1219 + Interview Questions in EQUILIBRIUM IN GENERAL AWARENESS Page 11 InterviewSolution

501.	Give two examples for Acedic buffer
Answer» A solution containing a weak acid and its salt with a strong base. (i) CH₃COOH + CH₃COONa (ii) HCOOH + HCOOK

501.

Give two examples for Acedic buffer

Answer»

A solution containing a weak acid and its salt with a strong base.

(i) CH₃COOH + CH₃COONa

(ii) HCOOH + HCOOK

Discussion

502.	Define the term Equilibrium state.
Answer» When rate of formation of a product in a process is in competition with rate of formation of reactants, the state is then named as "Equilibrium state". Equilibrium in physical processes: solid ⇌ liquid ⇌ gas H₂O(s )⇌ H₂O(l)⇌ H₂O(vap)

502.

Define the term Equilibrium state.

Answer»

When rate of formation of a product in a process is in competition with rate of formation of reactants, the state is then named as "Equilibrium state".

Equilibrium in physical processes:

solid ⇌ liquid ⇌ gas

H₂O(s )⇌ H₂O(l)⇌ H₂O(vap)

Discussion

503.	The pH of a solution is `1.30`. The number of signuficant figure isA. threeB. oneC. zeroD. two
Answer» Correct Answer - D In a pH value, only the digits after the decimal point are significant figures. The "I" in `1.30` is not a significant figurw as it comes form the power of ten.

Discussion

504.	Calculate (a) `DeltaG^(Θ)` and (b) the equilibrium constant for the formation of `NO` and `O_(2)` at `298 K` `NO(g)+1//2 O_(2)(g) hArr NO_(2)(g)` where `Delta_(f)G^(Θ)(NO_(2))=52.0 kJ mol^(-1)` `Delta_(f)G^(Θ)(NO)=87.0 kJ mol^(-1)` `Delta_(f)G^(Θ)(O_(2))=0 kJ mol^(-1)`
Answer» Correct Answer - a) `-35.0 kJ`, b) `1.365 xx 10^(6)` (a) For the given reaction, `DeltaG^(@) = DeltaG^(@)("Products")-DeltaG^(@)("Reactants")` `DeltaG^(@) = 52.0 - {87.0 + 0}` `= - 35.0 kJ "mol"^(-1)` (b) We know that `DeltaG^(@) = RT "log" K_(c)` `DeltaG^(@) = 2.303 RT "log" K_(c)` `K_(c) = (-35.0 xx 10^(-3))/(-2.303 xx 8.314 xx 298)` `= 6.134` `:. K_(c) = "antilog" (6.134)` `= 1.36 xx 10^(6)` Hence, the equilibrium constant for the given reaction `K_(c)` is `1.36 xx 10^(6)`

Discussion

505.	Does the number of moles of reaction products increase, decrease, or remain same when each of the following equilibrium is subjected to a decrease in pressure by increasing the volume? a. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` b. `CaO(s)+CO_(2)(g) hArr CaCO_(3)(s)` c. `3Fe(s)+4H_(2)O(g) hArr Fe_(3)O_(4)(s)+4H_(2)(g)`
Answer» (a) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase. (b) The number of moles of reaction products will decrease. (c) The number of moles of reaction products remains the same.

Discussion

506.	The solubility of Mg`(OH)_(2)` is `8.352xx10^(-3)` g/litre at `290^(@)` C. Find out its `K_(sp)` at this temperature.
Answer» `Mg(OH)_(2)` ionizes completely in the solution as : `Mg(OH)_(2)rarrMg^(2+)+2OH^(-)` `:. [Mg^(2+)]=[Mg(OH)_(2)] and [OH^(-)] = 2xx [ Mg (OH)_(2)]` But Molar mass of Mg`(OH)_(2) = 58 g "mol"^(-1)` `:. [Mg(OH)_(2)]=("Strength in g / litre")/("Molar mass") = (8.352xx10^(-3))/(58) = 1.44 xx 10^(-4)` moles/litre `:. [Mg^(2+)]=1.44xx10^(-4)` moles/litre and `[OH^(-)] = 2xx1.44xx10^(-4)=2.88xx10^(-4)` moles/litre `:. K_(sp) ` for `Mg(OH)_(2)=[Mg^(2+)][OH^(-)]^(2)=(1.44xx10^(-4))xx(2.88xx10^(-4))^(2)=1.194xx10^(-11)`

Discussion

507.	The solubility of AgCl in water at `25^(@)` C is found to be `1.06xx10^(-5)` moles per litre. Calculate the solubility product of AgCl at this temperature.
Answer» AgCl ionizes completely in the solution as : `Agcl rarr Ag^(+) + Cl^(-)` i.e., 1 mole of AgCl in the solution gives 1 moles 1 mole of `Ag^(+)` ions and 1 mole of `Cl^(-)` ions. Now, as the solubility of `AgCl = 1.6 xx 10^(-5)` moles per litre `:. [Ag^(+)] = 1.06xx10^(-5)` moles/litre and `[Cl^(-)] = 1.06 xx 10^(-5)` moles/litre `:. K_(sp) ` for ` AgCl = [Ag^(+)][Cl^(-)]=1.06xx10^(-5)xx1.06xx10^(_5)=1.1xx10^(-10)`

Discussion

508.	The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the `pK_(a)` of bromoacetic acid.
Answer» Correct Answer - `pH = 1.88, pK_(a) = 2.70` Degree of ionization, `alpha` = 0.132 Concentration, `c = 0.1 M` Thus, the concentration of `H_(3)O+ = c.alpha` `= 0.1 xx 0.132` `= 0.0132` `pH = -"log"[H^(+)] ` `= -log (0.0132)` `= 1.879 : 1.88` Now, `K_(a) = Calpha^(2)` `= 0.1 xx (0.132)^(2)` `K_(a) = .0017` `pK_(a) = 2.75`

Discussion

509.	The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the `pK_(a)` bromoacetic acid.
Answer» `{:(,CH_(2)(Br)CO OH ,hArr,CH_(2)(Br)CO O^(-) ,+,H^(+),,,),("Initial conc.",C,,0,,0,,,),("Conc. at eqm.",C-C alpha ,,C alpha ,,C alpha ,,,),(,,,,,,,,):}` `K_(a) = (C alpha . C alpha)/(C ( 1-alpha))=(C alpha^(2))/(1-alpha)~=C alpha^(2) = 0.1 xx (0.132)^(2) = 1.74xx10^(-3)` `pK_(a) = - log (1.74xx10^(-3))=3-0.2405=2.76` `[H^(+)]=C alpha = 0.1 xx 0.32 = 1.32 xx 10^(-2) M` `pH = - log (1.32xx10^(-2))=2-0.1206=1.88`

Discussion

510.	Which of the following salts will have the highest pH in water ?A. KClB. NaClC. `Na_(2) CO_(3)`D. `CuSO_(4)`
Answer» Correct Answer - C KCl and NaCl are salts of strong acid and strong base. Their aqueous solutions will be neutral with pH = 7. `Na_(2)CO_(3)` is a salt of strong base (NaOH) and weak acid `(H_(2)CO_(3))`. Its solution will be basic with `pH gt 7. CuSO_(4)` is a salt of weak base, `Cu(OH)_(2)` and strong acid, `H_(2)SO_(4)`. Its solution will be acidic with `pH lt 7`.

Discussion

511.	It has been found that the `pH` of a `0.01 M` solution of an organic acid is `4.15`. Calculate the concentration of the anion, the ionization constant of the acid and its `pK_(a)`.
Answer» `pH =- log [H^(+)] " or " log [H^(+)] =- pH =- 4.15 = bar(5).85` `[H^(+)] = " Antilog " (bar(5).85) =7.08 xx 10^(-5)M` `[A^(-)] = [H^(+)] =7.08 xx 10^(-5)M` `[K_(a)] =[[H^(+)][A^(-)]]/[[HA]] = ((7.08 xx 10^(-5))xx (7.08 xx 10^(-5)))/((0.01)) =5.0 xx 10^(-7)` `pK_(a) =- log K_(a)=- log (5.0 xx 10^(-7))= (7-log 5) =(7-0.699) =6.301`

Discussion

512.	According to Lewis concept acid isA. Proton donorB. Electron pair donorC. Electron pair acceptorD. Proton accetor
Answer» Correct Answer - C Conceptual question.

Discussion

513.	What would be the pH of a solution obtained by mixing 10 g of acetic acid and 15 g of sodium acetate and making the volume equal to 1L. Dissociation constant of acetic acid at `25^(@)`C is `1.75xx10^(-5)`.
Answer» `pH=pK_(a) + log. (["Salt"])/(["Acid"])=pK_(a) + log. ([CH_(3)CO ON a])/([CH_(3)CO OH])` `[CH_(3)CO OH]=(10)/(60) "mol L"^(-1)` (molar mass of `CH_(3)CO OH = 60 g "mol"^(-1)`) `[CH_(3)CO Ona]=(15)/(82) "mol" L^(-1)` (molar mass of `CH_(3)CO Ona = 82 g "mol" ^(-1)`) `pK_(a)=-log K_(a)=-log(1.75xx10^(-5))=5-0.2430 = 4.757` `:. pH = 4.757 + log (15//82)/(10//60) = 4.757+ log. ((15)/(82)xx(60)/(10))` `=4.757 + log 1.098 = 4.757 + 0.0406 = 4.80`

Discussion

514.	What is the minimum volume of water required to dissolve `1.0 g` of calcium sulphate at `298 K`? (For calcium sulphate , `K_(sp) is 9.1xx10^(-6))`.
Answer» Correct Answer - `2.43` litre of water `CaSO_(4(s)) harr Ca_((aq))^(2+) + SO_(4(aq))^(2-)` `K_(sp) = [Ca^(2+)][SO_(4)^(2-)]` Let the solubility of `CaSO_(4)` be s. Then, `K_(sp) = s^(2)` `9.1 xx 10^(-6) = s^(2)` `s = 3.02 xx 10^(-3) "mol"//L` Molecular mass of `CaSO_(4) = 136 g//"mol"` Solubility of `CaSO_(4)` in gram/L `= 3.2 xx 10^(-3) xx 136` `= 0.41 g//L` This means that we need 1L of water to dissolve 0.41g of `CaSO_(4)` Therefore, to dissolve 1g of `CaSO_(4)` we require `= (1)/(0.41) L = 2.44 L` of water.

Discussion

515.	Which of the following is a weak electrolyte ?A. `HF`B. `HCI`C. `HBr`D. `HI`
Answer» Correct Answer - A HF is a weak acid whereas HCI, HBr, and HI are strong acids.

Discussion

516.	In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains `Ag^(+) and Pb^(2+)` at a concentration of 0.10 M. Aqueous HCl is added to this solution until the `Cl^(-)` concentrations of `Ag^(+) and Pb^(2+)` at equilibrium ? `(K_(sp) "for" AgCl=1.8xx10^(-10), K_(sp) "for" PbCl_(2)=1.7xx10^(-5))`A. `[Ag^(+)]=1.8xx10^(-7) M , [Pb^(2+)]=1.7xx10^(-6)M`B. `[Ag^(+)]=1.8xx10^(-11) M , [Pb^(2+)]=8.5xx10^(-5)M`C. `[Ag^(+)]=1.8xx10^(-9) M , [Pb^(2+)]=1.7xx10^(-3)M`D. `[Ag^(+)]=1.8xx10^(-11) M , [Pb^(2+)]=8.5xx10^(-4)M`
Answer» Correct Answer - C `K_(sp) ` for `AgCl = [Ag^(+)][Cl^(-)]` `:. 1.8xx10^(-10)=[Ag^(+)][0.1]` or`[Ag^(+)]=1.8xx10^(-9)M` `K_(sp)` for `PbCl_(2) = [Pb^(2+)][Cl^(-)]^(2)` `:.1.7xx10^(-5)=[Pb^(2+)][0.1]^(2)` or `[Pb^(2+)]=1.7xx10^(-3)M`.

Discussion

517.	The equilibrium constant for the reaction `N_(2)(g)+O_(2)(g) hArr 2NO(g)` is `4.0xx10^(-4)` at `2000 K`. In the presence of a catalyst, the equilibrium is attained `10` times faster. Therefore, the equilibrium constant in presence of the catalyst at `2000 K` isA. `40xx10^(-4)`B. `4xx10^(-4)`C. `4xx10^(-2)`D. The data is insufficient
Answer» Correct Answer - B

Discussion

518.	For a gaseous reaction `xA+yB hArr lC +mD`A. `K_(p)=K_(c)`B. `K_(p)=K_(c)`C. `K_(p)=K_(c)(RT)^((I+m)-(x+y))`D. `K_(p)=(1)/(K_(c))`
Answer» Correct Answer - C

Discussion

519.	Which of the following statements are correct ?A. According to Bronsted Lowry concept, `H_(2)SO_(4)` can also act as a baseB. `SiF_(4)` is an acid according to Lewis conceptC. Stronger the acid, higher is its `pK_(a)` valueD. HCl, `HNO_(3) and H_(2)SO_(4)` act as equally strong acids in any solvent.
Answer» Correct Answer - A::B (a) is correct because in presence of `HClO_(4)`,`H_(2)SO_(4)` can accept a proton and hence can act as a base. (b) Si has empty d-orbitals to accept electrons. Hence, Si`F_(4)` is a Lewis acid.

Discussion

520.	Which of the following order of acidic strengths is incorrect ?A. `H_(3)PO_(4) lt HNO_(3)`B. `H_(2)SeO_(3) lt H_(2)SO_(3)`C. `H_(3)PO_(3) lt HNO_(2)`D. `H_(2)SO_(3) lt H_(2)SO_(4)`
Answer» Correct Answer - C Contrary to what we might expect, `H_(3)PO_(3)` is a stronger acid than `HNO_(2)`.

Discussion

521.	Which of the following statements are wrong?A. pH of neutral water is always 7.0B. When a base is titrated against an acid, the pH at the end point is 7.0C. Lesser is the pH than 7, more acidic is the solution and higher the pH than 7, less basic is the solutionD. AgCl is more soluble in `NH_(3) ` than in water.
Answer» Correct Answer - A::B::C (a) is wrong because PH of neutral water depends upon temperature (b) is wrong because at the end point, pH depends upon the nature of acid and base being titrated (c) is wrong because higher the pH than 7, more basic is the solution (d) is correct because AgCl dissolves in `NH_(3) ` to form a soluble complex.

Discussion

522.	A buffer solution can be prepared from a mixture ofA. sodium acetate and acetic acid in waterB. sodium acetate and HCI in waterC. ammonia and ammonia chloride in waterD. ammonia and sodium hydroxide in water.
Answer» Correct Answer - A::C (a,c) buffer will result in both the cases.

Discussion

523.	Assertion : For the reaction : `N_(2(g))+3H_(2(g))hArr2NH_(3(g)),K_(p)=K_(c)` Reason : Concentration of gaseous reactants and products is taken as unity.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 `K_(p)=K_(c)(RT)^(Deltan)` `N_(2(g))+3H_(2(g))hArr2NH_(3(g))` `Deltan=2-4=-2` `:." "K_(p)=K_(c)(RT)^(-2)` Concentration of solids and liquids is taken as unity.

Discussion

524.	Which of the following is the strongest acid ?A. `H_(3)PO_(4)`B. `H_(3)PO_(3)`C. `H_(3)PO_(2)`D. All are equally strong
Answer» Correct Answer - C Care must be excercised compare acids that have similar structures. `H_(3)PO_(2)`, which has two H atoms bonded to the P atom, is a stronger acid than `H_(3)PO_(3)` is a stronger acid than `H_(3)PO_(4)`, which has no H atoms bonded to the P atom. `H-overset(O)overset(\|\|)underset(H)underset(\|)(P)-O-HgtH-overset(O)overset(\|\|)underset(H)underset(\|)underset(O)underset(\|)(P)-O-HgtH-O-overset(O)overset(\|\|)underset(H)underset(\|)underset(O)underset(\|)(P)-O-H`

Discussion

525.	The pair(s) of ions where BOTH the ions are precipitated upon passing `H_(2)S` gas in presence of dilute HCl is (are)A. `Ba^(2+), Zn^(2+)`B. `Bi^(3+) , Fe^(3+)`C. `Cu^(2+), Pb^(2+)`D. `Hg^(2+), Bi^(3+)`
Answer» Correct Answer - C::D `Cu^(2+), Pb^(2+), Hg^(2+) and Bi^(3+)` all lie in Group II of qualitative analysis and are precipitated from the solution on passing `H_(2)S` gas in the solution acidified with dilute HCl.

Discussion

526.	A buffer solution can be prepared from a mixture ofA. sodium acetate and acetic acid in waterB. sodium acetate and hydrochloric acid in waterC. ammonia and ammonium chloride in waterD. ammonia and sodium hydroxide in water
Answer» Correct Answer - A::C A buffer solution is a mixture of weak acid/base and its salt with conjugate base/acid (strong base /acid)

Discussion

527.	Which of the following will have nearly equal pH ?A. 100 ml 0.1 M HCl mixed with 50 ml waterB. 50 ml `0.1 M H_(2)SO_(4)` mixed with 50 ml waterC. 50 ml of 0.1 M `H_(2)SO_(4) ` mixed with 100 ml waterD. 50 ml of 0.1 M HCl mixed with 50 ml of water
Answer» Correct Answer - A::C (a) 100 ml of 0.1 M HCl = 150 ml `xx` gt M HCl or `[HCl]=(1)/(15) M :. [H^(+)]=(1)/(15)M` (b) 50 ml of 0.1 M `H_(2)SO_(4) = 100 ml "of ? " MH_(2)SO_(4)` or `[H_(2)SO_(4)]=(1)/(20)M :. [H^(+)]=(1)/(10)M` (c) 50 ml of 0.1 `M H_(2)SO_(4) = 150 ml "of ?" M H_(2)SO_(4)` or `[H_(2)SO_(4)]=(1)/(30)M :. [H^(+)]=(1)/(15)M` (d) 50 ml of 0.1 M HCl = 100 ml of ? M HCl or `[HCl]=(1)/(20)M :. [H^(+)]=(1)/(20)M` (a) and (c) have same `[H^(+)]` and hence same pH.

Discussion

528.	When `CH_(3)COONa` is added to an aqueous solution of `CH_(3)COOH`A. pH value becomes zeroB. pH value remains unchangedC. pH value decreasesD. pH value increases
Answer» Correct Answer - D `CH_(3)COOH(aq.)hArr H^(+)(aq.)+CH_(3)COO^(-)(aq.)` The addition of sodium acetate which provides acetate ions shift the acetic acid ionization equilibrium towards left. This results in decrease in the concenreation of hydrogen ions `(C_(H^(+)))`. So the pH value of solution increass. This phenomenon is an example of common ion effect which is defined as a shift in equilibrium on adding a substence that provides more of an ionic species already present in the dissociation equilibrium.

Discussion

529.	The pH of a solution containing `0.20M CH_(3)COOH` and `0.30M CH_(3)COONa` isA. `2.89`B. `4.92`C. `5.04`D. `3.89`
Answer» Correct Answer - B Since we are dealing with a mixture of a weak acid and its salt with a strong base, we can calculate the pH of the solution by using the Henderson-Hasselbalch equation: `pH=pK_(a)+"log"(C_("salt"))/(C_("acid"))` In this case, we need to calculate `pK_(a)` and the acid first: `pK_(a)= -logK_(a)` `= -log(1.8xx10^(-5))` `=4.74` Now, we substitute the value of `pK_(a)` and concentrations of the acid and its salt: `pH=pK_(a)+"log"(C_("salt"))/(C_("acid"))` `=4.74+"log"(0.30)/(0.20)` `=4.74+log(1.5)` `=4.92`

Discussion

530.	A solution has been prepared by dissolving 0.63 g of nitric acid in 100 mL. What is its pH value ? Assume that the acid is completely dissociated.
Answer» Molar concentration of the solution `=("Amount of"HNO_(3)//"litre of solution")/("Molar mass of"HNO_(3)) = (0.63)/(100) xx (1000)/(63)` `=0.1 M` the complete dissociation of acid in solution is : `{:(HNO_(3)(l) ,+,H_(2)O(l) ,overset(aq)(to), H_(3)O^(+) ,+, NO_(3)^(-) (aq)),(,,,,0.1M,,0.1M):}` `pH=-log [H_(3)O^(+)] =- log[10^(-1)] =(-) (-log 10) =1`

Discussion

531.	Caculate the pH value of 0.20M solution of methyl amine `(CH_(3)NH_(2))` at 298 K, given that its ionisation constant `(K_(b))` is `4.4 xx 10^(-5).`
Answer» Let `alpha` be the degree of ionisation of methylamine. The concentration of the various species before the reaction and at the equilibrium point are: `{:(,CH_(3)NH_(2) +H_(2)O(l) ,hArr, CH_(3)NH_(3)^(+) (aq) ,+,OH^(-) (aq)),("Initial molar con".,0.2,,0,,0),("Equilibrium molar conc".,(0.2-alpha),,alpha,,alpha):}` Applying Law fo chemical equilibrium, `K_(b) =[[CH_(3)NH_(3)^(+)][OH^(-)]]/[[CH_(3)NH_(2)]] , 4.4 xx 10^(-5) = (alphaxx alpha)/((0.2 -alpha))` As methylamine is a weak base, `alpha` is very small. It can be negtected. Therefore ,0.2.This, `4.4 xx 10^(-5) (alpha^(2))/(0.2) or alpha = (0.2 xx 4.4 xx10^(-5))^(1//2) = 2.97 xx 10^(-3)` `[OH^(-)] = 2.97 xx 10^(-3) M` `[H_(3)O^(+)]=(K_(2))/[[OH^(-)]]=((1.0xx10^(-14)M^(2)))/((2.97xx10^(-3)M))=3.37 xx 10^(-12) M` `pH =- log [H_(3)O^(+)] =- log (3.37 xx 10^(-12))` `=12- log 3.37 =12 -0.53 =11.47`

Discussion

532.	The reaction quotient `(Q)` for thereaction `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` is given by `Q = ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))` The reaction will proceed from right to left if where `K_(C)` is the equilibrium constant.A. `Q=K_(C)`B. `Q lt K_(C)`C. `Q gt K_(C)`D. `Q = 0`
Answer» Correct Answer - C The reaction will proceed from right to left provided `[NH_(3)]^(2)gt[N_(2)][H_(2)]^(3)`. This implies that `QgtK_(c)`. This process will continnue until `Q = K_(C)`.

Discussion

533.	If the equilibrium constant for the given reaction is 0.25 `NOhArr(1)/(2)N_(2)+(1)/(2)O_(2)`, then the equilibrium constant for the reaction `(1)/(2)N_(2)+(1)/(2)O_(2)hArrNO` will beA. 1B. 2C. 3D. 4
Answer» Correct Answer - D `NOhArr(1)/(2)N_(2)+(1)/(2)O_(2),K_(c)=0.25` `(1)/(2)N_(2)+(1)/(2)O_(2)hArrNO,K_(c)=(1)/(0.25)=4`

Discussion

534.	When sulphur is heated at 900 K, `S_(8)` is converted to `S_(2)`. What will be the equilibrim constant for the reaction if initial pressure of 1 atm falls by `25%` at equilibrium ?A. `0.75 atm^(3)`B. `2.55atm^(3)`C. `25.0 atm^(3)`D. `1.33 atm^(3)`
Answer» Correct Answer - D `{:(,S_(8(g)),hArr,4S_(2(g)),),("Intial pressure",1-(25)/(100),,4xx(25)/(100),),(,=0.75,,=1,):}` `K_(p)=((P_(S_(2)))^(4))/(P_(S_(8)))=((1)^(4))/(0.75)=1.33" atm"^(3)`

Discussion

535.	1 mole of NO 1 mole of `O_(3)` are taken in a 10 L vessel and heated. At equilibrium, `50%` of NO (by mass) reacts with `O_(3)` according to the equation : `NO_((g))+O_(3(g))hArrNO_(2(g))+O_(2(g))`. What will be the equilibrium constant for this reaction ?A. 1B. 2C. 3D. 4
Answer» Correct Answer - A `{:(,NO_((g)),+,O_(3(g)),hArr,NO_(2(g)),+,O_(2(g))),("Intial conc.",1,,1,,0,,0),("At equilibrium",0.5,,0.5,,0.5,,0.5),("Conc.",(0.5)/(10),,(0.5)/(10),,(0.5)/(10),,(0.5)/(10)):}` `K_(c)=(0.05xx0.05)/(0.05xx0.05)=1`

Discussion

536.	At 473 K,`K_(c)` for the reaction `PCl_(5(g))rArrPCl_(3(g))Cl_(2(g))` is `8.3xx10^(-3)`. What will be the value of `K_(c)` for the formation of `PCl_(5)` at the same temperature ?A. `8.3xx10^(3)`B. `120.48`C. `8.3xx10^(-3)`D. 240.8
Answer» Correct Answer - B `PCl_(5(g))hArrPCl_(3(g))+Cl_(2(g)),K_(c)=8.3xx10^(-3)` `"For "PCl_(3(g))+Cl_(2(g))hArrPCl_(5(g)),K_(c)=(1)/(8.3xx10^(-3))` `K_(c)=(1)/(8.3)xx10^(3)=120.48`

Discussion

537.	The dimethyl ammonium ion, (CH3)2 NH2+, is a weak acid and ionizes to a slight degree in water what is its conjugate base?
Answer» (CH₃)₂ NH. .

Discussion

538.	`pK_(a)` of weak acid (HA) and `pK_(b)` of weak base (BOH) are 3.2 and 3.4 respectively. The pH of their salt (AB) solution isA. `7.0`B. `1.0`C. `7.2`D. `6.9`
Answer» Correct Answer - D `pH = 7 + (1)/(2) (pK_(a)-pK_(b))=7+(1)/(2)(3.2-3.4)` `=7-0.1=6.9`

Discussion

539.	`pK_(a)` of a weak acid is 5.76 and `pK_(b)` of a weak base is 5.25. What will be the pH of the salt formed by the two ?A. 7.255B. 7.005C. 10.225D. 4.255
Answer» Correct Answer - A `pH=(1)/(2)[pK_(w)+pK_(a)-pK_(b)]` `pH=7+(1)/(2)(5.76-5.25)=7.255`

Discussion

540.	Which of the following conditions will make the buffer most efficient?A. `pH=pK_(a)`B. `pH=pH_(a)+-1`C. `pH=pK_(a)+1`D. `pH=pH_(a)-1`
Answer» Correct Answer - A Since a buffer solution must maintaina relatively constant pH when either acid or base is added, it is usually desirable for the buffer to have roughly comparable the best case, when `[HA] = [A^(-)],1og[A^(-)]//[HA]=0`, and `pH=pK_(a)`.

Discussion

541.	The solubility of a salt of weak acid (AB) at pH 3 is `Y xx 10^(-3)` mol `L^(-1)`. The value of Y is ............ (Given that the value of solubility product of aB `(K_(sp))=2xx10^(-10)` and the value of ionization constant of HB `(K_(a))=1xx10^(-8)`)
Answer» Correct Answer - 4.47 `AB (s) hArr A^(+) (aq) + B^(-) (aq) ` ...(i) `B^(-)(aq)+ H^(+) (aq) hArrHB (aq) ` ...(ii) Adding eqns (i) and (ii), we get `{:(,AB(s),+,H^(+)(aq),hArr,A^(+)(aq),+,HB(aq)),("Initial conc",x,,10^(-3),,0,,0),("Final conc.",x-S,,10^(-3),,S,,S),(,,,,,,,):}` (S = solubility when pH = 3) From eqn. (i), `K_(sp)=[A^(+)][B^(-)]` From eqn. (ii), `K_(a)=([HB])/([B^(-)][H^(+)])` `:. (K_(sp))/(K_(a))=([A^(+)][HB])/([H^(+)])` `(2xx10^(-10))/(10^(-8))=(SxxS)/(10^(-3))` or `S^(2) = 2 xx10^(-5) = 20 xx 10^(-6) or S = 4.47 xx 10^(-3)` Hence , y = 4.47.

Discussion

542.	Consider the reaction equilibrium ` underset (("Greater volume "))"Ice"hArrunderset(("Lesser volume"))"Water"-2"kcal"` The favourable conditions for forward reaction areA. low temperature , high pressure and excess of iceB. low temperature ,low pressure and excess of iceC. high temperature , low pressure and excessD. high temperature , high pressure and excess of ice
Answer» Correct Answer - D Forward reaction is endothermic . Hence , forward reaction will be favoured by high temperature . Forward reaction is ccompanied by decrease in volume. Hence, forward reaction will be favoured by high prassure. Increases of concentration of reactant increases forward reaction . Hence, excess of ice will favour forward reaction .

Discussion

543.	Which of the following is arranged in the order of increseing ionization constants of `H_(3)PO_(4)` ?A. `K_(3) lt K_(1) lt K_(2)`B. `K_(1) lt K_(2) lt K_(3)`C. `K_(2) lt K_(1) lt K_(3)`D. `K_(3) lt K_(2) lt K_(1)`
Answer» Correct Answer - D Higher order ionization constants `(K_(2),K_(3))` are smaller than the lower order ionization constant `(K_(1))` of a polyprotic acid. It is more difficult to remove a positively charged proton form a negative ion `(H_(2)PO_(4)^(-),HPO_(4)^(2-))` due to electrostatic attraction.

Discussion

544.	Oxoacids are ___________ acids.A. binaryB. ternaryC. quatenaryD. secondary
Answer» Correct Answer - B Most ternary (three-element) acids are oxoacids. Such acids contain hydrogen, oxygen, and one other element usually a nonmetal which occupies a central position. Some typical examples are `underset("Carbonic acid")(H-O-overset(O)overset(\|\|)C-O-H)underset("Nitrous acid")(H-ON=O)underset("Nitric acid")(H-O-overset(O)overset(\|\|)N-O)` Thsu, oxacids are hydroxyl compounds of nometals that ionize to produce `H^(+)(aq.)`. As we can see from the above examples, oxiacids are characterized by the presence of one or more `O-H` bonds.

Discussion

545.	A patient is said to suffer from acidosis when the pH of his bloodA. falls below 7.35B. rises above 7.35C. shows sudden fall and riseD. has strong basic character
Answer» Correct Answer - A The fall in the pH of blood is called acidosis.

Discussion

546.	Which of the following pH curve represent the titration of weak acid and strong base (dotted line show equivalence point)?A. B. C. D.
Answer» Correct Answer - B

Discussion

547.	`10^(-6) ` M NaOH is diluted 100 times. The pH of the diluted base isA. between 5 and 6B. between 6 and 7C. between 10 and 11D. between 7 and 8
Answer» Correct Answer - D `[OH^(-)]` after dilution `= (10^(-6))/(100)=10^(-8)M` `:. OH^(-)` ions from `H_(2)O` cannot be neglected. Total `[OH^(-)]=10^(-8)+10^(-7)=10^(-8)xx11` `:. pOH=-log(11xx10^(-8))=8-1.04=6.96` `:. pH = 14 - 6.96 = 7.04`

Discussion

548.	0.023 g of sodium metal is reacted with `100 cm^(3)` of water. The pH of the resulting solution isA. 10B. 11C. 9D. 12
Answer» Correct Answer - D `Na+H_(2)OrarrNaOH=(1)/(2)H_(2)` 1 mole of Na produces 1 mole of NaOH `0.023 g Na = (0.023)/(23) "mole" = 10^(-3) "mole"` `:. ` NaOH produced `= 10^(-3)` mole `:. [NaOH]=(10^(-3))/(100)xx1000=10^(-2) M` `:. pOH=2 ` or pH `= 14-2=12`.

Discussion

549.	The solubility product of `AI(OH)_(3)` is `2.7 xx 10^(-11)`. Calculate its solubility in `mgL^(-1)`. (Atomic mass of `AI = 27 u)`.
Answer» Let the solubility of `Al(OH)_(3)` in water =S. `Al(OH)_(3)overset((aq))hArrAl^(3+)(aq)+ +3OH^(-)(aq)` Concentration of species at time (t) `= 0" " 1" "0" "0` Concentration of species at equilibrium point. `(1-S)" " S" "3S"` `K_(sp)=[Al^(3+)][OH^(-)]^(3)=[S][3S]^(3)=27S^(4)` `S^(4)=(K_(sp))/(27)=(2.7xx10^(-11))/(27)=1xx10^(-12)` `S^(4)=(1xx10^(-12))1//4=1xx10^(-3)mol L^(-1)` Solubility of `Al(OH)_(3) " in gram "//" litres " =(1xx10^(-3)mol L^(-1))xx[Molar Mass of Al (OH)_(3)]` `=(1xx10^(-3) mol L^(-1))xx(78g mol^(-1))` `78xx10^(-3)g L^(-1)=7.8xx10^(-2)gL^(-1)` Concentration of `(OH^(-)" in solution "=3S=3xx(1xx10^(-3)" mol " L^(-1))` `P_(OH)=-log [OH^(-)]` `=-log [3xx10^(-3)]=[3-log3]` `=(3-0.4771)=2.5229` `ph = 14 -P_(OH)= 14-2.5229=11.4771`

Discussion

550.	The `pH` of `0.1 M` solution of cyanic acid `(HCNO)` is `2.34`. Calculate the ionization constant of the acid and its degree of ionisation in the solution.
Answer» (i). Calculation of degree of dissociation `(alpha)` in soluiton `pH =2. 34 " or " - log [H^(+)] =2.34 " or " log [H^(+)] =- 2.34` `[H^(+)] = " Antilog " (-2.34) = " Antilog " (bar(3) .66) = 4.571 xx 10^(-3) M` `" Now"" " alpha [[H^(+)]]/(C) = ((4.571 xx 10^(-3)M))/((0.1M)) =4.571 xx 10^(-2) = 0.04571 M` `= 4.571 xx 10^(-2) =0.04571 M` (II). Calculation of ionisation constant of acid `[H^(+)] =(K_(a) xx C) ^(1//2) " or " [H^(+)]^(2) =K_(a) xx C` `K_(a) [[H^(+)]^(2)]/(C) =((.571 xx10^(-3) )^(2))/(0.1)= (20.89 xx 10^(-6))/(0.1) =20 .89 xx 10^(-5)` `K_(a) =2.089 xx 10^(-4) ~~ 2.09 xx 10^(-4)`

Discussion

Explore topic-wise InterviewSolutions in .

Give two examples for Acedic buffer

Define the term Equilibrium state.

The pH of a solution is `1.30`. The number of signuficant figure isA. threeB. oneC. zeroD. two

Calculate (a) `DeltaG^(Θ)` and (b) the equilibrium constant for the formation of `NO` and `O_(2)` at `298 K` `NO(g)+1//2 O_(2)(g) hArr NO_(2)(g)` where `Delta_(f)G^(Θ)(NO_(2))=52.0 kJ mol^(-1)` `Delta_(f)G^(Θ)(NO)=87.0 kJ mol^(-1)` `Delta_(f)G^(Θ)(O_(2))=0 kJ mol^(-1)`

The solubility of Mg`(OH)_(2)` is `8.352xx10^(-3)` g/litre at `290^(@)` C. Find out its `K_(sp)` at this temperature.

The solubility of AgCl in water at `25^(@)` C is found to be `1.06xx10^(-5)` moles per litre. Calculate the solubility product of AgCl at this temperature.

The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the `pK_(a)` of bromoacetic acid.

The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the `pK_(a)` bromoacetic acid.

Which of the following salts will have the highest pH in water ?A. KClB. NaClC. `Na_(2) CO_(3)`D. `CuSO_(4)`

It has been found that the `pH` of a `0.01 M` solution of an organic acid is `4.15`. Calculate the concentration of the anion, the ionization constant of the acid and its `pK_(a)`.

According to Lewis concept acid isA. Proton donorB. Electron pair donorC. Electron pair acceptorD. Proton accetor

What would be the pH of a solution obtained by mixing 10 g of acetic acid and 15 g of sodium acetate and making the volume equal to 1L. Dissociation constant of acetic acid at `25^(@)`C is `1.75xx10^(-5)`.

What is the minimum volume of water required to dissolve `1.0 g` of calcium sulphate at `298 K`? (For calcium sulphate , `K_(sp) is 9.1xx10^(-6))`.

Which of the following is a weak electrolyte ?A. `HF`B. `HCI`C. `HBr`D. `HI`

For a gaseous reaction `xA+yB hArr lC +mD`A. `K_(p)=K_(c)`B. `K_(p)=K_(c)`C. `K_(p)=K_(c)(RT)^((I+m)-(x+y))`D. `K_(p)=(1)/(K_(c))`

Which of the following order of acidic strengths is incorrect ?A. `H_(3)PO_(4) lt HNO_(3)`B. `H_(2)SeO_(3) lt H_(2)SO_(3)`C. `H_(3)PO_(3) lt HNO_(2)`D. `H_(2)SO_(3) lt H_(2)SO_(4)`

A buffer solution can be prepared from a mixture ofA. sodium acetate and acetic acid in waterB. sodium acetate and HCI in waterC. ammonia and ammonia chloride in waterD. ammonia and sodium hydroxide in water.

Which of the following is the strongest acid ?A. `H_(3)PO_(4)`B. `H_(3)PO_(3)`C. `H_(3)PO_(2)`D. All are equally strong

The pair(s) of ions where BOTH the ions are precipitated upon passing `H_(2)S` gas in presence of dilute HCl is (are)A. `Ba^(2+), Zn^(2+)`B. `Bi^(3+) , Fe^(3+)`C. `Cu^(2+), Pb^(2+)`D. `Hg^(2+), Bi^(3+)`

A buffer solution can be prepared from a mixture ofA. sodium acetate and acetic acid in waterB. sodium acetate and hydrochloric acid in waterC. ammonia and ammonium chloride in waterD. ammonia and sodium hydroxide in water

Which of the following will have nearly equal pH ?A. 100 ml 0.1 M HCl mixed with 50 ml waterB. 50 ml `0.1 M H_(2)SO_(4)` mixed with 50 ml waterC. 50 ml of 0.1 M `H_(2)SO_(4) ` mixed with 100 ml waterD. 50 ml of 0.1 M HCl mixed with 50 ml of water

When `CH_(3)COONa` is added to an aqueous solution of `CH_(3)COOH`A. pH value becomes zeroB. pH value remains unchangedC. pH value decreasesD. pH value increases

The pH of a solution containing `0.20M CH_(3)COOH` and `0.30M CH_(3)COONa` isA. `2.89`B. `4.92`C. `5.04`D. `3.89`

A solution has been prepared by dissolving 0.63 g of nitric acid in 100 mL. What is its pH value ? Assume that the acid is completely dissociated.

Caculate the pH value of 0.20M solution of methyl amine `(CH_(3)NH_(2))` at 298 K, given that its ionisation constant `(K_(b))` is `4.4 xx 10^(-5).`

The reaction quotient `(Q)` for thereaction `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` is given by `Q = ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))` The reaction will proceed from right to left if where `K_(C)` is the equilibrium constant.A. `Q=K_(C)`B. `Q lt K_(C)`C. `Q gt K_(C)`D. `Q = 0`

If the equilibrium constant for the given reaction is 0.25 `NOhArr(1)/(2)N_(2)+(1)/(2)O_(2)`, then the equilibrium constant for the reaction `(1)/(2)N_(2)+(1)/(2)O_(2)hArrNO` will beA. 1B. 2C. 3D. 4

When sulphur is heated at 900 K, `S_(8)` is converted to `S_(2)`. What will be the equilibrim constant for the reaction if initial pressure of 1 atm falls by `25%` at equilibrium ?A. `0.75 atm^(3)`B. `2.55atm^(3)`C. `25.0 atm^(3)`D. `1.33 atm^(3)`

1 mole of NO 1 mole of `O_(3)` are taken in a 10 L vessel and heated. At equilibrium, `50%` of NO (by mass) reacts with `O_(3)` according to the equation : `NO_((g))+O_(3(g))hArrNO_(2(g))+O_(2(g))`. What will be the equilibrium constant for this reaction ?A. 1B. 2C. 3D. 4

At 473 K,`K_(c)` for the reaction `PCl_(5(g))rArrPCl_(3(g))Cl_(2(g))` is `8.3xx10^(-3)`. What will be the value of `K_(c)` for the formation of `PCl_(5)` at the same temperature ?A. `8.3xx10^(3)`B. `120.48`C. `8.3xx10^(-3)`D. 240.8

The dimethyl ammonium ion, (CH3)2 NH2+, is a weak acid and ionizes to a slight degree in water what is its conjugate base?

`pK_(a)` of weak acid (HA) and `pK_(b)` of weak base (BOH) are 3.2 and 3.4 respectively. The pH of their salt (AB) solution isA. `7.0`B. `1.0`C. `7.2`D. `6.9`

`pK_(a)` of a weak acid is 5.76 and `pK_(b)` of a weak base is 5.25. What will be the pH of the salt formed by the two ?A. 7.255B. 7.005C. 10.225D. 4.255

Which of the following conditions will make the buffer most efficient?A. `pH=pK_(a)`B. `pH=pH_(a)+-1`C. `pH=pK_(a)+1`D. `pH=pH_(a)-1`

The solubility of a salt of weak acid (AB) at pH 3 is `Y xx 10^(-3)` mol `L^(-1)`. The value of Y is ............ (Given that the value of solubility product of aB `(K_(sp))=2xx10^(-10)` and the value of ionization constant of HB `(K_(a))=1xx10^(-8)`)

Which of the following is arranged in the order of increseing ionization constants of `H_(3)PO_(4)` ?A. `K_(3) lt K_(1) lt K_(2)`B. `K_(1) lt K_(2) lt K_(3)`C. `K_(2) lt K_(1) lt K_(3)`D. `K_(3) lt K_(2) lt K_(1)`

Oxoacids are ___________ acids.A. binaryB. ternaryC. quatenaryD. secondary

A patient is said to suffer from acidosis when the pH of his bloodA. falls below 7.35B. rises above 7.35C. shows sudden fall and riseD. has strong basic character

Which of the following pH curve represent the titration of weak acid and strong base (dotted line show equivalence point)?A. B. C. D.

`10^(-6) ` M NaOH is diluted 100 times. The pH of the diluted base isA. between 5 and 6B. between 6 and 7C. between 10 and 11D. between 7 and 8

0.023 g of sodium metal is reacted with `100 cm^(3)` of water. The pH of the resulting solution isA. 10B. 11C. 9D. 12

The solubility product of `AI(OH)_(3)` is `2.7 xx 10^(-11)`. Calculate its solubility in `mgL^(-1)`. (Atomic mass of `AI = 27 u)`.

The `pH` of `0.1 M` solution of cyanic acid `(HCNO)` is `2.34`. Calculate the ionization constant of the acid and its degree of ionisation in the solution.