1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 10 InterviewSolution

451.	Which of the following acids has the smallest dissociation constant?A. `BrCH_(2)CH_(2)COOH`B. `CH_(3)CHBrCOOH`C. `FCH_(2)CH_(2)COOH`D. `CH_(3)CHFCOOH`
Answer» Correct Answer - A The dissociation constant of acid is directly related to its acidic strength which increase with the negative inductive effect (electron withdrawing power) of the halogen. The more electronegative the halogen, the greater the negative inductive effect. The closer the halogen to the COOH group, the greater the influence of the inductive effct of the halogen. Thus, the order of acidic strength is `CH_(3)CHFCOOHgtCH_(3)CHBrCOOHgtFCH_(2)CH_(2)COOHgtBrCH_(2)CH_(2)COOH` Thus, `BrCH_(2)CH_(2)COOH` has the smallest dissociation constant.

Discussion

452.	Match the following equilibria with the corresponding condition(i) Liquid ⇄Vapour(a) Saturated solution(ii) Solid ⇄ Liquid(b) Boiling point(iii) Solid ⇄Vapour(c) Sublimation point(iv) Solute (s) ⇄ Solute (solution)(d) Melting point(e) Unsaturated solution
Answer» (i) → (b) (ii) → (d) (iii) → (c) (iv) → (a) (i) Liquid ⇌Vapour equilibrium exists at the boiling point. (ii) Solid ⇌Liquid equilibrium exists at the melting point. (iii) Solid ⇌ Vapour equilibrium exists at the sublimation point. (iv) Solute(s) ⇌Solute (solution) equilibrium exists in saturated solution.

452.

Match the following equilibria with the corresponding condition(i) Liquid ⇄Vapour(a) Saturated solution(ii) Solid ⇄ Liquid(b) Boiling point(iii) Solid ⇄Vapour(c) Sublimation point(iv) Solute (s) ⇄ Solute (solution)(d) Melting point(e) Unsaturated solution

Answer»

(i) → (b) (ii) → (d) (iii) → (c) (iv) → (a)

(i) Liquid ⇌Vapour equilibrium exists at the boiling point.

(ii) Solid ⇌Liquid equilibrium exists at the melting point.

(iii) Solid ⇌ Vapour equilibrium exists at the sublimation point.

(iv) Solute(s) ⇌Solute (solution) equilibrium exists in saturated solution.

Discussion

453.	Calculate the pH and concentration of all species present at equilibrium in 0.1 M `H_(3)PO_(4)` solution. `K_(a_(1))=7.5xx10^(-3), K_(a_(2))=6.2xx10^(-8), K_(a_(3))=4.2xx10^(-13)`
Answer» Correct Answer - `[H_(2)PO_(4)^(-)]=[H_(3)O^(+)]=0.024M, [H_(3)PO_(4)]=0.076M, [HPO_(4)^(2-)]=6.2xx10^(-8), [PO_(4)^(3-)]=1.30 xx 10^(-18)`.

Discussion

454.	Match the following equilibria with the corresponding condition `{:((i),"Liquid"hArr"Vapour",,(a),"Saturated solution"),((ii),"Solid" hArr"Liquid",,(b),"Boiling point"),((iii),"Solid" hArr"Vapour",,(c),"Sublimation point"),((iv) ,"Solute(s)"hArr"Solute (solution)",,(d),"Melting point"),(,,,(e),"Unsaturated solution"):}`
Answer» Correct Answer - (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a) (i) Liquid `hArr` Vapor equilibrium exists at the boiling point. Hence, (i) - (b). (ii) Solid ` hArr` Liquid equilibrium exists at the melting point. Hence, (ii)-(d). (iii) Solid `hArr`Vapour equilibrium exists at the sublimation point. Hence, (iii) - (c). (iv) Solute (s) `hArr` Solute (solute) equilibrium exists in a saturated solution . Hence, (iv) - (a).

Discussion

455.	The ionization constant of `HF` is `3.2xx10^(-4)`. Calculate the degree of ionization of HF in its `0.02M` solution. Calculate the concentration of all species present in the solution and its `pH`.
Answer» Strategy: Identify the primary reaction and write the reaction summary in terms of unknwon x. Solve for x, knowing the value of `K_a`. Solution: Step 1-2: The following ionization reactions are possible: `HF(aq.)=H^(+)(aq.)+F^(-)(aq.)` `K_a=3.2xx10^(-4)` `H_2O(1)=H^(+)(aq.)+OH^(-)(aq.)` `K_w=1.0xx10^(-14)` Step 3: As `K_w lt lt K_a`, the first reaction is the primary reaction. Step 4: Let x be the equilibrium concentration of `H^(+)` and `F^(-)` ions in `mol L^(-1)`. Then the equilibrium concentration of `HF` must be `(0.02-x)mol L^(-1)`. We can write the reaction summary as follows: `{:(,HF(aq.)hArrH^(+)(aq.)+F^(-)(aq.)),("Initial (M)"," 0.02 0 0"),("Change (M)"," -x +x +x"),("Equilibrium (M)",bar("(0.02-x) x x ")):}` Step 5: `K_a=(C_(H^+)C_(F^-))/(C_(HF))` `3.2xx10^(-4)=((x)(x))/(0.02-x)=(x^2)/(0.02-x)` This equation can be written as `x^2+3.2xx10^(-4)x-6.4xx10^(-6)=0` which fits the quadratic equation `ax^2+bx+c=0`. Using the quadratic formula `x=(-b+-sqrt(b^2-4ac))/(2a)` we get `x=2.4xx10^(-3)M` or `-2.4xx10^-3M` The second solution is physically impossible since the concentration of ions produced as a result of ionization cannot be negative. Thus, `x=2.4xx10^(-3)M` Step 6: Thus, at equilibrium, `C_(H^+)=2.4xx10^(-3)M` `C_(F^-)=2.4xx10^(-3)M` `C_(HF)=(0.02-2.4xx10^(-3))M` `=1.76xx10^(-2)M` `alpha=(C_(H^+))/(C_(HF))=(2.4xx10^(-3))/(0.02)=0.12` `pH=-"log"(C_(H^+))/(mol L^(-1))=-log(2.4xx10^(-3))` `=2.62`

Discussion

456.	What is the function of adding `NH_(4)OH` in group V ?
Answer» It converts any `NH_(4)HCO_(3) ` present into `(NH_(4))_(2)CO_(3)`.

Discussion

457.	pH of 0.08 mol `dm^(-3)` HOCl solution is 2.85. Calculate its ionisation constant.
Answer» pH of Hocl = 2.85, i.e, - log `[H^(+)]= 2.85 ` or log `[H^(+)]=-2.85= bar(3).15` or `[H^(+)]`= antilog ` bar(3) .15 = 1.413 xx 10^(-3)M` For weak monobasic acid, `HA hArr H^(+) + A^(-)` `K_(a) = ([H^(+)]^(2))/([HA])=((1.413xx10^(-3))^(2))/(0.08) = 2.4957xx10^(-5)`.

Discussion

458.	How is the acidic , basic character of acids and bases varies with respect to neutral water . Show these variations on the PH scale.
Answer» The PH of neutral water is 7 ,As the PH of solution deceases from 7 the acid characteristic of the solution increase so the acids having values 1 and 2 are strong acids and when the PH of solution increases from 7 the basic characteristic of the solution increase so the acids having values 11and 12 are strong bases.

Discussion

459.	A weak acid HX has `K_(a) =1 xx 10^(-5)M`. If forms a salt NaX on reaction with alkali. The perentage degree of dissociation of 0.1 M solution of NaX isA. 1.0E-5B. 0.01C. 0.1D. 0.15
Answer» Correct Answer - B `underset(("Salt"))(X^(-)) +H_(2)O hArr HX+OH^(-)` For the salt `K_(h) =(K_(w))/(K_(a)) =Calpha^(2)` `:. (10^(-4))/(10^(-5)) =0.1 alpha^(2)` `" or "" "alpha^(2) =(10^(-14))/(10^(-6)) =10^(-8)` `alpha =(10^(-8))^(1//2) =10^(-4)` `% " degree of dissociation "underset(=10^(-2) =0.01%)(=10^(-4) xx 100)`

Discussion

460.	A solution is found to contain 0.63 g of nitric acid per 100 ml of solution. What is the PH of solution if acid is completely dissociated?
Answer» M=n/V =w₂/M₂ x V = 0.63x1000/63 mol^-1L^-1 =0.01 molL^-1 [H⁺] =[HNO₃] =10^-2 PH = - log(H⁺) = - log(10^-2 ) = -(-2)log10 PH =2

460.

A solution is found to contain 0.63 g of nitric acid per 100 ml of solution. What is the PH of solution if acid is completely dissociated?

Answer»

M=n/V

=w₂/M₂ x V

= 0.63x1000/63 mol^-1L^-1

=0.01 molL^-1

[H⁺] =[HNO₃] =10^-2

PH = - log(H⁺)

= - log(10^-2 )

= -(-2)log10

PH =2

Discussion

461.	What is the difference between dissociation and ionization?
Answer» The term ionization is used for the compounds which are covalent and when dissolve in the solution undergo dissociation where as for the ionic compounds which are in ionic form in the compound when dissolve into the solution undergo dissociation.

Discussion

462.	The vapour density of completely dissociated `NH_(4)Cl` isA. double than that of `NH_(4)Cl`B. half than that of `NH_(4)Cl`C. same as that of `NH_(4)Cl`D. determined by the amount of solid `NH_(4)Cl` taken
Answer» Correct Answer - B `NH(4)Cl to NH_(3) + HCl .` Thus , 1 mole of `NH_(4)Cl` on dissociation produces 2 moles of the vapour (1 mole `NH_(3)` and 1 mole HCl) which will have double the volume at the same temperature and pressure . As density `prop 1//volume` , therefore density will be halved .

Discussion

463.	Discuss the Application of Solubility Product and common ion Effect?
Answer» (1) Calculation of solubility of sparingly soluble salts. (2) Predicting ionic or precipitation reaction. (3) In explaining salting out or precipitation of soluble salts. (4) In qualitative analysis. (5) In fractional precipitation.

463.

Discuss the Application of Solubility Product and common ion Effect?

Answer»

(1) Calculation of solubility of sparingly soluble salts.

(2) Predicting ionic or precipitation reaction.

(3) In explaining salting out or precipitation of soluble salts.

(4) In qualitative analysis.

(5) In fractional precipitation.

Discussion

464.	Which of the following salts will give highest `pH` in water?A. `CuSO_(4)`B. `KCI`C. `NaCI`D. `Na_(2)CO_(3)`
Answer» Correct Answer - D `Na_(2)CO_(3) overset((aq))(to) underset("Strong base")(2NaOH) + underset("Weak acid")(H_(2)CO_(3))` Since the solution is highly basic the pH will be maximum.

Discussion

465.	Which has the highest `pH`?A. `CH_(3)COOK`B. `Na_(2)CO_(3)`C. `NH_(4)CI`D. `NaNO_(3)`
Answer» It is a case of salt hydrolysis. `NaNO_(3)` [salt of strong acid `(HNO_(3))` and strong base `(NaOH)`] does not undergo hydrolysis. Thus, its aqueous solution is neutral `(pH=7)`. `NH_(4)Cl` [salt of strong acidf `(HCl)` and weak base `(NH_(3)`] undergoes cationic hydrolysis. Thus, its aqueous solution is acidic `(pHlt7)`. Both `CH_(3)COOK` and `Na_(2)CO_(3)` are salts of strong base and weak acid. They undergo anion hydrolysis to provide basic solution `(pHgt7)`. However, `CO_(3)^(2-)` ion is a relatively stronger base than `CH_(3)COO^(-)` ion a (as `H_(2)CO_(3)` is a weaker acid than `CH_(3)COO`). Hence, `Na_(2)CO_(3)` solution is most basic and has the highest pH.

Discussion

466.	`K_(a)` for butyric acid is `2.0xx10^(-5)`. Calculate pH and hydroxyl ion concentration in `0.2M` aqueous solution of sodium butyate.
Answer» (I). Calculation of pH of the solution. sodium butyrate is a salt of strong base and weak acid. `pH =- 1/2 [log K_(w) +logK_(a) - logC]` According to available data , `K_(w) =1.0 xx 10^(-14), K_(a) = 2.0 xx10^(-5) , C= 0.2 M` `pH =- 1/2 [log 10^(-14) + log (2.0xx10^(-5))- log (0.2)]` `=- 1/2 [(-14)- (5 -log2) +0.698] =-1/2 [-14-4 .699+0.698] =9.00` (II). Calculation of hydroly ion concentration. `pH =9 , pOH = pK_(w) -pH =14 -9 =5 :. [OH^(-)] =10^(-5) M`

Discussion

467.	Calculate the degree of ionisation and `[H_(3)O^(+) ]` of 0.01 M acetic acid solution . `K_(a)` for acetic at 298 K is `1.8 xx 10^(-5)`
Answer» The ionisation of acetic acid is represented as : `{:(,CH_(3)COOH(aq) ,hArr, CH_(3)COO^(-) (aq) ,+, H^(+) (aq)),("Initial molar concentration",0.01,,0,,0),("Equilibrium molar concentration",0.01(1-alpha),,0.01alpha,,0.01alpha):}` Applying Law of chemical equilibrium, `K_(a) = [[CH_(3)COO^(-)(aq)][H^(+)(aq)]]/[[CH_(3)COOH(aq)]] , 1.8 xx 10^(-5) = ((0.01alpha)(0.01alpha))/(0.01(1-alpha))` Since `alpha` is very small for weak electrolyte `(CH_(3)COOH) ," therefore " (1-alpha) ~~ 1`. `:. " "1.8 xx 10^(-5) = ((0.01alpha)(0.01alpha))/(0.01) = 0.01alpha^(2)` `"or "" "alpha^(2) =(1.8xx 10^(-5))/(0.01) =1.8 xx 10^(-3) = 18 xx 10^(-3) = 18 xx 10^(-4)` ` " or "" "alpha sqrt(18xx10^(-4)) = 4.24 xx 10^(-2)` `"or "" "[H_(3)O^(+)]=[H^(+)]Calpha = 0.01 xx 4.24 xx 10^(-2) = 4.24 xx 10^(-4)M`

Discussion

468.	Calculate the concentration of hydroxyl ions in 0.2 M solution of ammonium hydroxide having `K_(b) =1.8 xx 10^(-5)`.
Answer» `NH_(4)OH(aq) hArr NH_(4)^(+) (aq) + OH^(-) (aq)` `K_(b) =[[NH_(4)^(+)(aq)][OH^(-) (aq)]]/[[NH_(4)OH(aq)]]` For the base, `[NH_(4)^(+) (aq)] = [OH^(-) (aq)]` `:. " "K_(b) =[[OH^(-) (aq)]^(2)]/[[NH_(4) OH (aq)]]` ` " or "" "[OH^(-) (aq)]^(2) = K_(b) xx [NH_(4)OH(aq) ] = 1.8 xx 10^(-5) xx 0.2 = 3.6 xx 10^(-6)` `:. " " [OH^(-) (aq)] = (3.6 xx 10^(-6))^(1//2) = 1.897 xx 10^(-3) M`

Discussion

469.	The principal buffer present in human blood isA. `NaH_(2)PO_(4)+Na_(2)HPO_(4)`B. `H_(3)PO_(4)+NaH_(2)PO_(4)`C. `CH_(3)COOH+CH_(3)COONa`D. `H_(2)CO_(3)+HCO_(3)^(-)`
Answer» Buffer present in human blood consists of weak acid `(H_(2)CO_(3))` and weak base `(HCO_(3)^(-))`.

Discussion

470.	A reaction between ammonia and boron tri fluoride is given by:Identify the acid and base in this reaction. Which theory explain it?
Answer» Ammonia donate the lone pair of electron so it is a Lewis base and boron trifluoride accept the lone pair of electron so it is a Lewis acid .This is Lewis acid base concept.

Discussion

471.	Describe the effect of the following (i) Addition OfH2 (ii) Addition of CH3OH (iii) Removal of CO in the following equilibrium reaction: 2H2 (g) +CO (g) ------> CH3OH (l)
Answer» (i) equilibrium will shift in forward direction (ii) Equilibrium will shift in backward direction. (iii) Equilibrium will shift in backward direction.

471.

Describe the effect of the following (i) Addition OfH2 (ii) Addition of CH3OH (iii) Removal of CO in the following equilibrium reaction: 2H2 (g) +CO (g) ------> CH3OH (l)

Answer»

(i) equilibrium will shift in forward direction

(ii) Equilibrium will shift in backward direction.

(iii) Equilibrium will shift in backward direction.

Discussion

472.	Hydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction.CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g)Write an expression for Kp for the above reaction.
Answer» K_p = \(\frac{p_{CO(g)}\times p^3_{H_2(g)}}{p_{CH_4(g)}\times pH_2(g)}\)

Discussion

473.	Hydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reactionCH4(g) + H2O(g) ⇌ CO(g) + 3H2(g)How will the value of Kp and composition of equilibrium mixture be affected by(i) increasing the pressure(ii) increasing the temperature(iii) using a catalyst
Answer» (i) The reaction proceeds with an increase in volume. Therefore, an increase in pressure will shift the equilibrium in the backward direction. Therefore, the reaction mixture will now contain lesser amounts of CO and H₂ and larger amounts of CH₄ and H₂O(g). As a result, K_p value for the reaction will decrease. (ii) The reaction is an endothermic reaction. So, an increase in temperature will shift the equilibrium in the forward direction. Therefore, the reaction mixture will now contain larger amounts of the products, (CO and H₂), and lesser amounts of the reactants (CH₄ and H₂O(g)). The value of K_p will increase. (iii) A catalyst has no effect on the position of the equilibrium. Therefore, the composition of the reaction mixture and the value of K_p will remain unchanged.

473.

Hydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reactionCH4(g) + H2O(g) ⇌ CO(g) + 3H2(g)How will the value of Kp and composition of equilibrium mixture be affected by(i) increasing the pressure(ii) increasing the temperature(iii) using a catalyst

Answer»

(i) The reaction proceeds with an increase in volume. Therefore, an increase in pressure will shift the equilibrium in the backward direction. Therefore, the reaction mixture will now contain lesser amounts of CO and H₂ and larger amounts of CH₄ and H₂O(g).

As a result, K_p value for the reaction will decrease.

(ii) The reaction is an endothermic reaction. So, an increase in temperature will shift the equilibrium in the forward direction. Therefore, the reaction mixture will now contain larger amounts of the products, (CO and H₂), and lesser amounts of the reactants (CH₄ and H₂O(g)). The value of K_p will increase.

(iii) A catalyst has no effect on the position of the equilibrium. Therefore, the composition of the reaction mixture and the value of K_p will remain unchanged.

Discussion

474.	An aqueous solution of ferric chloride gives a brown precipitate upon standing . Justify.
Answer» An aqueous solution of ferric chloride gets hydrolysed on standing to form ferric hydroxide and hydrochloric acid. The brown precipitate is due to the formation of ferric hydroxide. `FeCI_(3) +3H_(2)O to Fe(OH)_(3) +3HCI` (Brown ppt.)

Discussion

475.	following equilibrium is studied by taking 1 mole of `N_(2)` and 3 moles of `H_(2)` in a 1L flask at a given temperature? `N_(2)(g) +3H_(2) (g) hArr 2NH_(3)(g)` `NH_(3)(g)` formed at equilibrium is neutralised by 200 mL of 1M HCl. Calculate equilibrium constant.
Answer» `200 mL of 1M HCl = (1)/(1000) XX 200 = 0.2 M` `NH_(3) ` and HCl react in equimolar Proporation to form `NH_(4)Cl` `:. NH_(3)` evolved n the reaction = 0.2 M `N_(2)(g) + 3H_(2) (g) hArr 2NH_(3)(g)` No. of moles of `NH_(3)(g)` formed =0.2M No. of moles of `N_(2)(g) " reacted " =(0.2)/(2) = 0.1M` No. of moles of `H_(2)(g) " reacted " =0.2xx(3)/(2) =0.3M` The molar concentration of various species at the equilibrium point is : `{:(,N_(2)(g),+,3YH_(2)(g),hArr,2NH_(3)(g)),("Initial moles/litre",1,,3,,0),("Moles/litre at eqm. point",1-0.1,,3-0.3,,0.2),(,=0.9,,2.7,,):}` Applying Law of chemical equilibrium : `K_(c) = [[NH_(3)(g)]^(2))/[[N_(2)(g)][H_(2)(g)]^(3))= (0.2)^(2)/(0.9xx(2.7)^(3))=(0.04)/(17.71)=2..26xx 10^(-3)`

Discussion

476.	The reaction , `CO(g) + 3 H_(2) (g) hArr CH_(4) (g) + H_(2)O (g),` is at equilibrium at 1300 K in a 1 L flask. It also contains 030 mol of CO, 010 mol of `H_(2) and 002 " mol of " H_(2)O` and an unknown amount of `CH_(4) " in the flask. Determine the concentration of " CH_(4)` in the mixture. The equilibrium constant, `K_(c)`, for the reaction at the given temperature is` 390`.
Answer» ` K_(c) = ([ CH_(4)][H_(2)O])/([CO][H_(2)]^(3))` ` :. 3* 90 = ([CH_(4)][H_(2)O])/([CO][H_(2)]^(3)) " " ("Molar conc = No. of moles because volume of flask = 1 L")` `[CH_(4)] = 00585 M = 585 xx 10^(-2) M`

Discussion

477.	The reaction, `CO(g)+3H_(2)(g) hArr CH_(4)(g)+H_(2)O(g)` is at equilibrium at `1300 K` in a `1 L` flask. It also contains `0.30 mol` of `CO, 0.10 mol` of `H_(2)` and `0.02` mol of `H_(2)O` and an unknown amount of `CH_(4)` in the flask. Determine the concentration of `CH_(4)` in the mixture. The equilibrium constant `K_(c )` for the reaction at the given temperature us `3.90`.
Answer» Correct Answer - `5.85 xx 10^(-3)` Let the concentration of methane at equilibrium be x. `CO_(g) + 3H_(2(g)) harr CH_(4(g)) + H_(2)O_((g))` At equalibrium `(0.3)/(1)= 0.3M (0.1)/(1) = 0.1M , x , (0.02)/(1) = 0.02M` It is given that `K_(c) = 3.90` Therefore, `([CH_(4(g))][H_(2)O_((g))])/([CO_((g))][H_(2(g))]) K_(c)` `rArr (x xx 0.02)/(0.3 xx (0.1)^(3)) = 3.90` `rArr x = (3.90 xx 0.3 xx (0.1)^(3))/(0.02)` `= (0.00117)/(0.02)` `= 0.0585 M` `= 5.85 xx 10^(-2) M` Hence, the concentration of `CH_(4)` at equilibrium is `5.85 xx 10^(-2) M`.

Discussion

478.	A mixtue of `H_(2) and I_(2) ` (vapour) in molecular proportion of 2: 3 was heated at ` 449^(@)C` till the reaction `H_(2) + I_(2) hArr 2 HI` reached equilibrium state . Calculate the percentage of iodine converted into `HI (K_(c) "at" 440^(@)C " is " 0*02).`
Answer» ` {:(,H_(2)(g),+,I_(2)(g),hArr,2HI(g)),(" Intial",2,,3,,0 "moles"),("Molar concs.",(2-x)/V,,(3-x)/V,,(2x)/V),(,,,,,):}` ` K_(c) = (2x//V)^(2)/([(2-x)//V][3-x)//V]=(4x^(2))/((2-x)(3-x))=002 " "` (Given) This on solving gives `x = 01615` ` :. % " of iodine converted into HI" = (01615)/3 xx 100 = 5 38%`

Discussion

479.	Calculate pH of 0.1 M of H2SO4 (concentration 0.1 × 2 hydrogen = 0.2).
Answer» pH = -log₁₀[H⁺] = -log₁₀[0.1 × 2] = -log₁₀[0.2] = -log₁₀[2 × 10^-1] = 1 – log2 pH = 1 – 0.3010 = 0.699

479.

Calculate pH of 0.1 M of H2SO4 (concentration 0.1 × 2 hydrogen = 0.2).

Answer»

pH = -log₁₀[H⁺]

= -log₁₀[0.1 × 2]

= -log₁₀[0.2]

= -log₁₀[2 × 10^-1]

= 1 – log2 pH

= 1 – 0.3010

= 0.699

Discussion

480.	Calculate pH of 0.00025 M HNO3.
Answer» pH = -log₁₀[H⁺] = -Iog₁₀[0.00025] – log₁₀[2.5 × 10^-4] = 4 – log2.5 pH = 4 – 0.3979 = 3.6021

480.

Calculate pH of 0.00025 M HNO3.

Answer»

pH = -log₁₀[H⁺] = -Iog₁₀[0.00025] – log₁₀[2.5 × 10^-4] = 4 – log2.5

pH = 4 – 0.3979 = 3.6021

Discussion

481.	In the following questions, two statements are given; one in assertion (A) column while the other in reason (R ) column. Examine the statements carefully and mark the correct answer according to the instructions given below: If both (A) and (R) are correct and (R) is the correct explanation of (A) If both (A) and (R) are correct but (R) is not the correct explanation of (A)If (A) is correct but (R) is wrongIf (A) is wrong but (R) is correct(A) PH of a neutral solution is always 7(R) The PH of a solution depends upon temperature
Answer» If both (A) and (R) are correct and (R) is the correct explanation of (A)

Discussion

482.	The pH of 0.1 M solution of an organic acid is 4.0. Calculate its dissociation constant.
Answer» Correct Answer - `1xx 10^(-7)` `pH =4 " or " - log [H_(3)O^(+)] =4` `" or " " "log[(1)/(H_(3)O)] =4 " or " [H_(3) O^(+)] =10^(-4) m` Let the acid HA ionise as : `HA overset((aq))(hArr) H_(3)O^(+) + A^(-) (aq)` `K_(a) =[[H_(3)O^(+)][A^(-)(aq)]]/[[HA]] =[[H_(3)O^(+)]^(2)]/[[HA]]` `" Now "" "[H_(3)O^(+)] =10^(-4) M, HA =0.1 M` `:." " K_(a) =((10^(-4))/((0.1)) =1xx 10^(-7)`

Discussion

483.	the solubility of `A_(2)B_(3)` is y mol `dm^(-3)`. Its solubility product isA. `6y^(4)`B. `64y^(4)`C. `36Y^(5)`D. `108y^(5)`
Answer» Correct Answer - D `A_(2)B_(3) hArr 2A^(+3) +3B^(-2)` `K_(sp) =[A^(+3)]^(2)[B^(-2)]^(3)=(2y)^(2)(3y)^(3)` `K_(sp) =108 y^(5)`

Discussion

484.	At a certain temperature, the solubility of the salt `A_(x)B_(y)` is S moles per litre. The general expression for the solubility product will beA. `K_(sp)=X^(y)Y^(x)S^(x+y)`B. `K_(sp)=(XY)^(x+y)S^(x+y)`C. `K_(sp)=(X^(x)Y^(y))S^(x+y)`D. `K_(sp)=X^(y)Y^(x)S^(xy)`
Answer» Correct Answer - C `A_(x)B_(y)(s)hArr xA^(y+)(aq.)+yB^(x-)(aq.)` `K_(ap)=C_(A^(y+))^(x)C_(B^(x-))^(y)` `= (xS)^(x)(yS)^(y)` `= (x^(x)S^(x))(y^(y)S^(y))` `x^(x)y^(y)S^(x+y)`

Discussion

485.	Assertion : Higher order ionization constants `(K_(a_(2)),K_(a_(3)))` are smaller than the lower order ionization constant `(K_(a_(1)))` of polyprotic acid. Reason : Polyprotic acid solutions contain a mixture of acids.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - B The reason for samaller values of high order ionixation constants is that it is more difficult to remove a positively charged proton from a negative ion due to electrostatic forces.

Discussion

486.	Conjugate acid of `SO_(4)^(2-)` isA. `HSO_(4)^(-)`B. `H^(+)`C. `H_(2)SO_(4)`D. `SO_(4)^(2-)`
Answer» Correct Answer - A Conjugate acid of `SO_(4)^(2-)` is `HSO_(4)^(-)`

Discussion

487.	Which of the following species can act both as an acid as well as a base ?A. `SO_(4)^(2-)`B. `HSO_(4)^(-)`C. `PO_(4)^(3-)`D. `OH^(-)`
Answer» Correct Answer - B `underset(SO_(4)^(2-))underset(" "darr-H^(+))(HSO_(4)^(-))overset(+H^(+))(to)H_(2)SO_(4)` `HSO_(4)^(-)` can accept or give a proton.

Discussion

488.	For a reaction, `A_(x),B_(y)hArrxA^(y+)+yB^(x-),K_(sp)` xan be represented asA. `[A^(y+)]^(x)[B^(x-)]^(y)`B. `[A]^(y)[B]^(x)`C. `[A]^(x)[B]^(y)`D. `[A]^(x+y)[B]^(x-y)`
Answer» Correct Answer - A `A_(x)B_(y)hArrxA^(y+)+yB^(x-)` `K_(sp)=[A^(y+)]^(x)[B^(x-)]^(y)`

Discussion

489.	Assertion : The strength of haloacids increases in the order : `HIltltBHrltltHClltltHF` Reason : Strength of acid HA depends only on the electronegatively difference between H and A.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - D The correct order of acidic strength is `HFltltHClltltHBrltltHI`. While comparing elements in the same group of the periodic table, H-A bond strength is a more important factor in determining acidity than its polar nature. As the size of A increases down the group, H-A bond strength decreases and so the acid strength increases.

Discussion

490.	Consider the following gaseous equilibrium with equilibrium constants `K_(1) and K_(2)` respectively ` SO_(2) (g) + 1/2 O_(2) (g) hArr SO_(3) (g)` `2 SO_(3) (g) hArr 2 SO_(2)(g) + O_(2)` The equilibrium constants are related asA. ` K_(1)^(2) = 1/K_(2)`B. ` 2k_(1) = K_(2)^(2)`C. ` K_(2) = 2/K_(1)^(2)`D. ` K_(2)^(2) = 1/ K_(1)`
Answer» Correct Answer - A `K_(1) = [SO_(3)]/([SO_(2)] [ O_(2)]^(1//2)) K_(2)= ([SO_(2)]^(2)[O_(2)])/([SO_(3)]^(2))` Obviously `K_(1)^(2)= 1/K_(2)`.

Discussion

491.	Hydrolysis of sucrose gives `"Sucrose" +H_(2)OhArr"Glucose + Fructose"` Equilibrium constant `K_(c)` for the reaction is `2xx10^(13)` at `300 K`. Calculate `DeltaG^(ɵ)` at `300 K`.
Answer» `DeltaG^(ɵ) = -RT "ln"K_(c)` `DeltaG^(ɵ) = - 8.314J "mol"^(-1)K^(-1) xx 300 K xx "ln"(2 xx 10^(13))` `DeltaG^(ɵ) = -7.64 xx 10^(4) J "mol"^(-1)`

Discussion

492.	At 1127 K and 1 atm pressure , a gaseous mixture of `CO and CO_(2)` in equilibrium with solid carbon has 90*55 % CO by mass in the reaction, `C(s) + CO_(2) (g) hArr 2 CO (g)`. Calculate `k_(c) ` for the reaction at the above temperature.
Answer» If total mass of the mixture of `CO and CO_(2)` is 100 g , then ` :. "At eqm., " [BrCl] = ( 330 xx 10^(-3) - 30 xx 10^(-3))= 030 xx 10^(-3) = 30 xx 10^(-4) "mol"L^(-1)` Number of moles of ` :. "At eqm., " [BrCl] = ( 330 xx 10^(-3) - 30 xx 10^(-3))= 030 xx 10^(-3) = 30 xx 10^(-4) "mol"L^(-1)` ` :. p_(CO) = (3234)/(3234 + 0215 ) xx 1" atm " = 0938 " atm" , p_(CO_(2)) = (0215)/(3234 + 0215 ) xx 1 "atm " = 0062 "atm "` ` k_(p) = (p_(CO)^(2))/p_(CO_(2)) = (0938 )^(2)/(0062 )= 14 19` . ` Delta n_(g) = 2-1 =1 :. K_(p) = K_(c) (RT) or K_(c) = (K_(p))/(RT) = (1419)/(00821 xx 1127)=0153 ` .

Discussion

493.	At what temperature the solid, and liquid are in equilibrium under 1 atm. pressure?
Answer» Melting point or freezing point.

Discussion

494.	50 mL of 0.01 M solution of `Ca(NO_(3))_(2)` is added to 150 mL of 0.08 M solution of `(NH_(4))_(2)SO_(4)`. Predict Whether `CaSO_(4)` will be precipitated or not. `K_(sp) " of " CaSO_(4) =4 xx 10^(-5)`
Answer» Correct Answer - Yes , ionic product `=1.5 xx 10^(-3)` the precipitate of `CaSO_(4)` is to be formed . The solubility equilibrium may be represented as : `CaSO_(4) (s) hArr Ca^(2+) (aq) +SO_(4)^(2-) (aq)` Now `Ca^(2+)` ions are to be provided by `Ca(NO_(3))_(2)` solution while `SO_(4)^(2-)` ions by `(NH_(4))_(2)SO_(4)` solution as a result of dissociation. `Ca(NO_(3))_(2) overset((aq))(to) Ca^(2+)(aq) + 2NO_(3)^(-)(aq) , (NH_(4))_(2)SO_(4) (s) overset((aq))(to) 2NH_(4)^(+) (aq) + SO_(4)^(2-) (aq)` The total volume of the solution after mixing =(50 +150) = 200 mL In this solution , the concentration of `Ca^(2+)` ions after mixing will be `1//4 (50//200)` while that `SO_(4)^(2-)` ions will be `3//4(150//200)` Thus `[Ca^(2+)]` before mixing =0.01 M` `[Ca^(2+)]` after mixing `=0.01 xx 1/4 =0.025 M` Similarly , `[SO_(4)^(2-)]` before mixing =0.08 M `[SO_(4)^(2-)]` after mixing `=(0.08 xx 150)/(200) =0.06 M` The ionic product `=[Ca^(2)][SO_(4^(2-)]=(0.025)xx (0.06) =1.5 xx 10^(-3)` `K_(sp)` value of `CaSO_(4) =4 xx 10^(-5)` (Given) Since the ionic product is more than the solubility product this means that `CaSO_(4)` will get precipitaed.

Discussion

495.	What is Hydrolysis? Give an examples.
Answer» Process of interaction between water and cations/anions or both of salts is called hydrolysis. The cations (e.g., Na⁺ , K⁺ ,Ca²⁺, Ba²⁺, etc.) of strong bases and anions(e.g., Cl^– , Br^– , NO^3– , ClO^4–etc.) of strong acids simply get hydrated but do not hydrolyse, and therefore the solutions of salts formed from strong acids and bases are neutral i.e., their pH is 7

495.

What is Hydrolysis? Give an examples.

Answer»

Process of interaction between water and cations/anions or both of salts is called hydrolysis.

The cations (e.g., Na⁺ , K⁺ ,Ca²⁺, Ba²⁺, etc.) of strong bases and anions(e.g., Cl^– , Br^– , NO^3– , ClO^4–etc.) of strong acids simply get hydrated but do not hydrolyse, and therefore the solutions of salts formed from strong acids and bases are neutral i.e., their pH is 7

Discussion

496.	Which of the following will not affect the value of equilibrium constant of a reaction?A. Change in concentration of reactantsB. Change in temperatureC. Change in pressureD. Addition of catalyst.
Answer» Correct Answer - A::C::D (a,c,d) will not affect the equilibrium constant for the reaction

Discussion

497.	What is ionic products of water?
Answer» Kw = [H⁺] [OH^-]

Discussion

498.	What do you mean by reversible process?
Answer» Reversible process is a process in which products can react to form reactants back.

Discussion

499.	Define equilibrium.
Answer» Equilibrium is a stage at which rate of forward reaction is equal to rate of backward reaction.

Discussion

500.	What is physical equilibrium? Give one example.
Answer» Physical equilibrium is am equilibrium between two different physical states of same substance, e.g., H₂O(s) ⇌ H₂O(l)

500.

What is physical equilibrium? Give one example.

Answer»

Physical equilibrium is am equilibrium between two different physical states of same substance, e.g.,

H₂O(s) ⇌ H₂O(l)

Discussion

Explore topic-wise InterviewSolutions in .

Which of the following acids has the smallest dissociation constant?A. `BrCH_(2)CH_(2)COOH`B. `CH_(3)CHBrCOOH`C. `FCH_(2)CH_(2)COOH`D. `CH_(3)CHFCOOH`

Match the following equilibria with the corresponding condition(i) Liquid ⇄Vapour(a) Saturated solution(ii) Solid ⇄ Liquid(b) Boiling point(iii) Solid ⇄Vapour(c) Sublimation point(iv) Solute (s) ⇄ Solute (solution)(d) Melting point(e) Unsaturated solution

Calculate the pH and concentration of all species present at equilibrium in 0.1 M `H_(3)PO_(4)` solution. `K_(a_(1))=7.5xx10^(-3), K_(a_(2))=6.2xx10^(-8), K_(a_(3))=4.2xx10^(-13)`

The ionization constant of `HF` is `3.2xx10^(-4)`. Calculate the degree of ionization of HF in its `0.02M` solution. Calculate the concentration of all species present in the solution and its `pH`.

What is the function of adding `NH_(4)OH` in group V ?

pH of 0.08 mol `dm^(-3)` HOCl solution is 2.85. Calculate its ionisation constant.

How is the acidic , basic character of acids and bases varies with respect to neutral water . Show these variations on the PH scale.

A weak acid HX has `K_(a) =1 xx 10^(-5)M`. If forms a salt NaX on reaction with alkali. The perentage degree of dissociation of 0.1 M solution of NaX isA. 1.0E-5B. 0.01C. 0.1D. 0.15

A solution is found to contain 0.63 g of nitric acid per 100 ml of solution. What is the PH of solution if acid is completely dissociated?

What is the difference between dissociation and ionization?

The vapour density of completely dissociated `NH_(4)Cl` isA. double than that of `NH_(4)Cl`B. half than that of `NH_(4)Cl`C. same as that of `NH_(4)Cl`D. determined by the amount of solid `NH_(4)Cl` taken

Discuss the Application of Solubility Product and common ion Effect?

Which of the following salts will give highest `pH` in water?A. `CuSO_(4)`B. `KCI`C. `NaCI`D. `Na_(2)CO_(3)`

Which has the highest `pH`?A. `CH_(3)COOK`B. `Na_(2)CO_(3)`C. `NH_(4)CI`D. `NaNO_(3)`

`K_(a)` for butyric acid is `2.0xx10^(-5)`. Calculate pH and hydroxyl ion concentration in `0.2M` aqueous solution of sodium butyate.

Calculate the degree of ionisation and `[H_(3)O^(+) ]` of 0.01 M acetic acid solution . `K_(a)` for acetic at 298 K is `1.8 xx 10^(-5)`

Calculate the concentration of hydroxyl ions in 0.2 M solution of ammonium hydroxide having `K_(b) =1.8 xx 10^(-5)`.

The principal buffer present in human blood isA. `NaH_(2)PO_(4)+Na_(2)HPO_(4)`B. `H_(3)PO_(4)+NaH_(2)PO_(4)`C. `CH_(3)COOH+CH_(3)COONa`D. `H_(2)CO_(3)+HCO_(3)^(-)`

A reaction between ammonia and boron tri fluoride is given by:Identify the acid and base in this reaction. Which theory explain it?

Describe the effect of the following (i) Addition OfH2 (ii) Addition of CH3OH (iii) Removal of CO in the following equilibrium reaction: 2H2 (g) +CO (g) ------&gt; CH3OH (l)

Hydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction.CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g)Write an expression for Kp for the above reaction.

An aqueous solution of ferric chloride gives a brown precipitate upon standing . Justify.

following equilibrium is studied by taking 1 mole of `N_(2)` and 3 moles of `H_(2)` in a 1L flask at a given temperature? `N_(2)(g) +3H_(2) (g) hArr 2NH_(3)(g)` `NH_(3)(g)` formed at equilibrium is neutralised by 200 mL of 1M HCl. Calculate equilibrium constant.

A mixtue of `H_(2) and I_(2) ` (vapour) in molecular proportion of 2: 3 was heated at ` 449^(@)C` till the reaction `H_(2) + I_(2) hArr 2 HI` reached equilibrium state . Calculate the percentage of iodine converted into `HI (K_(c) "at" 440^(@)C " is " 0*02).`

Calculate pH of 0.1 M of H2SO4 (concentration 0.1 × 2 hydrogen = 0.2).

Calculate pH of 0.00025 M HNO3.

The pH of 0.1 M solution of an organic acid is 4.0. Calculate its dissociation constant.

the solubility of `A_(2)B_(3)` is y mol `dm^(-3)`. Its solubility product isA. `6y^(4)`B. `64y^(4)`C. `36Y^(5)`D. `108y^(5)`

At a certain temperature, the solubility of the salt `A_(x)B_(y)` is S moles per litre. The general expression for the solubility product will beA. `K_(sp)=X^(y)Y^(x)S^(x+y)`B. `K_(sp)=(XY)^(x+y)S^(x+y)`C. `K_(sp)=(X^(x)Y^(y))S^(x+y)`D. `K_(sp)=X^(y)Y^(x)S^(xy)`

Conjugate acid of `SO_(4)^(2-)` isA. `HSO_(4)^(-)`B. `H^(+)`C. `H_(2)SO_(4)`D. `SO_(4)^(2-)`

Which of the following species can act both as an acid as well as a base ?A. `SO_(4)^(2-)`B. `HSO_(4)^(-)`C. `PO_(4)^(3-)`D. `OH^(-)`

For a reaction, `A_(x),B_(y)hArrxA^(y+)+yB^(x-),K_(sp)` xan be represented asA. `[A^(y+)]^(x)[B^(x-)]^(y)`B. `[A]^(y)[B]^(x)`C. `[A]^(x)[B]^(y)`D. `[A]^(x+y)[B]^(x-y)`

Hydrolysis of sucrose gives `"Sucrose" +H_(2)OhArr"Glucose + Fructose"` Equilibrium constant `K_(c)` for the reaction is `2xx10^(13)` at `300 K`. Calculate `DeltaG^(ɵ)` at `300 K`.

At 1127 K and 1 atm pressure , a gaseous mixture of `CO and CO_(2)` in equilibrium with solid carbon has 90*55 % CO by mass in the reaction, `C(s) + CO_(2) (g) hArr 2 CO (g)`. Calculate `k_(c) ` for the reaction at the above temperature.

At what temperature the solid, and liquid are in equilibrium under 1 atm. pressure?

50 mL of 0.01 M solution of `Ca(NO_(3))_(2)` is added to 150 mL of 0.08 M solution of `(NH_(4))_(2)SO_(4)`. Predict Whether `CaSO_(4)` will be precipitated or not. `K_(sp) " of " CaSO_(4) =4 xx 10^(-5)`

What is Hydrolysis? Give an examples.

Which of the following will not affect the value of equilibrium constant of a reaction?A. Change in concentration of reactantsB. Change in temperatureC. Change in pressureD. Addition of catalyst.

What is ionic products of water?

What do you mean by reversible process?

Define equilibrium.

What is physical equilibrium? Give one example.

Describe the effect of the following (i) Addition OfH2 (ii) Addition of CH3OH (iii) Removal of CO in the following equilibrium reaction: 2H2 (g) +CO (g) ------> CH3OH (l)