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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1051. |
For the following equilibrium, `K_(c )=6.3xx10^(14) at 1000 K` `NO(g)+O_(3)(g) hArr NO_(2)(g)+O_(2)(g)` Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is `K_(c )`, for the reverse reaction? |
| Answer» For the reverse reaction `K_(c) = (1)/(K_(c)) =(1)/(6.3 xx 10^(14)) = 1.59 xx 10^(-15).` | |
| 1052. |
The `pH` of `10^(-8)M` solution of `HCl` in water is |
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Answer» `2H_(2)O (I) hArr H_(3)O^(+) (aq) + OH^(-)(aq)` `K_(w) = [OH^(-)][H_(3)O^(+)]` `= 10^(-14)` Let, `x = [OH] = [H_(3)O^(+)]` from `H_(2)O`. The `H_(3)O^(+)` concentration is generated `(i)` from the inonization of HCl dissolved i.e., `HCl(aq) + H_(2)(I) hArr H_(3)O^(+)(aq)+Cl^(-)(aq)` and (ii) from ionization of `H_(2)O`. In these very dilute solutions. both sources of `H_(3)O^(+)` must be considered. `[H_(3)O^(+)] = 10^(-8)+x` `K_(w) = (10^(-8) + x)(x) = 10^(-14)` or `x^(2) + 10^(-8) x - 10^(-14) = 0` `[OH^(-)] = x = 9.5 xx 10^(-8)` So, `pOH = 7.02` and `pH = 6.98` |
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| 1053. |
Reaction between nitrogen and oxygen takes place as following: `2N_(2(g))+O_(2)hArr2N_(2)O_((g))` If a mixture of `0.482 "mole" N_(2)` and `0.933 "mole"` of `O_(2)` is placed in a reaction vessel of volume `10 litre` and allowed to form `N_(2)O` at a temperature for which `K_(c)=2.0xx10^(-37)litre mol^(-1)`. Determine the composition of equilibrium mixture. |
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Answer» Let x moles of `N_(2) (g)` take part in the reaction. According to the equation, `x//2` moles of `O_(2) (g)` will react to form x moles of `N_(2)O(g)`. The molar concentration per litre of different species before the reaction nad at the equilibrium point is : `2N_(2)(g) +O_(2)(g) hArr 2N_(2)O(g)` `" Initial moles"//"litre : "" " (0.482)/(10) " " (0.933)/(10)" " "Zero"` `" Moles"//"litre at eqm. point : "" " (0.482 -x)/(10) " " (0.933 -(x)/(2))/(10) " " (x)/(10)` The value of equilibrium constant `(2.0xx 10^(-37))` is extremely small. This means that only small amounts of reactants have reacted. Therefore , x is extremely small and can be omitted as far as the reactants are concerned. Applying Law of chemical equilibrium ,`K_(c) =[[N_(2)O(g)]^(2)]/[[N_(2)(g)]^(2)[O_(2)(g)]^(2)]` `2.0 xx 10^(-37) =((x)/(10))^(2)/(((0.482)/(10))^(2) xx((0.933)/(10)))=(0.01 x^(2))/(2.1676xx10^(-4))` `x^(2) =43. 352 xx 10^(-40) or x =6.6 xx 10^(-20)` As x is extremely small , it can be neglected. Thus in the equilibrium mixture Molar conc. of `N_(2) = 0.0482 mol L^(-1)` Molar conc. of `O_(2) =0.0933 mol L^(-1)` Molar conc. of `N_(2)O= 0.1 xx X = 0.1 xx 6.6 xx 10^(-20) mol L^(-1)` `=6.6 xx 10^(-21) mol L^(-1)` |
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| 1054. |
How much volume of 0.1 M Hac should be added to 50 mL of 0.2 M NaAc solution if we want to prepare a buffer solution of pH 4.91. Given `pK_(a)` for acetic acid is 4.76. |
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Answer» Correct Answer - 70.92 mL `pH=pK_(a) + log. (["Salt"])/(["Acid"]), i.e., 4.91 = 4.76 + log . (["Salt"])/(["Acid"])` or `log. (["Salt"])/(["Acid"])=0.15 or (["Salt"])/(["Acid"])=Antilog 0.15 = 1.41` `("Moles of Salt")/("Moles of Acid")=1.41," i.e.," ((0.2)/(1000)xx50)/((0.1)/(1000)xxV)=1.41 or (0.01)/(1000V)=1.41 or V= 100//1.41 = 70.92 mL` |
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| 1055. |
In which of the following equilibrium, change in the volume of the system does not alter the Number of mole(a) N2(g) + O2(g) ⇋ 2NO(g)(b) PCl5(g) ⇋ PCl3(g) + Cl2(g) (c) N2(g) + 3H2(g) ⇋ 2NH3(g)(d) SO2Cl2 ⇋ SO2(g) + Cl2(g) |
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Answer» (a) N2(g) + O2(g) ⇋ 2NO(g) |
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| 1056. |
When rain is accompanied by a thunderstorm, the collected rain water will have pH valueA. slightly higer than that when the thunder storm is not thereB. uninfluenced by the thunder stormC. which depends on the amount of dust in airD. slightly lower than that of rain water without thunder storm |
| Answer» Correct Answer - A::C::D | |
| 1057. |
`NH_(4)CN` is a salt of weak acid `HCN(K_(a)=6.2xx10^(-10))` and a weak base `NH_(4)OH(K_(b)=1.8xx10^(-5))`. 1 molar solution of `NH_(4)CN` will be :-A. neutralB. strongly acidicC. strongly basicD. weakly basic. |
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Answer» Correct Answer - D Since `K_(b)gtK_(a)`, the solution will be slightly basic. |
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| 1058. |
Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed together. Will it lead to precipitation of copper iodate ? For copper iodate, `K_(sp)=7.4xx10^(-8)`. |
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Answer» `2 NaIO_(3)+CuCrO_(4) rarr Na_(2)CrO_(4)+Cu(IO_(3))_(2)` After mixing, `[NaIO_(3)]=[IO_(3)^(-)]=(2xx10^(3))/(2)=10^(-3)M` `[CuCrO_(4)]=[Cu^(2+)]=(2xx10^(-3))/(2) = 10^(-3)M` Ionic product of `Cu(IO_(3))_(2)=[Cu^(2+)][IO_(3)^(-)]^(2)=(10^(-3))(10^(-3))^(2)=10^(-9)` As ionic product is less than `K_(sp)`, no precipitation will occur. |
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| 1059. |
The ionization constant of HF, HCOOH and HCN at 298 K are `6.8xx10^(-4), 1.8xx10^(-4) and 4.8xx106(-9)` respectively. Calculate the ionization constant of the corresponding conjugate base. |
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Answer» For `F^(-) , " " K_(b) = K_(w)//K_(a) = 10^(-14)//(6.8xx10^(-4))=1.47 xx 10^(-11) ~= 1.5 xx 10^(-11)`. For `HCO O^(-) , " " K_(b) = 10^(-14)//(1.8xx10^(-4))=5.6xx10^(-11)` For `CN^(-), " " K_(b) = 10^(-14) // (4.8xx10^(-9))=2.08 xx 10^(-6)` |
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| 1060. |
Equal volumes of 0.002 M solution of sodium iodate and cupric chlorate are mixed togather. Will it lead to precipitation of copper iodate? `("for cupric iodate" K =7.4xx10^(-8))`. |
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Answer» The solubility equilibrium of copper iodate may be represented as : `Cu(IO_(3))_(2) hArr Cu^(2+) (aq) +2IO_(3)^(-)(aq)` `Cu^(2+)(aq)` are to be provided from copper chlorate solution while `IO_(3)^(-) (aq)` ions from sodium iodate solutions `Cu(CIO_(3))_(2)(s) overset((aq))(to) Cu^(2+) (aq) +2CIO_(3)^(-) (aq)` `NaIO_(3)(s)overset((aq))(to) Na^(+) (aq) +IO_(3)^(-) (aq)` Since equal volumes of the solutions have been mixed therefore , concentration of both `Cu^(2+) (aq)` and `IO_(3)^(-)(aq)` ions in solution after mixing will be reduced to half i.e., `(0.002M)/(2) =0.001 M` The ionic product `=[Cu^(2+)][IO_(3)^(-)]^(2) =(0.001) xx (0.001)^(2) = 1.0 xx 10^(-9)` `K_(sp) " value of " Cu(IO_(3))_(2) =7.4 xx 10^(-8)` (given) Since the ionic product is less than the `K_(sp)` value copper iodate will not be precipitated. |
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| 1061. |
The `pH` of a sample of vinegar is `3.76`, Calculate the concentration of hydrogen ion in it. |
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Answer» Correct Answer - `1.7 xx 10^(-4) M` Given `pH = 3.76` It is known that, `pH = -"log"[H^(+)]` `rArr "log"[H^(+)] = - pH` `{:(rArr[H^(+)],="antilog"(-pH)),(,="antilog"(-3.76)),(,=1.74xx10^(-4)M):}` |
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| 1062. |
The pH of a sample of vinegar is 3.76, calculate the concentration of hydrogen ion in it. |
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Answer» pH of the solution = 3.96 log\(\frac{1}{[H_3O^+]}\) = 3.96 or, \(\frac{1}{[H_3O^+]}\) = Antilog 3.96 = 5754 [H3O+] = \(\frac{1}{5754}\) = 1.7 x 10-4 M |
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| 1063. |
The concentration of hydrogen ion in a sample of soft drink is `3.8xx10^(-3) M`. What is its pH ? |
| Answer» `pH = - log [H^(+)]=- log (3.8xx10^(-3))= - log 3.8 + 3 = 3 - 0. 5798 = 2.4202 = 2.42` | |
| 1064. |
The correct order of decreasing acidic nature of `H_(2)O, ROH, CH -= CH and NH_(3) ` isA. `CH-=Ch gt H_(2) O gt ROH gt NH_(3)`B. `H_(2)O gt ROH gt CH -= CH gt NH_(3)`C. `ROH gt NH_(3) gt CH -= CH gt H_(2)O`D. `H_(2)O gt rOH gt NH_(3) gt CH -= CH` |
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Answer» Correct Answer - B `{:(,H_(2)O,ROH,CH-=CH,NH_(3),),(pK_(a),15.7,16-18," "26,38,):}` Smaller the `pK_(a)` value, stonger is the acid. Hence, decreasing order of acidic nature is `H_(2)O gt ROH gt CH-=CH gt NH_(3)`. |
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| 1065. |
A sample of pure `PCl_(5)` was introduced into an evacuted vessel at `473 K`. After equilibrium was attained,concentration of `PCl_(5)` was found to be `0.5xx10^(-1)mol litre^(-1)`. If value of `K_(c)` is `8.3xx10^(-3) mol litre^(-1)`. What are the concentrations of `PCl_(3)` and `Cl_(2)` at equilibrium ? |
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Answer» Correct Answer - `0.02"molL"^(-1)` for both. Let the concentrations of both `PCl_(3)` and `Cl_(2)` at equilibrium be x `"molL"^(-1)`. The given reaction is: `{:(,PCl_(5(g)),harr,PCl_(3(g)),+,Cl_(2(g))),("At equilibrium",0.5xx10^(-1)"mol L"^(-1),,x"mol"L^(-1),,x "mol" L^(-1)):}` It is given that the value of equilibrium constant. `K_(c)` is `8.3 xx 10^(-3)`. now we can write the expression for equilibrium as : `([PCl_(2)][Cl_(2)])/([PCl_(5)]) = K_(c)` `rArr (x xx x)/(0.5 xx 10^(-1)) = 8.3 xx 10^(-3)` `rArr x^(2) = 4.15 xx 10^(-4)` `rArr x = 2.04 xx 10^(-2)` `= 0.0204` `= 0.2` (approximately) Therefore, at equilibrium, Therefore at equilibrium `[PCl_(3)] = [Cl_(2)] = 0.02 "mol" L^(-1)`. |
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| 1066. |
The pH of a sample of pure water is 7 at room temperature. What is its pH when a pinch of solid sodium bicarbonate is dissolved in it? (a) vary near to 7 (b) less than 7 (c) more than 7 (d) exactly 7 |
| Answer» (c) more than 7 | |
| 1067. |
There are acid base theories that define an acid as any species that can (a) donate a proton (b) donate an electron (c) accept a proton (d) none of these |
| Answer» (a) donate a proton | |
| 1068. |
The colour of neutral litmus solution is (a) red (b) blue (c) purple (d) yellow |
| Answer» (b) The colour of neutral litmus solution is blue | |
| 1069. |
The equilibrium constant `(K_(p))` for the decomposition of gaseous `H_(2)O` `H_(2)O(g)hArr H_(2)(g)+(1)/(2)O_(2)(g)` is related to the degree of dissociation `alpha` at a total pressure P byA. `K_(p)=(alpha^(3)P^(1//2))/((1+alpha)(2+alpha)^(1//2))`B. `K_(p)=(alpha^(3)P^(3//2))/((1-alpha)(2+alpha)^(1//2))`C. `K_(p)=(alpha^(3//2)P^(2))/((1-alpha)(2+alpha)^(1//2))`D. `K_(p)=(alpha^(3//2)P^(1//2))/((1-alpha)(2+alpha)^(1//2))` |
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Answer» Correct Answer - D Consider the equilibrium expression `{:(,H_(2)O(g),hArrH_(2)(g)+,(1)/(2)O_(2)(g)),("Inital moles",1,0,0),("Equilibrium moles",1-alpha,alpha,alpha//2):}` Total moles at equilibrium `=(1-alpha)+(alpha)+(alpha)/(2)` `=1+alpha//2` Note that is always convenient to consider 1 mol whenever we del with the degree of dissociation. At equilibrium, the partial pressure of gas can be calculated as follows: Partial pressure = (Mole fraction) x (Total pressure) `:. P_(H_(2)O)= (1-alpha)/(1+alpha//2)P` `= (2(1-alpha))/(2+alpha)P` `P_(H_(2))=(alpha)/(1+alpha//2)P=(2alpha)/(2+alpha)P` `P_(O_(2))=(alpha//2)/(1+alpha//2)P=(alpha)/(2+alpha)P` According to the law of chemical equilibrium, `K_(p)=(P_(H_(2))P_(O_(2))^(1//2))/(P_(H_(2)O))` `= (((2alpha)/(2+alpha)P)((alphaP)/(2+alpha))^(1//2))/((2(1-alpha))/(2+alpha)P)` `=(alpha)/(1-alpha)((alphaP)/(2+alpha))^(1//2)` `=(alpha^(3//2)P^(1//2))/((1-alpha)(2+alpha)^(1//2))` |
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| 1070. |
In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?(i) H2 (g) + I2 (g) ⇄ 2HI (g)(ii) PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)(iii) N2 (g) + 3H2 (g) ⇄ 2NH3 (g)(iv) The equilibrium will remain unaffected in all the three cases. |
| Answer» (iv) The equilibrium will remain unaffected in all three cases on addition of small amount of inert gas at constant volume. | |
| 1071. |
For the reaction `N_(2)O_(4)(g) hArr 2 NO_(2) (g)`, the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct ?A. The reaction is endothermicB. The reaction is exothermicC. If `NO_(2) (g) and N_(2) O_(4)(g)` are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more `N_(2)O_(4) (g)` will be formed,D. The entropy of the system increases. |
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Answer» Correct Answer - A::C::D (a) As the value of K increases with increase of temperature and `K=(k_(f))/(k_(b))`, this means that `k_(f)` increases, i.e., formard reaction is favoured. Hence, reaction is endothermic. (c) At 400 K, `Q=(p_(NO_(2))^(2))/(p_(N_(2)O_(4)))=((20)^(2))/(2) = 200 `. Thus `Q gt K`. Equilibrium will shift backward to form more `N_(2)O_(4)`. (d) As reaction is accompanied by increase in the number of moles , entropy increases. |
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| 1072. |
Calculate the hydrolysis constant, degree of hydrolysis and pH of 0.10 M KCN solution at `15^(@)`C . For HCN, `K_(a)=6.2xx10^(-10)`. |
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Answer» As KCN is a salt of strong base and weak acid, Hydrolysis constant, `K_(h) = (K_(w))/(K_(a))=(10^(-14))/(6.2xx10^(-10))=1.6xx10^(-5)` Degree of hydrolysis, `h=sqrt((K_(h))/(c))=sqrt((1.6xx10^(-5))/(0.1))=1.26xx10^(-2)` The hydrolysis reaction will be : `{:(,CN^(-),+,H_(2)O,hArr,HCN,+,OH^(-)" ...(i)"),("Initial conc.",cM,,,,0,,0),("At eqm.",c-x,,,,x,,x " (x-No. of moles of" CN^(-) "reacted)"),(,,,,,,,):}` `K(h)=([HCN][OH^(-)])/([CN^(-)])=(x xx x)/(c-x)=(x^(2))/(c) or x=sqrt(K_(h)xxc)=sqrt((1.6xx10^(-5))(0.1))=1.26xx10^(-3)` i.e., `[OH^(-)]=1.26xx10^(-3)` `:. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(1.26xx10^(-3))=7.94xx10^(-12)` `pH = - log [H^(+)]=-log (7.94xx10^(-12))=12-0.90=11.1` Alternatively, reaction (i) is the hydrolysis of the base `(CN^(-))`. We need `K_(b)` for this reaction. We are given `K_(a)` of the conjugate acid (HCN) . Hence, `K_(b)=(K_(w))/(K_(a))=(10^(-14))/(6.2xx10^(-10))=1.61xx10^(-5)` For equilibrium (i), `K_(b)=(x xx x)/(c-x) ~= (x^(2))/(c)` `:. 1.61xx10^(-5)=(x^(2))/(0.1) or x^(2)xx1.61xx10^(-6) or x = 1.26xx10^(-3)` i.e., `[OH^(-)]=1.26xx10^(-3)` `pOH = log (1.26xx10^(-3))=2.9` `:. pH = 14-2.9 = 11.1` Alternatively, pH can be calculated directly by applying the formula, `pH = 7 +(1)/(2) [pK_(z)+log c]` |
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| 1073. |
The pH of `10^(-2)` M NaOH solution is .............. Times the pH of `10^(-2)` M HCl solution |
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Answer» Correct Answer - 6 pH of `10^(-2)` M NaOH solution = 12. pH of `10^(-2)` M HCl = 2. |
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| 1074. |
Which of the following salts is neutral in water ?A. `NH_(4)CH_(3)COO`B. `NH_(4)NO_(3)`C. `NH_(4)CN`D. `NH_(4)F` |
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Answer» Correct Answer - A `NH_(4)CH_(3)COO` is a salt of weak base and weak acid for which `K_(b)=K_(a)`. |
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| 1075. |
For cationic hydrolysis, pH given byA. `pH=(1)/(2)pK_(w)+(1)/(2)pK_(a)+(1)/(2)log C`B. `pH=(1)/(2)pK_(w)-(1)/(2)pK_(a)-(1)/(2)log C`C. `pH=(1)/(2)pK_(w)+(1)/(2)pK_(a)-(1)/(2)pK_(b)`D. `pH=(1)/(2)pK_(w)+(1)/(2)pK_(b)+(1)/(2)log C` |
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Answer» Correct Answer - B `B^(+)+H_(2)O hArr BOH+H^(+)` `[H^(+)] = ((K_(w).C)/(K_(b)))^(1//2)` `pH= - log[H^(+)]= -"log" ((K_(w).C)/(K_(b)))^(1//2)` `= (-1)/(2)[log K_(w)-logK_(b)+log C]` `= (1)/(2)[pK_(w)-pK_(b)-logC]` |
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| 1076. |
The degree of dissociation of an acid HA is 9 times the degree of dissociation of acid `HA_(2)` of the same concentration, then the ratio of the strengths of acid `HA_(1)` to `HA_(2)` will be |
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Answer» Correct Answer - 9 `("Strength of "HA_(1))/("Strength of "HA_(2))=(alpha_(1))/(alpha_(2))=(9)/(1) = 9`. |
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| 1077. |
What is pOH of an aqueous solution with hydrogen ion concentration equal to `3xx10^(-5)" mol L"^(-1)`?A. 9.47B. 4.52C. 12.69D. 11.69 |
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Answer» Correct Answer - A `pH=-log[H^(+)]=-log(3xx10^(-5))=4.5229pOH=14-pH,14-4.5229=9.47` |
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| 1078. |
The degree of ionisation of an acid HA is 0.00001 at 0.1 M concentration. Its dissociation constant will beA. `10^(-9)`B. `10^(-11)`C. `10^(-8)`D. `10^(-7)` |
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Answer» Correct Answer - B `K_(a)=calpha^(2)=0.1xx(10^(-5))^(2)=10^(-11)` |
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| 1079. |
Which of the following salts has the minimum pH value? (a) (NH4)2SO4 (b) NaHCO3 (c) K2SO4 (d) NaCl |
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Answer» (a) The minimum pH value is (NH4)2SO4 |
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| 1080. |
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) H2SO4 acts as a base in presence of HClO4.(R) Perchloric acid is stronger acid than H2SO4. |
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Answer» If both (A) and (R) are correct and (R) is the correct explanation of (A) |
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| 1081. |
An alkali is titrated against an acid with methyl orange as indicator. Which of the following is a correct combination ?A. `{:(,"Base","Acid","End point",),((a),"Weak","Strong","Colourless to pink",):}`B. `{:(,"Base","Acid","End point",),((b) , "Strong","Strong","Pinkish red to yellow",):}`C. `{:(,"Base","Acid","End point",),((b) , "Weak ","Strong","Yellow to pinkish red",):}`D. `{:(,"Base","Acid","End point",),((b) , "Strong","Strong","Pink to colourless",):}` |
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Answer» Correct Answer - C pH range of methyl orange is 3.1 - 4.5 . Below 3.1, it is pinkish red and above 4.5, it is yellow. Weak base has `pH gt 7`. When methyl orange is added to weak base solution, the solution becomes yellow. When the solution is titrated with a strong acid, after the end point, solution is acidic and pH is `lt ` 3.1 . Therefore solution becomes pinkish red. |
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| 1082. |
The `pK_(a)` of acteylsalicylic acid (aspirin) is `3.5`. The pH of gastric juice in human stomach is about `2-3` and the pH in the small intestine is about 8. Aspirin will be:A. Unionised in the small intestine and in the stomach .B. Completely ionised in the small intestine and in the stomach.C. Ionised in the stomach and almost unionised in the small intestine.D. Ionised in the small intestine and almost unionised in the equilibrium |
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Answer» Correct Answer - D (d) Gastric juice is acidic . Hence the ionisation of aspirin is suppressed whereas small intestine has basic medium. Therefore aspirin is more ionised here. |
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| 1083. |
Thermal decomposition of gaseous `X_(2)` to gaseous X at 298 K takes place according to following equation `X_(2) (g) hArr 2X(g)` The standard reaction Gibbs energy `Delta_(r)G^(@)` of this reaction is positive . At the start of the reaction there is one mole of `X_(2) " and no. " X`. As the reaction proceeds the number of moles of X formed is given by `beta`. Thus `beta_("equilibrium")` is the the number of moles of X formed at equilibrium . The reaction is carried out a constant total pressure of 2 bar. Consider the gases to behave ideally. `("Given" : R =0.083 L " bar " K^(-1) " mol"^(-1))` The incorrect statement among the following for this reaction isA. Decrease in the total pressure will result in formation of more moles of gaseous XB. At the start of the reaction , dissocition of gaseous `X_(2)` takes place spontaneouslyC. `beta_("equilibrium") =0.7`D. `K_(c)lt 7` |
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Answer» Correct Answer - C (C) It is incorrect statement among the (a) If the pressure on the system is decreased , the equilibrium will shift in the direction in which pressure increases i.e.,increases in no. of moles takes place i.e., in forward idrection . (b) At the start of the reaction `Q lt K` thus the reaction will proceed in the forward direction i.e., reaction is spontaneous. (C) `" if "b_(eq) =0.7 " then " K_(p) =(8xx(0.7)^(2))/(4-(0.7)^(2)) gt 1` `DeltaG^(@) =- RT" In " K_(p) SO, DeltaG^(@) =- " ve but ginen " DeltaG^(@) = + " ve so, " K_(p) " should be less than " 1 " Hence " beta_(eq) ne 0.7` (d)`K_(p) =K_(c) (RT)^(Dn).` `K_(c) lt K_(p) ne (":." RT gt 1)` `" If " K_(p) lt1 " then " K_(c) lt 1` |
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| 1084. |
Which of the following options will be correct for the stage of half completion of the reaction A ⇄ B.(i) ΔGΘ = 0(ii) ΔGΘ > 0(iii) ΔGΘ < 0(iv) ΔGΘ = –RT ln2 |
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Answer» (i) ΔGΘ = 0 Justification : ΔGΘ = – RT lnK At the stage of half completion of reaction [A] = [B], Therefore, K = 1. Thus, ΔGΘ = 0 |
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| 1085. |
`K_(a)` for an acid HA is `4.9 xx 10^(-8)`. Calculate percentage dissocitation `H^(+)` ion concentration for its 0.1 M aqueous solution. |
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Answer» Correct Answer - `alpha = 7xx 10^(-4) ,[H^(+)]=7 xx 10^(-5) "mol " L^(-1)` Degree of dissociation of acid `(alpha) = sqrt((K_(a))/(C))= ((4.9xx 10^(-8))/(0.1))^(1//2) =7xx 10^(-4)` `H^(+)" ion concentration i.e.," [H^(+)] =Calpha =0.1 xx7xx10^(-4) =7xx10^(5) " mol "L^(-1)` |
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| 1086. |
Derive the expression Kb = [M+ ][OH- ] / [MOH] for dissociation of a weak base MOH in water , which is undergo complete dissociation in the solution. |
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Answer» MOH -----> M+ + OH- Initial conc Cx1 0 0 C-Cα Cα Cα Kb = [H+ ][X- ] / [HX] = Cαx Cα/C- Cα =C2 α2 /C(1-α) Kb=Cα2 /1-α |
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| 1087. |
Ka for CH3COOH is 1.8 × 10–5 and Kb for NH4OH is 1.8 × 10–5 . The pH of ammonium acetate will be(i) 7.005(ii) 4.75(iii) 7.0(iv) Between 6 and 7 |
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Answer» The correct answer is (iii) 7.0 |
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| 1088. |
`K_(a)` for ascorbic acid `("HA sc" ) ` is `5xx10^(-5)`. Calculate the hydrogen ion concentration and percentage of hydrolysis in an aqueous solution in which the concentration of `Asc^(-)` ions is 0.02 M. |
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Answer» As ascorbic acid is a weak acid, aqueous solution of `Asc^(-)` means the solution of a salt of weak acid with strong base. For such a salt , degree of hydrolysis is given by `h=sqrt((K_(w))/(K_(a)xxc))=sqrt((10^(-14))/(5xx10^(-5)xx2xx10^(-2))=10^(-4) = 10^(-4) xx 100 % = 0.01 % ` `{:(Asc^(-),+,H_(2)O,hArr,HAsc,+,H^(-),),(c " mol" L^(-1),,,,,,,),(c -ch ,,,,ch,,ch,):}` `[OH^(-)]=ch=0.02xx10^(-4)=2xx10^(-6) "mol " L^(-1) :. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(2xx10^(-6))=5xx10^(-9) "mol" L^(-1)` |
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| 1089. |
Does the number of moles of reaction products increase, decrease or remain same when3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)equilibria is subjected to a decrease in pressure by increasing the volume? |
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Answer» 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g) Number of moles of reaction products remain same when the equilibria is subjected to a decrease in pressure by increasing the volume. |
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| 1090. |
What are buffer solution? |
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Answer» A solution which resist the change in pH on dilution or with the addition of small amounts of an acid or alkali is called buffer solution. |
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| 1091. |
(a) Define buffer solution. (b) Give one example each of acidic and basic buffer. |
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Answer» (a) The solution which resists change in pH on dilution or with the addition of small amounts of acid or alkali is called buffer solution. (b) Example of acidic buffer- CH3COOH + CH3COONa (pH = 4.75) Example of basic buffer- NH4OH + NH4Cl (pH = 9.25) |
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| 1092. |
Give an example for acidic /basic buffer |
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Answer» Acidic buffer example: Mixture of acetic acid and sodium acetate. Basic buffer example: Mixture of ammonium hydroxide and ammonium chloride. |
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| 1093. |
What happens when the temperature of a reversible reaction at equilibrium is increased, if enthalpy change is positive? |
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Answer» Equilibrium shifts to the right. |
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| 1094. |
Give an example for a reversible reaction in which Kp = KcRT. |
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Answer» PCI5(g) ⇌ PCI3(g) + CI2(g) |
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| 1095. |
What is the percentage hydrolysis of NaCN in N/80 solution when the dissociation constant for HCN is `1.3xx10^(-9) and K_(w)=1.0xx10^(-14)`A. 2.48B. 5.26C. 8.2D. 9.6 |
| Answer» Correct Answer - A::C | |
| 1096. |
The pH of a solution obtained by dissolving 0.1 mole of an acid HA is 100 ml of the aqueous solution was found to be 3.0 . Calculate the dissociation constant of the acid. |
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Answer» `pH = - log [H_(3)O^(+)] :. Log [H_(3)O^(+)]=-pH = - 3.0 ` [pH = 3.0 , given] or `[H_(3)O^(+)]` = antilog (-3) = antilog `bar(3) = 10^(-3)` g ions/litre = 0.001 g ion/litre Original conc.of the acid HA = 0.1 mole in 100 ml. = 1 mole/litre `{:("HA dissociates as :",HA" "+" "H_(2)O,hArr,H_(3)O^(+)+,A^(-)),("Initial conc. :",1 M,,0,0),("Conc. at equilibrium:",1-0.001M,,0.001M,0.001M),(,,,,):}` `:.` Dissociation constant (K) will be given by `K=([H_(3)O^(+)][A^(-)])/([HA])=(0.001xx0.001)/(1-0.001)=(10^(-6))/(1)=10^(-6)` (Neglecting 0.001 in comparison to 1) |
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| 1097. |
An acid HA ionises as `HA hArr H^(+)+A^(-)` The pH of 1.0 M solution is 5. Its dissociation constant would beA. `1xx10^(-10)`B. 5C. `5xx10^(-8)`D. `1xx10^(-5)` |
| Answer» Correct Answer - A | |
| 1098. |
Would you expect equilibrium constant for the reaction `l_2(g)hArr2l(g)` to increase or decrease as temperature increases. Assign reson. |
| Answer» In the forward reaction, energy is needed to bring about the dissociation of `l_2(g)` molecules. This menas that the forward reaction is endothermic in nature. The increase in temperature will favour the forward. Therefore, the equilibrium constant will increase with rise in temperature. | |
| 1099. |
What pH do you expect for `10^(-8)` M solution of an acid ? |
| Answer» Close to 7 but `lt 7`. | |
| 1100. |
STATEMENT-1: HCl is a strong acid and true electrolyte. STATEMENT-2: Liquid HCl is bad conductor of electricity.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-12B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-12C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - D |
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