes when dissolved in water, they dissociated almost completely into cation or anions. If anion interacts with water it is called as anionic hydrolysis. A + H2O  HA + OH- Akaline solution (pH increases). If cation intereacts with water it is called as cationic hydrolysis. B+ + 2H2O  B(OH) + H3O+ A c i d i c solution (pH lowers down). "The phenomenon of the interaction of anions and cations of the salt with H+ and OH- ions furnished by water yielding acidic or alkaline solution is known as salt hydrolysis. For the study of hydrolysis, salts are divided into four groups. * 1) Hydrolysis of salt of strong Acid and weak base: NH4Cl is a salt of weak base (NH4OH) and strong acid (HCl). After hydrolysis resultant solution will be acidic due to presence of strong acid HCl. NH Cl H O NH OH HCl 4 2 4     NH Cl H O NH OH H Cl 4 2 4           4 2 4 (acidic) NH H O NH OH H       4 h 4 [NH OH][H ] K [NH ]    * Relation between K ,Kh b and Kw : NH OH NH OH 4 4      4 b 4 [NH ][OH ] K [NH OH]    ... (A) H O H OH 2      K [H ][OH ] w    ... (B) Dividing (B) by (A) w 4 b 4 K [H ][OH ] [NH OH] K [NH ][OH ]       w h b K K K 