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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.
| 1151. |
If `S_(0), S_(1), S_(2)`, and `S_(3)` are the solubility of AgCI in water, `0.01M CaCI_(2), 0.01M NaCI`1, and `0.5M AgNO_(3)` solutions, respectively, then which of the following is true ?A. `S_(0)gtS_(2)gtS_(1)gtS_(3)`B. `S_(0)=S_(2)=S_(1)=S_(3)`C. `S_(3)gtS_(1)gtS_(2)gtS_(0)`D. `S_(0)gtS_(2)gtS_(3)gtS_(1)` |
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Answer» Correct Answer - A Solubility equilibrium is `AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)` `K_(sp)=C_(Ag^(+))C_(Cl^(-))` The solubility of AgCI is maximum in water as there is no common ion effect. In other solutions, due to the presence of common ions, solubility is less than in pure water. The larger the concentration of common ions, the lower is the solubility. `0.01M CaCl_(2)` provides `Cl^(-)` with `C_(Cl^(-))=2(0.01M) = 0.02M` `0.01M NaCI` provides `Cl^(-)` with `C_(Cl^(-))= 0.01M` `0.5M AgNO_(3)` provides `Ag^(+)` with `C_(Ag^(+)) = 0.5M` Thus, the solubility of AgCI will be minimum in `0.5M AgNO_(3)` where the concentration of common ions is maximum. The correct order of solubility is, thus, `S_(0)gtS_(2)gtS_(1)gtS_(3)` |
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| 1152. |
Let the solubilities of `AgCI` in `H_(2)O`, and in `0.01M CaCI_(2), 0.01M NaCI`, and `0.05M AgNO_(3)` be `S_(1),S_(2),S_(3),S_(4)`, respectively. What is the correct relationship between these quantites.A. `S_(1)lt S_(2)lt S_(3)ltS_(4)`B. `S_(1)gtS_(3)gtS_(2)gtS_(4)`C. `S_(1)gt S_(2)=S_(3)gt S_(4)`D. `S_(1)gt S_(3)gtS_(4)gt S_(2)` |
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Answer» Correct Answer - B |
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| 1153. |
Predict whether a precipitate will be formed or not on mixing 20 mL of 0.001 M NaCI with 80 mL of 0.01 M `AgNO_(3)` solution `(K_(sp) " for " AgCI =1.5 xx 10^(-10))` |
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Answer» Correct Answer - Yes , ionic product `=1.6 xx 10^(-6)` The precipitate of AgCI is to be formed . the solubility equilibrium may be represented as `AgCI (s) hArr Ag^(+) (aq) +CI^(-) (aq)` Now `Ag^(+)` ions are to be provided by `AgNO_(3)` solution while `CI^(-)` ions by NaCI solution as a result of dissociation. `AgNO_(3)overset((aq))(to) Ag^(+) (aq) + NO_(3)^(-) (aq) , NaCI (s) overset((aq))(to) Na^(+) (aq) +CI^(-) (aq)` the total volume of the solution after mixing `=(20 +80) =100 mL` In the solution , the concentration of `Ag^(+)` ions after mixing will be `4//5 (80//100)` while that of `CI^(-)` ions after mixing will be `1//5 (20// 100)` `" Thus "" "[Ag^(+)] " before mixing " =0.01 M , [Ag^(+)] " Aftre mixing " =(0.01 xx 4)/(5) =0.008 M` Similarly, `[CI^(-)]` before mixing =0.001 M `[CI^(-)]` after mixing `=(0.001 xx 1)/(5) =0.0002 M` The ionic product = `Ag^(+)][CI^(-)] = (0.008) xx (0.0002) =8 xx 10^(-3) xx 2xx 10^(-4) =1.6 xx 10^(-6)` `K_(sp)` value of AgCI =`1.5 xx 10^(-10)` (given) As the ionic product is more than the solubility product this means AgCI will get precipitated |
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| 1154. |
The `[Ag^(+)] =10^(-5)` in a solution . The `[CI^(-)]` to precipitate `AgCI (K_(sp) " of " AgCI =2 xx 10^(-12))` is :A. `10^(-5)`B. `10^(-7)`C. `10^(-8)`D. `10^(-9)` |
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Answer» Correct Answer - A `[CI^(-)] =(K_(sp))/([Ag^(+)]] = (2xx 10^(-12))/(10^(-5))` Thus `[CI^(-)]` must be more than `10^(-7)` |
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| 1155. |
Ka1 , Ka2 and Ka3 are the respective ionisation constants for the following reactions.H2S ⇄ H+ + HS–HS– ⇄ H+ + S2–H2S ⇄ 2H+ + S2–The correct relationship between Ka1 , Ka2 and Ka3 is(i) Ka3 = Ka1 × Ka2(ii) Ka3 = Ka1 + Ka2(iii) Ka3 = Ka1 - Ka2(iv) Ka3 = Ka1 / Ka2 |
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Answer» (i) Ka3 = Ka1 × Ka2 |
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| 1156. |
Ka1 Ka2 and Ka3 are the respective ionization constants for the following reactions-H2S ⇌ H+ + HS-H2S- ⇌ H+ + S2-H2S ⇌ 2H+ + S2-Write the correct relationship between Ka1, Ka2 and Ka3. |
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Answer» For the reaction, H2S ⇌ H+ + HS− Ka1 = \(\frac{[H^+][HS^-]}{[H_2S]}\) For the reaction, HS-⇌ H+ + S2- Ka2 = \(\frac{[H^+][S^{2-}]}{[HS^{-}]}\) When above reactions are added, their equilibrium constants are multiplied. So, for the reaction, H2S ⇌ 2H+ + S2- Ka3 = \(\frac{[H^+]^2[S^{2-}]}{[H_2S]}\) = Ka1 x ka2 \(\therefore\) Ka3 = Ka1 x ka2 |
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| 1157. |
Study the figure below and mark the correct statement about `K_(c)` and dependence of extent of reaction on it. A. `{:(X,Y,Z),("Reaction does not occur","Reaction processds to completion","Reaction does not occur"):}`B. `{:(X,Y,Z),("Reaction completes","Reaction does not occur","Reactants and products are at equilibrium"):}`C. `{:(X,Y,Z),("Reaction hardly occurs","Reactants and products are at equilibrium","Reaction proceed to completion" ):}`D. `{:(X,Y,Z),("Reaction proceeds to completion","Reactants and products are at equilibrium","Reaction hardly occurs"):}` |
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Answer» Correct Answer - C Value of `K_(c)` predicts the extent of the reaction. |
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| 1158. |
`K_(a1)`,`K_(a2)`and `K_(a3)` are the respective ionisation constants for the following reactions. `H_(2)ShArrH^(+)+HS^(-)`,`HS^(-)hArrH^(+)S^(-2)` `H_(2)ShArr2H^(+)+S^(2-)` The correct relationship between `K_(a1)`,`K_(a2)`and `K_(a3)` isA. `K_(a_(3))=K_(a_(1))xxK_(a_(2))`B. `K_(a_(3))=K_(a_(1))+K_(a_(2))`C. `K_(a_(3))=K_(a_(1))-K_(a_(2))`D. `K_(a_(3))=K_(a_(1))//K_(a_(2))` |
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Answer» Correct Answer - A For the reaction : `H_(2)ShArrH^(+)+HS^(-),K_(a_(1))` `K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])` . . . (i) For the reaction : `HS^(-)hArrH^(+)+S^(2-),K_(a_(2))` `K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)])` . . .(ii) For the reaction : `H_(2)ShArr2H^(+)+S^(2-),K_(a_(3))` `K_(a_(3))=([H^(+)]^(2)[S^(2-)])/([H_(2)S])` . . . (iii) By multiplying eqn. (i) and eqn. (ii), we get `K_(a_(1))xxK_(a_(2))=([H^(+)][Hs^(-)])/([H_(2)S])xx([H^(+)][S^(2-)])/([HS^(-)])` `=([H^(+)]^(2)[S^(2-)])/([H_(2)S])=K_(a_(3))` |
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| 1159. |
`{:(,"Column I(Types of titration)",,"Column II (Indicator used)",),((A),"Strong acid vs strong base",(p),"Methyl orange",),((B),"Strong acid vs weak base",(q),"Methyl red",),((C),"Weak acid vs strong base",(r),"Phenolphthalein",),((D),"Weak acid vs weak base",(s),"Bromothymol blue",):}` |
| Answer» Correct Answer - (A-p, q, r, s ; B-p, q, s ; C-r, s ; D-s) | |
| 1160. |
Define Buffer solution. |
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Answer» The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions. |
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| 1161. |
What is the relationship between `pK_(ln) and pH` at the equivalence point ? |
| Answer» `pK_(ln) ` = pH at the equivalence point. | |
| 1162. |
`1 cm^(3)` of 0.01 N HCl solution is added to one litre of sodium chloride solution . Calculate the pH of the resulting solution. |
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Answer» Since NaCI is a neutral salt, its solution as such will not give any `H^(+)` ions. It will dilute HCI solution. Concentration of HCI before dilution =0.01 N Concentration of HCI after dilution `=0.01//1000 = 10^(-5) N=10^(-5) M.` lt brgt (HCI is a monobasic acid. Its molarity and normality are the same) `[H_(3)O^(+)] =10^(5) , pH =- log (10^(-5) M) =(-) (-5) log 10 =5` |
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| 1163. |
Which indicator should preferably be used for titration of `NH_(4)OH ` with HCl solution? |
| Answer» Correct Answer - Methyl orange. | |
| 1164. |
Assertion :- A solution of `NH_(4)Cl` in water is acidic in nature. Reacon : - Ammonium ions undergo hydroysis to from `NH_(4)OH` .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. |
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Answer» Correct Answer - A `NH_(4)Cl_((aq))toNH_(4(aq))^(+)+Cl_((aq))^(-)` `NH_(4)^(+)` ions formed, undergo hydrolysis with water to from `NH_(4)OH and H^(+)` ions. `NH_(4)OH` is a weak base and thus remains almost unionised in solution. As a result, `[H^(+)]` ions increase in solution, making the solution acidic. |
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| 1165. |
Application of Le-Chatelier’s principal indicates that synthesis of ammoniaN2 + 3H2 ⇋ 2NH3, ΔH = -92.4kJ mol-1 Is favored by(a) Low pressure. (b) High temperature.(c) Pressure of catalyst. (d) Removal of ammonia. |
| Answer» (d) Removal of ammonia. | |
| 1166. |
For the reaction, H2 + I2 ⇋ 2HI, the equilibrium constant changes with(a) Total pressure (b) Catalyst(c) Temperature. (d) The amount of H2 and I2 present. |
| Answer» (c) Temperature. | |
| 1167. |
For the reaction H2(g) + I2(g) ⇋ 2HI(g), the equilibrium constant Kp changes with –(a) Temperature (b) Total pressure (c) Catalyst (d) amount of H2 and I2 |
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Answer» (a) Temperature |
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| 1168. |
The melting of ice is favoured by ………….. pressure and ………….. temperature.A. Low T and PB. High T and PC. Low T high PD. Low P high T |
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Answer» Correct Answer - B |
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| 1169. |
The value of K for the reaction;H2 + I2 ⇋ 2HI is 49. The value of K for the reaction,2HI ⇋ H2 + I2 will be(a) 49 (b) 1/49(c) 7 (d) 1/7 |
| Answer» (b) The value of K is 1/49. | |
| 1170. |
Solubility of a substance which dissolve with a decrease in volume and absorption of heat will be favoured by(a) High pressure and high temperature.(b) Low pressure and high temperature.(c) Low pressure and low temperature.(d) High pressure and low temperature. |
| Answer» (a) High pressure and high temperature. | |
| 1171. |
In a chemical reaction equilibrium is established when(a) Opposing reaction ceases(b) Concentration of reactants and product are equal.(c) Velocity of the opposing reaction is same as that of forward reaction.(d) Forward reaction ceases. |
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Answer» (c) Velocity of the opposing reaction is same as that of forward reaction. |
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| 1172. |
Study the given figure and label X,Y, and Z. A. `{:(X,Y,Z),("Backword reaction","Forward reacton","Products"):}`B. `{:(X,Y,Z),("Backword reaction","Forward reacton","Equilibrium"):}`C. `{:(X,Y,Z),("Reversible reaction","Irreversible reacton","Equilibrium"):}`D. `{:(X,Y,Z),("Forward reaction","Forward reaction","Backward reaction"):}` |
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Answer» Correct Answer - B The concentration of reactants decreacse and that of products increases with time. Rate of reaction increases with time. At equilibrium, `R_(f)=R_(b)` |
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| 1173. |
For the reaction : `PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g)`:A. Equal volumes of `PCl_(5),PCl_(3)andCl_(2)` are present.B. Equal masses of `PCl_(5),PCl_(3)andCl_(2)` are present.C. The concentrations of `PCl_(5),PCl_(3)andCl_(2)` become constant.D. Reaction comes to a stop. |
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Answer» Correct Answer - C The concentrations of reactants and products become constant at equilibrium. |
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| 1174. |
If the value of equilibrium constant `K_(c)` for the reaction, `N_(2)+3H_(2)hArr2NH_(3)` is 7. The equilibrium constant for the reaction `2N_(2)+6H_(2)hArr4NH_(3)` will beA. 49B. 7C. 14D. 28 |
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Answer» Correct Answer - A If reaction is multiplied by 2, the equilibrium constant becomes square of the previous value. `K=7^(2)=49` |
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| 1175. |
Which of the following correctly depicts the attainment of equilibrium for the reaction: `N_2(g)+3H_2(g)hArr2NH_3(g)` beginning with stoichiometric amounts of `N_2(g)` and `H_2(g)` and no `NH_3(g)`.A. B. C. D. |
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Answer» Correct Answer - B In the beginning (when time is zero), the concentrations of `N_2` and `H_2` are maximum and the concentration of `NH_3` is zero (because no `NH_3` is present in the container). When `N_2` and `H_2` react, their concentrations decrease with time while the concentration of `NH_3` increase due to its formation. When equilibrium is attained, there is no net reaction. Thus, the concentrations of `N_2`, `H_2`, and `NH_3` become constant. |
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| 1176. |
Equilibrium constant `K_(c)` is expreseed as the concentration of products divided by reactants , each term raised to the stoichiometric coefficient for reaction `aA+bB hArr cC+ dD` `K_(c) = [[C]^(c)[D]^(d)]/[[A]^(a)[B]^(b)] ("unit of concentration are " mol//L)` Answer the following on the basis of above paragrah (i) Write the equilibrium constnat for the reaction `N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g)` (ii) Write the units of equilibrium constant for the reaction `N_(2) (g) +O_(2) (g) hArr 2NO (g)` (iii) What will be the effect of catalyst on the euilibrium constant : |
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Answer» `(i) K_(c) =[[NH_(3)]^(2)]/[[N_(2)][H_(2)]^(3)]` `(ii) K_(c) =((mol L^(-1))^(2))/((mol L^(-1))(mol L^(-1)))` Equilibrium constant has no units . (iii) A catalyst does not influence the reaction in a state of equilibrium . Therefore equilibrium constant has no units. |
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| 1177. |
Calculate the PH of solution in which 0.37 g of Ca(OH)2 dissolve into water to give 500 ml of solution. |
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Answer» moality of Ca(OH)2 = 0.37/74x1000/500 = 0.05M PH = - log(H+ ) =- log( 0.05) =0.301 |
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| 1178. |
For the reaction `H_(2)(g)+I_(2)(g)hArr2HI(g)` the equilibrium constant `K_(p)` changes withA. CatalystB. TemperatueC. Amounts of `H_(2)and I_(2)`D. Amount of HI |
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Answer» Correct Answer - B |
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| 1179. |
The initial molar concentration of the reactants A and B were 0.1 M and 0.2 M respectively in the following reaction `A+B hArr 2C` When equilibrium was attained the concentration of A in the reaction mixture was found to be 0.06M . Calculate the equilibrium constant. |
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Answer» Correct Answer - `K_(c) =0.67` the number of moles of A initially present =0.1 M `The number of moles of A left =0.06 M `:.` The number of moles of A reacted =0.1 -0.06 =0.04 M According to the equation 0.04 M of A will react with 0.04 M of B to form 0.08 M of C. Therefore , the molar concentration of different species before the reaction and at the equilibrium point is : `{:(,A ,+,B,hArr,2C),("Initial molar conc.",0.1 -0.04,,0.2-0.04,,2xx0.04),("Equilibrium molar conc.",0.06,,0.16,,0.08):}` Applying Law of chemical equilibrium `K_(C)=[[C]^(2)]/[[A][B]]=(0.08 xx 0.08)/(0.06xx0.16)=0.67` |
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| 1180. |
The aqueous solution of which of the salts has pH close to 7 ?A. `FeCl_(3)`B. `CH_(3)CO ON a`C. `CH_(3)COONH_(4)`D. KCN |
| Answer» Correct Answer - C | |
| 1181. |
Calculate the `pH` of the following solutions: a. `2 g` of `TlOH` dissolved in water to give `2` litre of solution. b. `0.3 g` of `Ca(OH)_(2)` dissolved in water to give `500 mL` of solution. c. `0.3 g` of `NaOH` dissolved in water to give `200 mL` of solution. d. `1 mL` of `13.6 M HCl` is duluted with water to give `1` litre of solution. |
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Answer» Correct Answer - a) `11.65` , b) `12.21`, c) `12.57` , d) `1.87` (a) For 2g of TlOH dissolved in water to give 2 L of solution: `[TIOH_((aq))] = (2)/(2)g//L` `= 2/2 xx 1/(221) M` `= 1/(221) M` `TlOH_((aq)) rarr Tl_((aq))^(+) + OH_((aq))^(-)` `[OH_((aq))^(-)] = [TlOH_((aq))] = 1/(221) M` `K_(w) = [H^(+)] [OH^(-)]` `10^(-4) = [H^(+)] ((1)/(221))` `221 xx 10^(-14) = [H^(+)]` `rArr pH = - "log" [H^(+)] = - "log" (221 xx 10^(-14))` `= - "log" (221 xx 10^(-12))` `= 11.65` (b) For `0.3 g` of `Ca(OH)_(2)` dissolved in water to give mL of solution: `Ca(OH)_(2) rarr Ca^(2) + 2OH^(-)` `[Ca(OH)_(2)] = 0.3 x (1000)/(500) = 0.6 M` `[OH_((aq))^(-)] = 2 xx [Ca(OH)_(2aq)] = 2 xx 0.6` `= 1.2 M` `[H^(+)] = (K_(w))/([OH_(aq)^(-)])` `= (10 - 14) M` `= 0.833 xx 10^(-14)` `pH = - "log" (0.833 xx 10^(-14))` `= - "log" (8.33 xx 10^(-13))` `= (-0.902 + 13)` `= 12.098` (c) For `0.3 g` of `NaOH` dissolved in water to give 200 mL of solution. `NaOH rarr Na_((aq))^(+) + OH_((aq))^(-)` `[NaOH] = 0.3 xx (1000)/(200) = 1.5 M` `[OH_(aq)^(-)] = 1.5 M` Then, `[H^(+)] = (10^(-14))/(1.5)` `= 6.66 xx 10^(-13)` `pH =- "log" (6.66 xx 10^(-13))` `= 12.18` (d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution: `13.6 ×x 1 mL = M_(2) × 1000 mL` (Before dilution) (After dilution) `13.6 ×x 10^(-3) = M_(2) × 1L` `[H^(+)] = 1.36 × 10^(-2)` `pH = - "log" (1.36 × 10^(-2))` `= (– 0.1335 + 2)` `= 1.866 ∼ 1.87` |
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| 1182. |
Give the mathematical expression for the equlibrium constant `(K_(c) " and " K_(p))` for the ractions `(i) H_(2)(g) + I_(2)(g) hArr2HI(g)` `(ii) N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g)` `(iii) PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` |
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Answer» `(i) " "K_(c) =([HI]^(2))/([H_(2)][I_(2)]) , K_(p) = K_(c)(RT)Deltang=K_(c)(RT)^(2-2)=K_(c)` `(ii)" "K_(c)= ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3)), K_(p)=K_(c)(RT)Deltang=K_(c)(RT)^(2-4) = K_(c)(RT)^(-2)` `(iii) " "K_(c) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]), K_(p)=K_(c)(RT)Deltang=K_(c)(RT)Deltang = K_(c)(RT)^(2-1)=K_(c)(RT)` |
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| 1183. |
Which of the following salts in aqueous solution has the lowest pH value?A. `NaCIO`B. `NaCIO_(4)`C. `NaCIO_(4)`D. `NaCIO_(2)` |
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Answer» Correct Answer - B All the salt dissolve in water to form corresponding acids `NaCIO +H_(2)O to NaOH +HCIO` `NaCIO_(4) +H_(2)O to NaOH +HCIO_(4)` `NaCIO_(3) +H_(2)O to NaOH + HCIO_(3)` `NaCIO_(2) +H_(2)O to NaOH + HCIO_(2)` Since `HCIO_(4)` is the strongest acid the pH of the aqueous solution of the salt is the lowest . |
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| 1184. |
Acetic acid is highly soluble in water but still a weak electrolyte. Why? |
| Answer» The strength of the electrolyte does not depend upon its amount present in solution but on its ionisation in solution Since acetic acid is a weak (organic acid) , it is ionised only to small extent. Terefore it is a weak electrolyte. | |
| 1185. |
How will you account for the following : (a) Clothes dry quicker on a windy day We sweat more on a humid day? |
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Answer» (a) When wet clothes are spread on a stand the water evaporates and the surrounding air tends to get saturated thus hampering the process of drying . On a windy day when breeze blows the nearby wet air is replaced by dry air which helps the process of evaporation further. Thus clothes dry quicker when there is a bredze. (b) the explanation is the same as given above. On a humid day the air is already saturated with water vapour. This means that the water that comes out of the pores of the body as sweat does not vaporise. This will result in greater sweating on a humid day. |
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| 1186. |
The aqueous solution of following salt will have the lowest pHA. NaClOB. `NaClO_(4)`C. `NaClO_(3)`D. `NaClO_(2)` |
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Answer» Correct Answer - B `NaClO+H_(2)O rarr NaOH+HClO` `NaClO_(4) +H_(2)O rarr NaOH+HClO_(4)` `NaClO_(3) +H_(2)O rarr NaOH+HClO_(3)` `NaClO_(2) +H_(2)O rarr NaOH+HClO_(2)` As `HClO_(4)` is the strongest, its solution will have the lowest pH . |
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| 1187. |
2 g of NaOH has been dissolved into 500 ml of solution. Find: (a) Molarity of the solution (b) OH- ion concentration (c) PH of the solution |
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Answer» (a) M = Wx1000/MxV = 2x1000/40x500 [OH- ]=0.1M (b) [H+ ]] = 10-13 ( c) PH= 13 |
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| 1188. |
The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is `1.0xx10^(-9)M`. If 10 mL of this solution is added to 5 mL of 0.04 M solution of `FeSO_(4), MnCl_(2), ZnCl_(2) and CdCl_(2)`, in which solutions precipitation will take place ? Given `K_(sp)` for `Fes=6.3xx10^(-18), MnS = 2.5xx10^(-13), ZnS=1.6xx10^(-24) and CdS=8.0xx10^(-27)`. |
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Answer» Precipitation will take place in the solution for which ionic product is greater than solubility product. As 1o mL of solution containing `S^(2-)` ion is mixed with 5 mL of metal salt solution, after mixing `[S^(2-)]=1.0xx10^(-19)xx(10)/(15)=6.67 xx 10^(-20)`. `[Fe^(2+)]=[Mn^(2+)]=[Zn^(2+)]=[Cd^(2+)]=(5)/(15)xx0.04 = 1.33xx10^(-2)M` `:.` Ionic product for each of these will be `=[M^(2+)][S^(2-)]=(1.33xx10^(-2))(6.67xx10^(-20))=8.87xx10^(-22)` As this is greater than the solubility product of ZnS and CdS, therefore , `ZnCl_(2) and CdCl_(2)` solutions will be precipitated. |
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| 1189. |
If a neutral solution has `pK_(w) =13.36` at `50^(@)C`, then pH of the solution isA. `6.63`B. `7.0`C. `7.13`D. `6.0` |
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Answer» Correct Answer - A For pure water or any netural aqueous solution, `pH=(pK_(w))/(2)=(13.26)/(2)=6.63` |
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| 1190. |
The pH of a solution of hydrochloric acid is 4. The molarity of this solution isA. `4.0`B. `0.4`C. `0.0001D. `0.04` |
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Answer» Correct Answer - C `(H^(+)) = 10^(-4) M` Molarity =0.0001 M |
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| 1191. |
`BF_(3) ` does not have proton but still acts as an acid and reacts with `NH_(3)`. Why is it so ? What type of bond is formed between the two ? |
| Answer» `BF_(3)` is electron deficient and hence acts as Lewis acid. : `NH_(3)` has one lone pair which it can donate to `BF_(3)` and form a coordinate bond. Hence, `NH_(3)` acts as a Lewis base `(H_(3)N : rarr BF_(3))`. | |
| 1192. |
The concentration of suphide ion in `0.1 M HCl` solution saturated with hydrogen sulphide is `1.0xx10^(-19)M`. If `10 mL` of this is added to `5 mL` of `0.04 M` solution of the following: `FeSO_(4), MnCl_(2), ZnCl_(z)` and `CdCl_(2)`. In which of these solutions precipitation will take place? |
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Answer» Precipitation will take place in the solution for which ionic product is greater than solubility product. As 10 mL of solution containing `S^(2-)` ion is mixed with 5 mL of metal salt solution , after mixing. `[S^(2-)] =1.0 xx 10^(-19) xx (10)/(15) =6.67 xx 10^(-20)` `[M^(2+)] i.e., [Fe^(2+)] =[Mn^(2+)] =[Zn^(2+)] =[Cd^(2+)] =(5)/(15) xx 0.04` `=1.33 xx 10^(-2)M` lt brgt Ionic product of `[M^(2+)] " and "[S^(2-)] =(1.33 xx 10^(-2)) xx (6.67 xx 10^(-20)) =8.87 xx 10^(-22)` As this is more than the `K_(sp)` of ZnS and CdS, both `ZnCI_(2) " and " CdCI_(2)` solution be precipitated. |
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| 1193. |
If a neutral solution has `pK_(w) = 13.36 ` at `50^(@) C`, then pH of the solution isA. 6.68B. 7C. 7.63D. None of these |
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Answer» Correct Answer - A `pK_(w)=pH+pOH`. As `[H^(+)]=[OH^(+)]=[OH^(-)], pK_(w)=2xxpH` or, `pH = 13.36//2=6.68` |
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| 1194. |
Which one of the following will decrease the pH of 50 mL of 0.01 M hydrochloric acid ?A. Addition of 50 mL of 0.01 M HClB. Addition of 50 mL of 0.002 M HClC. Addition of 150 mL of 0.002 M HClD. Addition of 5 mL of 1 M HCl |
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Answer» Correct Answer - D For 0.01 M HCl, `[H^(+)]=10^(-2)M`, pH=2 (a) `N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))` `50xx0.01+50xx0.01=N_(3)(50+50)` or `N_(3)=(0.5+0.5)/(100)=10^(-2), pH=2` (No change) (b) `50xx0.01+50xx0.02=N_(3)(50+50)` or, `N_(3)=(0.5+0.1)/(100)=6xx10^(-3)` `pH=-log(6xx10^(-3))=2.22` (pH increases) (c) `50xx0.01+150xx0.002=N_(3)(50+150)` or, `N_(3)=(0.5+0.3)/(200)=4xx10^(-3)` `pH=-log(4xx10^(-3))=2.39` (pH increases) (d) `50xx0.01+5xx1=N_(3)(50+5)` or, `N_(3)=(0.5+5)/(55)=(55)/(5)=10^(-1)` pH = 1 (pH decreases) |
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| 1195. |
`BF_(3)` does not have proton but still acts as an acid and reacts with `NH_(3)`. Why is it so? What type of bond is formed between the two ? |
| Answer» `BF_(3)` is an electron deficient compound and is therfore, a Lewis acid. It takes up an electron pair form `NH_(3)` which is a Lewis base. The two combine to form a co-ordinate or dative bond `NH_(3)toBF_(3)` | |
| 1196. |
Assuming complete dissociation, calculate the pH of the following solutions : (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH |
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Answer» (a) `HCl + aq rarr H^(+) + Cl^(-) , :. [H^(+)]=[HCl]=3xx10^(-3)M, pH = - log (3xx10^(-3))=2.52` (b) `NaOH + aq rarr Na^(+) + OH^(-)= 5xx10^(-3) M, [H^(+)]=10^(-14)//(5xx10^(-3))=2xx10^(-12)M` `pH = - log (2xx10^(-12))=11.70` (c) `HBr + aq rarr H^(+) + Br^(-), :. [H^(+)]=2xx10^(-3)M, pH = - log (2xx 10^(-3))=2.70` (d) `KOH + aq rarr K^(+) + OH^(-) , :. [OH^(-)]=2xx10^(-3)M,[H^(+)]=10^(-14)//(2xx10^(-3))=5xx10^(-12)` `pH = - log (5xx10^(-12))=11.30` |
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| 1197. |
The `pH` of `0.005 M` codenine `(C_(18)H_(21)NO_(3))` solution is `9.95`. Calculate its ionisation constant and `pK_(b)`. |
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Answer» Correct Answer - `K_(b) = 1.6 xx 10^(-6), pK_(b) = 5.8` `c = 0.005` `pH = 9.95` `pOH = 4.05` `pH = – log (4.105)` `4.05 = -"log" [OH^(-)]` `[OH^(-)] = 8.91 xx 10^(-5)` `calpha = 8.91 xx 10^(-5)` `alpha = (8.91 xx 10^(-5))/(5 xx 10^(-3)) = 1.782 xx 10^(-2)` Thus, `K_(b) = calpha^(2)` `= 0.005 xx (1.782)^(2) xx 10^(-4)` `= 0.005 xx 3.1755 xx 10^(-4)` `= 0.0158 xx 10^(-4)` `K_(b) = 1.58 xx 10^(-6)` ltbr gt `Pk_(b) = - "log" K_(b)` `= -"log"(1.58 xx 10^(-6))` `= 5.80` |
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| 1198. |
Assuming complete dissociation, calculate the `pH` of the following solutions, a. `0.003 M HCl, b. 0.005 M NaOH`, c. `0.002 M HBr, d. 0.002 M KOH` |
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Answer» Correct Answer - a) `2.52` , b) `11.70` , c) `2.70` , d) `11.30` (i) `0.003 MHCl`: `H_(2)O + HCl harr H_(3)O^(+) + Cl^(-)` Since HCl is completelly ionized `[H_(3)O^(+)] = [HCl]` `rArr [H_(3)O^(+)] = 0.003` Now, `pH = - "log"[H_(3)O^(+)] = -"log" (.003)` `= 2.52` Hence, the pH of the solution is 2.52. (ii) 0.005MNaOH: `NaOH_((aq)) harr Na_((aq))^(+) + HO_((aq))^(-)` `[HO^(-)] = [NaOH]` `rArr [HO^(-)] = .005` `pOH = -"log"[HO^(-)] = -"log"(.005)` `pOH = 2.30` `:. pH = 14-2.30` `= 11.70` Hence, the pH of the solution is `11.70` (ii) `0.002 HBr`: `HBr + H_(1)O harr H_(3)O^(+) + Br^(-)` `[H_(3)O^(+)] = [HBr]` `rArr [H_(3)O^(+)] = .002` `:. pH = -"log" [H_(3)O^(+)]` `= - "log" (0.002)` `= 2.69` Hence the pH of the solution is `2.69`. (iv) `0.002 M KOH`: `KOH_((aq)) harr K_((aq))^(+) OH_((aq))^(-)` `[OH^(-)] = [KOH]` `rArr [OH^(-)] = .002` Now, `pOH = -"log" [OH^(-)]` `= 2.69` `:. pH = 14 - 2.69` `= 11.31` Hence the pH of the solution `11.31` |
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| 1199. |
Passing `H_(2)S` gas into a mixture of `Mn^(2+), Ni^(2+), Cu^(2+) and Hg^(2+)` ions in an acidified aqueous solution precipitatesA. CuS and HgSB. MnS and CuSC. MnS and NiSD. NiS and HgS |
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Answer» Correct Answer - A `H_(2)S` in acidic medium precipitates radicals of Group 2 of quantitative analysis, i.e. `Cu^(2+), Hg^(2+)`, (and `Pb^(2+), Bi^(3+), Cd^(2+), As^(3+), Sb^(3+), Sn^(2+)`). |
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| 1200. |
The pH of a `0.005 M H_(2)SO_(4)` solution isA. `3.3`B. `5.0`C. `2.0`D. `4.0` |
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Answer» Correct Answer - C Since `H_(2)SO_(4)` is a strong acid and 1 mol of `H_(2)SO_(4)` gives 2 mol of `H^(+)` ions, we have `C_(H^(+))=2xxC_H_(2)SO_(4))=(2)(0.005 M)` `=0.01` `=1xx10^(-2)M` `pH= -"1og"C_(H^(+))/(mol L^(-1))= -1og(1xx10^(-2))` `= 2.0` |
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