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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.
| 101. |
The solubility of `Pb(OH)_(2)` in water is `6.7xx10^(-6)M`. Calculate the solubility of `Pb(OH)_(2)` in a buffer solution of pH = 8 |
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Answer» `Pb(OH)_(2) hArr Pb^(2) + 2 OH^(-)` `:. K_(sp)=[Pb^(2+)][OH^(-)]^(2)=s xx (2s)^(2)=4 s^(3) = 4xx(6.7xx10^(-6))^(3) = 1.20 xx 10^(-15)` In a solution with pH = 8, `[H^(+)]=10^(-8) and [OH]^(-) = 10^(-6)` `:. 1.2xx10^(-15)=[Pb^(2+)]xx(10^(-6))^(2) or [Pb^(2+)]=(1.2xx10^(-15))/((10^(-6))^(2))=1.2xx10^(-3)M` |
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| 102. |
Calculate the solubility product of silver bromide if the solubility of the salt in saturated solution is `5.7xx10^(-7)` moles/litre. |
| Answer» Correct Answer - `3.25xx10^(-13)` | |
| 103. |
The solubility product of `Al(OH)_(3) ` is `2.7xx10^(-11)`. Calculate its solubility in g `L^(-1) ` and also find out pH of this solution . (Atomic mass of Al = 27 u ). |
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Answer» Suppose the solubility is S mol `L^(-1)`. Then `{:(Al(OH)_(3),hArr,Al^(3+),+,3OH^(-),,),(,,S,,3 S,,):}` `K_(sp)= S xx (3 s)^(3) = 27 S^(4)` `:. 27S^(4)=2.7xx10^(-11) or S^(4) = 10^(-12) or S=10^(-3) ` mol `L^(-1)` Molar mass of `Al(OH)_(3) ` in g `L^(-1) = 10^(-3)xx78=7.8xx10^(-2) g L^(-1)` `[OH^(-)]=3 S = 3 xx 10^(-3) ` mol `L^(-1)` `:. pOH = - log (3xx10^(-3))=3-0.4771=3.5229` pH = 14 - 2.5229 = 11.4771. |
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| 104. |
`pH=7.40, K_(1) "of" H_(2)CO_(3)=4.5xx10^(-7)`. What will be the ratio of `[HCO_(3)^(-)]` to `[H_(2)CO_(3)]` ? |
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Answer» `H_(2)CO_(3) hArr H^(+) + HCO_(3)^(-)` `K_(1) = ([H^(+)][HCO_(3)^(-)])/([H_(2)CO_(3)])` `:. ([HCO_(3)^(-)])/([H_(2)CO_(3)])=(K_(1))/([H^(+)])` pH = 7.40 means - log `[H^(+)] = 7.4` or `log [H^(+)] = - 7.4 bar(8) . 6 or [H^(+)] = 3.981 xx 10^(-8)` `:. ([HCO_(3)^(-)])/([H_(2)CO_(3)]) = (4.5xx10^(-7))/(3.981xx10^(-8))=11.3` |
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| 105. |
Write the conjugate bases for the following Brddotonsted acids (a) `HF` (b) `H_(2)SO_(4)` (c) `HCO_(3)^(Θ)` |
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Answer» Correct Answer - `F^(-),HSO_(4^(-)),CO_(3^(2-))` The table below lists the conjugate bases for the given Bronsted acids. `{:("Bronsted acid",,"Conjugate base"),(HF,,F^(-)),(H_(2)SO_(4),,HSO_(4)^(-)),(HCO_(3)^(-),,CO_(3)^(2-)):}` |
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| 106. |
A buffer solution is prepared in which the concentration of `NH_(3)` is 0.30 M and the concentration of `NH_(4)^(+)` is 0.20 M. If the equilibrium constant, `K_(b)` for `NH_(3)` equals `1.8xx10^(-5)`, what is the pH of this solution ?A. 8.73B. 9.08C. 9.43D. 11.72 |
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Answer» Correct Answer - C `p_(OH)=pK_(b)+log.(["Salt"])/(["Base"])=4.74+log.(0.20)/(0.30)` `=4.74+(0.301-0.4.74)=4.74-0.176` `=4.56` `:. pH = 14 - 4.56 = 9.44` |
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| 107. |
Blood is a buffer of `H_(2)CO_(3) and[HCO_(3)^(-)] ` with pH = 7.40. Given `K_(1) ` of `H_(2)CO_(3) = 4.5xx10^(-7)`. What will be the ratio of `[HCO_(3)^(-)]` to `[H_(2)CO_(3)]` in the blood ? |
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Answer» `H_(2)CO_(3)hArr H^(+)+HCO_(3)^(-), K_(1)=([H^(+)][HCO_(3)^(-)])/([H_(2)CO_(3)]) or ([HCO_(3)^(-)])/([H_(2)CO_(3)])=(K_(1))/([H^(+)])` `pH = 7.40` means - log `[H^(+)] = 7.4` or log `[H^(+)] = -7.4= bar(8).6 or [H^(+)]=3.981 xx 10^(-8)` `([HCO_(3)^(-1)])/([H_(2)CO_(3)])=(4.5xx10^(-7))/(3.981xx10^(-8))=11.3`. |
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| 108. |
A weak acid of dissociation constant `01^(-5)` is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralization of the acid will beA. 5 log 2 - log 3B. 5 - log 2C. 5 - log 3D. 5 - log 6 |
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Answer» Correct Answer - B On partial neutralization of the weak acid, salt is formed. Hence, it becomes a buffer `pH = pK_(a) + log. (["Salt"])/(["Acid"])` 1/3rd neutralisation of the acid means out of 1 mole of the acid , salt formed = 1/3 mole and acid left = 2/3 mole Hence, `pH = - log(10^(-5))+log.(1//3)/(2//3)` `=5+ log .(1)/(2) = 5 - log2`. |
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| 109. |
The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and equilibrium is represented as: `CH_(3)COOH_((l))+C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5_((aq)))+H_(2)O_((l))` (`a`) Write the concentration ratio (reaction quotient), `Q_(e)`, for this reaction. Note that water is not in excess and is not a solvent in this reaction. (`b`) At `293 K`, if one starts with `1.00 "mole"` of acetic acid and `0.180` of ethanol, there is `0.171 "mole"` of ehtyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (`c`) Starting with `0.500 "mole"` of ethanol and `1.000 "mole"` of acetic acid and maintaining it at `293 K`, `0.214 "mole"` of ethyl acetate is found after some time. Has equilibrium been reached? |
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Answer» Correct Answer - (i) `[CH_(3)COOC_(2)H_(5)][H_(2)O]//[CH_(2)COOH][C_(2)H_(5)OH]` (ii) `3.92` , (iii) value of `Q_(c)` is less than `K_(c)` therefore equilibrium is not attained. (i) Reaction quotient, `Q_(c) = ([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])` (ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess. The given reaction is : `{:(,CH_(3)COOH_((l)),+,C_(2)H_(5)OH_((l)),harr,CH_(3)COOC_(2)H_(5(l)),,H_(2)O_((l))),("Initial conc.",1/(V)M,,(0.18)/(V)M,,0,,0),("At equilibrium",(1-0.171)/(V),,(0.18-0.171)/(V),,(0.171)/(V)M,,(0.171)/(V)M),(,=(0.829)/(V)M,,=(0.009)/(V) M,,,,):}` Therefore, equilibrium costant for the given reaction is : `K_(c) = ([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])` `= ((0.171)/(V)xx (0.171)/(v))/((0.829)/(V) xx (0.009)/(V)) = 3.919` `= 3.92` (approximately) (iii)Let the volume of the reaction mixture be V. `{:(,CH_(3)COOH_((l)),+,C_(2)H_(5)OH_((l)),harr,CH_(3)COOC_(2)H_(5(l)),+,H_(2)O_((l))),("Initial conc.",(1.0)/(V)M,,(0.5)/(V)M,,0,,0),("After some time",(10-0.214)/(V),,(0.5-0.214)/(V),,(0.214)/(V)M,,(0.214)/(V)M),(,=(0.786)/(V)M,,=(0.286)/(V)M,,,,):}` Therefore, the reaction quotient is, `Q_(c) = ([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])` `= ((0.214)/(V)xx (0.214)/(V))/((0.786)/(V) xx (0.286)/(V))` `= 0.2037` `= 0204` (approximately) Since `Q_(c) lt K_(c)` equilibrium has not been reached. |
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| 110. |
For the reaction, A (soild) + 2B (Gas) ➝ 3C (solid) + 2D (gas) :(a) Kp = Kc(RT)°(b) Kp = Kc(R2T2)(c) Kp = Kc(RT)(d) Kp = Kc(R-2T-2) |
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Answer» (a) Kp = Kc(RT)° |
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| 111. |
What will be the amount of `(NH_(4))_(2)SO_(4)` (in g) which must be added to 500 mL of 0.2 M `NH_(4)OH` to yield a solution of pH 9.35? `["Given," pK_(a)"of "NH_(4)^(+)=9.26,pK_(b)NH_(4)OH=14-pK_(a)(NH_(4)^(+))]`A. 5.35B. 6.47C. 10.03D. 7.34 |
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Answer» Correct Answer - A `pK_(a)"of"NH_(4)^(+)=9.26` Hence, `pK_(b)"of"NH_(4)OH=14-9.26=4.74` `pOH=14-9.35=4.65` Each 1 mol of `(NH_(4))_(2)SO_(4)` furnishes 2 moles of `NH_(4)^(+)` in solution. Let `[(NH_(4))_(2)SO_(4)]=x"mol L"^(-1)` `[NH_(4)^(+)]"2 x mol L"^(-1)` `[NH_(4)OH]=0.2"mol L"^(-1)` using Henderson - Hasselbalc equation, `pOH=pK_(b)+"log"([NH_(4)^(+)])/([NH_(4)OH])` `4.65=4.74+log((2x)/(0.2))` This gives `x=0.081"mol L"^(-1)` `=0.0405` mol in 500 mL (as required) `=0.0405xx132=5.35` g |
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| 112. |
How much KOH must be dissolved in one litre of solution to get a pH of 12 at `25^(@)C`? |
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Answer» Correct Answer - `0.56 g L^(-1)` `pH =- log [H_(3)O^(+)] " or " 12 =- log [H_(3)O^(+)]` `log[1//H_(3)O^(+)] =12 " or " [1//H_(3)O^(+)] = " Antilog " 12 =10^(12)` `:. " "[H_(3)O^(+)] =10^(-12) M` `[OH^(-)] =(K_(w))/[[H_(3)O^(+)]] =((1xx 10^(-14) M^(2)))/((10^(-12)M)) =10^(-2) M` Amount of KOH`//`litre of the solution `=10^(-2) " mol "L^(-1) xx `(Molar mass of KOH) `=(10^(-2) " mol " L^(-1)) xx ( 56 g " mol"^(-1)) =0.56 g L^(-1)` |
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| 113. |
Calculate the amount of `(NH_(4))_(2)SO_(4)` in g which must be added to 500 mL of 0.2 M `NH_(3)` to yield a solution of pH = 9.35, `K_(b)` for `NH_(3) = 1.78xx10^(-5)`. |
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Answer» As it is a basic buffer, `pOH=pK_(b) + log. (["Salt"])/(["Base"])=-log K_(b) + log. ([NH_(4)^(+)])/([NH_(4)OH])` As `pH = 9.35, :. pOH = 14 - 9.35 = 4.65` Millimoles of `NH_(4)OH` in solution `= 0.2xx500 = 100` Suppose millimoles of `NH_(4)^(+)` to be added = x `:. 4.65 = - log (1.78xx10^(-5))+ log. (x//500)/(100//500)= (5 - 0.2504) + log .(x)/(100)` or `log. (x)/(100) = - 0.0996 = bar(1).0004 ~= 0.1 or log x = 2.1 or x = 125.9` `:.` Millimoles of `(NH_(4))_(2)SO_(4)` to be added `=(125.9)/(2) = 62.95` (`:. 1 "millimole of " (NH_(4))_(2)SO_(4)-=2 "millimoles of " NH_(4)^(+)`) `:.` Mass of `(NH_(4))_(2)SO_(4)` to be added `=(62.95xx10^(-3) "moles" ) (132 g "mol"^(-1))=8.3094 g`. |
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| 114. |
How many grams of NaOH must be dissolved in one litre of the solution to give it a pH value of 12 ? |
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Answer» `pH=-log[H_(3)O^(+)] :. Log[H_(3)O^(+)]=-pH=-12` `:. [H_(3)O^(+)]`= antilog (-12) = antilog `(bar(12))=10^(-12)` g ions/litre But we know that `[H_(3)O^(+)][OH^(-)]=K_(w)=10^(-14) :. [OH^(-)]=(K_(w))/([H_(3)O^(+)])=(10^(-14))/(10^(-12))=10^(-2)` g ions/litre Since NaOH is a strong electrolyte, it undergoes complete ionization as : `NaOH rarr Na^(+)+OH^(-)` `:. [OH^(-)]=[NaOH]=10^(-2)M = 10^(-2) "mol" L^(-1)` Mol. mass of NaOH = 40 `:.` Amount of NaOH dissolved per litre `=10^(-2) xx 40 = 0.4 g` |
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| 115. |
The solubility of `BaSO_(4)` in water is `2.42xx10^(-3)g L^(-1)` at 298 K. The value of its solubility product `(K_(sp))` will be (Given molar mass of `BaSO_(4)=233 g L^(-1)`)A. `1.08xx10^(-10) "mol"^(2)L^(-2)`B. `1.08xx10^(-12) "mol"^(2)L^(-2)`C. `1.08xx10^(-14) "mol"^(2)L^(-2)`D. `1.08xx10^(-8) "mol"^(2)L^(-2)` |
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Answer» Correct Answer - A Solubility (S) `=2.42xx10^(-3) g L^(-1)` `=(2.42xx10^(-3))/(233) "mol" L^(-1)` `=0.0104xx10^(-3) "mol" L^(-1)` `=1.04xx10^(-5) "mol" L^(-1)` `{:(BaSO_(4) ,hArr,Ba^(2+),+,SO_(4)^(2-)),(,,S,,S):}` `K_(sp)=[Ba^(2+)][SO_(4)^(2-)]=SxxS=S^(2)` `=(1.04xx10^(-5) "mol" L^(-1))^(2)` `=1.08xx10^(-10) "mol"^(2)L^(-2)` |
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| 116. |
What is minimum concentration of `SO_(4)^(2-)` required to precipitate `BaSO_(4)` in solution containing `1 xx 10^(-4)` mole of `Ba^(2+)` ? (`K_(sp)` of `BaSO_(4) = 4 xx 10^(-10)`)A. `4xx10^(-10)M`B. `2xx10^(-10)M`C. `4xx10^(-6)M`D. `2xx10^(-3)M` |
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Answer» Correct Answer - C `underset(Conc.: )(BASO_(4))hArrunderset(1xx10^(-4))(Ba^(2+))+underset(s)(SO_(4)^(2-))` `K_(sp)=4xx10^(-10)` `4xx10^(10)=1xx10^(-4)xxs` `(s=4xx10^(-10))/(1xx10^(-4))=4xx10^(-6)M` |
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| 117. |
On adding 0.1 M solution each of `Ag^(+), Ba^(2+), Ca^(2+)` ions in a `Na_(2)SO_(4)` solution, species first precipitated is `(K_(sp)BaSO_(4)=10^(-11), K_(sp) CaSO_(4)=10^(-6), K_(sp)Ag_(2)SO_(4)=10^(-5))`A. `Ag_(2)SO_(4)`B. `BaSO_(4)`C. `CaSO_(4)`D. all of these |
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Answer» Correct Answer - B `K_(sp)` for `Ag_(2)SO_(4)=[Ag^(+)]^(2)[SO_(4)^(2-)]` `:. [SO_(4)^(2-)]` for precipitation of `Ag_(2)SO_(4) lt (K_(sp))/([Ag^(+)]^(2))=(10^(-5))/((0.1)^(2))=10^(-3)M` Similarly, `[SO_(4)^(2-)]` for precipitation of `BaSO_(4) lt (K_(sp))/([Ba^(2+)])=(10^(-11))/(0.1)=10^(-10)M` `[SO_(4)^(2-)]` for precipitation of `CaSO_(4) lt (K_(sp))/([Ca^(2+)])=(10^(-6))/(0.1)=10^(-5)M` Thus, minimum `[SO_(4)^(2-)]` reuired for precipitaion is for `BaSO_(4)`. |
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| 118. |
Mixing of solutions of `BaCl_(2) and Na_(2)SO_(4)` results in the formation of a precipitate of `BaSO_(4)` only if.......greater than......... . |
| Answer» Correct Answer - ionic product, solubility product | |
| 119. |
STATEMENT-1: Solubility of `BaSO_(4)` in 0.1 M `Na_(2)SO_(4)is 10^(-9)` M hence its `K_(sp)` is `10^(-18).` STATEMENT-2: In aqueous solution, solubility product of `BaSO_(4)=S^(2).` (Where S is solubility of `BaSO_(4)`)A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-7B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-7C. 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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| 120. |
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 10-8M HCl is not equal 8(R) HCl does not dissociates properly in very dilute solution |
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Answer» If both (A) and (R) are correct and (R) is the correct explanation of (A) |
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| 121. |
If 20 ml of `2xx10^(-5) BaCl_(2)` solution is mixed with 20 ml of `1xx10^(-5) M Na_(2)SO_(4)` solution, will a ppt. form ? `(K_(sp) "for" BaSO_(4) " is" 1.0xx10^(-10))` |
| Answer» Correct Answer - No | |
| 122. |
The equilibrium constant of given reaciton will be `HCO_(3)^(-)+H_(2)OhArrH_(2)CO_(3)+OH^(-)`A. `sqrt(K_(w))`B. `sqrt((K_(w))/(K_(a_(1))))`C. `(K_(w))/(K_(a_(2)))`D. `K_(w)K_(a_(1))` |
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Answer» Correct Answer - B |
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| 123. |
The value of `K_(c)` for the following equilibrium is `CaCO_(3(s))hArrCaO_((s))+CO_(2(g))`. Given `K_(p)=167` bar at 1073 K.A. `1.896" mol L"^(-1)`B. `4.38xx10^(-4)" mol L"^(-1)`C. `6.3xx10^(-4)" mol L"^(-1)`D. `6.626" mol L"^(-1)` |
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Answer» Correct Answer - A `K_(p)=K_(c)(RT)^(Deltan)," "Deltan=1` `K_(p)=167" bar"`, `K_(c)=(167" bar")/(0.0821"L bar K"^(-1)"mol"^(-1)xx1073K)=1.896" mol L"^(-1)` |
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| 124. |
Assertion : Benzoic acid is stronger acid than acetic acid. Reason : `K_(a)` for benzoic acid is `6.5xx10^(-5)` and for acetic acid is `1.74xx10^(-5)`.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 - A | |
| 125. |
In the equilibrium, ` CaCO_(3) (s) hArr CaO (s) + CO_(2) (g) , " at 1073 K,the pressure of " CO_(2) " is found to be " 2*5 xx 10^(4) Pa.` What is the equilibrium constant of this reaaction at 1073 K ? |
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Answer» With reference to the standard state pressure of 1 bar, 1.e., `10^(5)` Pa, ` K_(p) = p_(cO_(2)) = (2*5 xx 10^(4) Pa)/(P^(0)) = ( 2.5 xx 10^(4) pa)/(10^(5) Pa) = 0.25` |
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| 126. |
Solubility product of silver bromide is `5.0xx10^(-13)`. The quantity of potassium bromide (molar mass taken as `120g mol^(-1)`) to be added to `1 L` of `0.05 M` solution of silver nitrate to start the precipitation of `AgBr` isA. `5.0 xx 10^(-8) g`B. `1.2 xx 10^(-10) g`C. `1.2 xx 10^(-9) g`D. `6.2 xx 10^(-5) g` |
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Answer» Correct Answer - C `K_(sp) =[Ag^(+)][Br^(-)] =5.0 xx 10^(-13) M^(2)` `[Ag^(+)] =0.05 M` `[Br^(-)] =((5.0xx 10^(-13)M^(2)))/((0.5 M)) =10^(-11)M` Moles of KBr =`10^(-11)M` Moles of KBr = `(10^(-11) "mol") xx (120.0 g "mol"^(-1))` `=1.20 xx 10^(-9) g` |
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| 127. |
The ionization constant of acetic acid `1.74xx10^(-5)`. Calculate the degree of dissociation of acetic acid in its `0.05 M` solution. Calculate the concentration of acetate ion in the solution and its `pH`. |
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Answer» `CH_(3)COOH hArr CH_(3)COO^(-)+H^(+)` `K_(a) [[CH_(3)COO^(-)][H^(+)]]/[[CH_(3)COOH]] =[[H^(+)]^(2)]/[[CH_(3)COOH]]` `[H^(+)] =(K_a)xx [CH_(3)COOH])^(1//2)` `=(1.74 xx 10^(-5) xx 5 xx 10^(-2))^(1//2) =9.33 xx 10^(-4)M` `pH =- log [H^(+)]=- log (9.33 xx 10^(-4))` `=(4 - log 9.33 ) = 4- 0.97 =3.03` |
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| 128. |
Calculate the partial pressure of carbon monoxide from the following data `:` `CaCO_(3)overset(Delta)(hArr) CaO(s)+CO_(2) uarr ,K(p)=8xx10^(-2)` `CO_(2)(g)+C(s) hArr 2CO(g), K_(p)=2`A. `0*2`B. `0*4`C. `1*6`D. 4 |
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Answer» Correct Answer - B For reaction (i), `K_(p) = p_(CO_(2)) = 8 xx 10^(-2) ` (Given) For reaction (ii), `K_(p) = (p_(CO)^(2))/(p_(CO_(2)))=2 ` ( Given) ` :. 2= (p_(CO)^(2))/(8 xx 10^(-2) ) or p_(CO) = 0.4 ` |
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| 129. |
The solubility product of `Ni(OH)_(2)` is `2.0xx10^(-15)`. The molar solubility of `Ni(OH)_(2)` in `0.1 M NaOH` solution is |
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Answer» Let the solubility of `Ni(OH)_(2)` be equal to S. Dissolution of `S "mol"// L` of `Ni(OH)_(2)` provides `S "mol"//L` of `Ni^(2+)` and `2S "mol"//L` of `OH^(-)`. But the total concentration of `OH^(-) = (0.10 + 2S) "mol"//L` because the solution already contains `0.10 "mol"//L` of from `NaOH`. `K_(sp) = 2.0 xx 10^(-15) = [Ni^(2+)][OH^(-)]^(2)` `= (S) (0.10 + 2S)^(2)` As `K_(sp)` is small, `2S lt lt 0.10` thus, `(0.10 + 2S) = 0.10` Hence, `2.0 xx 10^(-18) = S (0.10)^(2)` `S = 2.0 xx 10^(-13) M = [Ni^(2+)]` |
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| 130. |
One of the reactions that take place in producing steel from ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and `CO_(2)` `Fe O(s) + CO(g)hArr Fe (s) + CO_(2) (g) , K_(p) = 0*265 " at " 1050 K ` What are the equilibrium partial pressures of CO and `CO_(2)` at 1050 K if the intial pressures are : `p_(co) = 1*4 " atm " and p _(CO_(2))= 0*80 " atm" ?` |
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Answer» ` {:(,FeO(s),+,CO(g),hArr,Fe(s),+,CO_(2)) ,("Intial pressures",,,1*4 "atm",,,,0*80"atm"),(,,,,,,,):}` ` Q_(p) = p_(CO_(2))/ (p_(CO))= = (0*80)/1*4 = 0*571 ` As ` Q_(p) gt K_(p) ,` reaction will move in the backward direction , i.e., pressure of `CO_(2)` will decrease and that of CO will increase to attain equilibrim . Hence , if p is the decrease in pressure of ` CO_(2)` increase in pressure of CO= p ` :. "At equilibrium " , p_(CO_(2)) = (0*80- p) "atm" , p_(CO) = (1*4 + p)"atm "` ` K_(p) = p_(CO_(2))/(p_(CO)) :. 0*265 = (0*80 - p)/(1*4 +p) or 0*265 (1*4 +p)= 0*80 - p ` or ` 0* 371 + 0*265 p = 0*80 - p or 1 * 265 p = 0*49 or p = 0*339 "atm"` ` :. (p_(CO))_(eq) = 1*4 + 0*339 " atm" = 1*739 " atm" , (p_(CO_(2)))_(eq) = 0*80 - 0*339 "atm" = 0*46` |
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| 131. |
Anhydrous `AlCl_(3)` is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous solution (Ionization energy for `Al=5137 kJ "mol"^(-1), Delta H_("hydration") "for" Al^(3+) = - 4665 kJ "mol"^(-1), Delta H_("hydration")` for `Cl^(-1)=-381 kJ "mol"^(-1))`. |
| Answer» Total energy released due to hydration of ions = - 4665`-3xx381=-5808 kJ "mol"^(-1)`. This energy is greater than the ionisation energy of Al which is 5137 kJ `"mol"^(-1)`. Hence , it would be ionic in the solution . | |
| 132. |
One of the reaction that takes plece in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and `CO_(2)`. `FeO(s)+CO(g) hArr Fe(s)+CO_(2)(g), K_(p)=0.265` atm at `1050 K` What are the equilibrium partial pressure of `CO` and `CO_(2)` at `1050 K` if the partical pressure are: `p_(CO)=1.4 atm` and `p_(CO_(2))=0.80 atm`? |
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Answer» `FeO(s)+CO(g)hArr Fe(s)+CO_(2)(g)` `" Initial pressure : " " "1.4 atm " "0.8atm ` `Q_(p)=(Pco_(2))/(Pco) = ((0.8atm))/((1.4 atm)) =0.571` Since `Q_(p) gt K_(p) (0.265),` this means that the reaction will move in the backward direction to attain the equilibrium. Therefore , partial pressure of `CO_(2)` will decrease while that of CO will increase so that the equilibrium may be attained again. Let P atm be teh decrease in the partial pressure of `CO_(2)`. Therefore, the partial pressure of CO will increase by the same magnitude i.e.,p atm. `Pco_(2) =(0.8 -p) " atm " , Pco(g) =(1.4 +p ) " atm "` `"At equilibrium "" "K_(p) =(Pco_(2))/(Pco)= ((0.8-p) atm)/((1.4 +p) atm) = ((0.8 -p))/((1.4+p))` `" or "" "0.265 =((0.8 -p))/((1.4 +p))` `0.371 +0.265p= 0.8 -P or 1.265p = 0.8 -0.371 =0.429` `p=0.429//1.265 =0.339 " atm "` `(Pco)_(eq) = (1.4+0.339)=1.739 " atm "` `(Pco_(2))_(eq)=(0.8-0.339)=0.461` atm |
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| 133. |
The value of `DeltaG^(ɵ)` for the phosphorylation of glycose in glycolysis is `13.8 kJ mol^(-1)`. Find the value of `K_(c)` at `298 K` |
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Answer» `DeltaG^(ɵ) = 13.8 kJ//"mol" = 13.8 xx 10^(3) J//"mol"` Also, `DeltaG^(ɵ) = - RT"ln"K_(c)` Hence, In `K_(c) = - 13.8 xx 10^(3)J//"mol"` `(8.314 K "mol"^(-1)K^(-1) xx 298 K)` In `K_(c) = -5.569` `K_(c) = e^(-5.569)` `K_(c) = 3.81 xx 10^(-3)` |
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| 134. |
The values of `K_(sp)` of two sparingly soluble salts, Ni`(OH)_(2)` and AgCN are `2.0xx10^(-15) and 6.0xx10^(-17)`, respectively. Which salt is more soluble ? Explain. |
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Answer» Correct Answer - `S(Ni (OH)_(2))=5.8xx10^(-5) M, S (AgCN) = 7.8xx10^(-9) M, Ni (OH)_(2) ` is more soluble `{:(Ni(OH)_(2),hArr,Ni^(2+),+,2OH^(-),,,),(s,,s,,2s,,,):}` `K_(sp) = s (2s)^(2) = 4 s^(3 ) = 2.0 xx 10^(-15) `. This gives `s=5.8xx10^(-5)` mol `L^(-1)` `{:(AgCN,hArr,"Ag"^(+),+,CN^(-),,,),(s,,s,,s,,,):}` or `s=sqrt(K_(sp))=sqrt(6.0xx10^(-17))=7.8xx10^(-9) ` mol `L^(-1)` |
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| 135. |
The values of `K_(sp)` of two sparingly solubles salts, `Ni(OH)_(2)` and `AgCN` are `2.0 xx 10^(-15)` and `6 xx 10^(-7)` respectively, which salt is more soluble? Explain |
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Answer» `AgCN hArr Ag^(+) + CN^(-)` `K_(sp) = [Ag^(+)][CN^(-)] = 6 xx 10^(-17)` `Ni(OH)_(2) hArr Ni^(2+) + 2OH^(-)` `K_(sp) = [Ni^(2+)][OH^(-)]^(2)=2xx10^(-15)` Let `[Ag^(+)] = S_(1),` then `[CN] = S_(1)` Let `[Ni^(2+)] = S_(2)`, then `[OH^(-)] = 2S_(2)` `S_(1^(2)) = 6 xx 10^(-17), S_(1) = 7.8 xx 10^(-9)` `(S_(2))(2S_(2))^(2) = 2 xx 10^(-15), S_(2) = 0.58 xx 10^(-4)` `Ni(OH)_(2)` is more soluble than `AgCN`. |
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| 136. |
The values of Ksp of two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 x 10-15 and 6.0 x 10-17 respectively. Which salt is more soluble? Explain. |
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Answer» AgCN ⇌ Ag+ + CN− Ksp = [Ag+][CN−] = 6 × 10−17 Ni(OH)2 ⇌ Ni2+ + 2OH- Ksp = [Ni2+][OH−] 2 = 2 × 10−15 Let [Ag+] = S,then [CN−] = S1 Let [Ni2+] = S2,then [OH−] = 2S2 ∴ From eq. (i) S1 × S1 = 6 × 10−17 S12 = 6 × 10−17 S1 = 7.8 × 10−9 ∴ From eq. (ii) (S1) × (2S2)2 = 2 × 10−15 4S23 = 2 × 10−15 S23 = \(\frac{2\times10^{-15}}{4}\) = 0.5 × 10−15 or S2 = 0.58 × 10−4 Since solubility of Ni(OH)2 is more than AgCN, so, Ni(OH)2 is more soluble than AgCN. |
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| 137. |
The pH of pure water at `25^(@)C and35^(@)C` are 7 and 6 respectively. Calculate the heat of dissociation of `H_(2)O` into `H^(+) and OH^(-)` ions. |
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Answer» The dissociation reaction is : `H_(2)O hArr H^(+)+OH^(-)` At `25^(@)C`, pH = 7 means `[H^(+)]=10^(-7)M :. K_(w)=10^(-14)` At `35^(@)C`, pH =6 means `[H^(+)]=10^(-6)M :. K_(w)=10^(-12)` As equilibrium constant for the dissociation of `H_(2)O` are in the same ratio as ionic products of water, we can apply the relation `log.(K_(w_(2)))/(K_(w_(1)))=(Delta H)/(2.303 R)((1)/(T_(1))-(1)/(T_(2))):. log. (10^(-12))/(10^(-14))=(Delta H)/(2.303xx8.314 JK^(-1)"mol"^(-1))((1)/(298K)-(1)/(3-8K))` or `DeltaH=52898 J "mol"^(-1) = 52.898 kJ "mol"^(-1)`. |
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| 138. |
An aqueous solution of a metal bromide `MBr_(2)` (0.05 M) is saturated with `H_(2)S`. What is the minimum pH at which MS will be precipitated ? `(K_(sp) "for MS" = 6.0 xx 10^(-21), "concentration of saturated " H_(2)S=0.1M, K_(1)=10^(-7) and K_(2)=1.3xx10^(-13) "for " H_(2)S)`. |
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Answer» `[M^(2+)]=[MBr_(2)]=0.05 M ` For the precipitation of MS, minimum `[S^(2-)]` can be calculated from `[S^(2-)][M^(2+)]=K_(sp)` i.e., `[0.05][S^(2-)]=6.0xx10^(-21) or [S^(2-)]=1.2xx10^(-19)` `S^(2-)` ions are obtained from following dissociations `H_(2)S overset(K_(1)) hArr H^(+)+HS^(-)` `HS^(-) overset(K_(2))hArr H^(+)+S^(2-)` `K_(1) = ([H^(+)][HS^(-)])/([H_(2)S]), K_(2) = ([H^(+)][S^(2-)])/([HS^(-)])` `:. K_(1)K_(2)=([H^(+)]^(2)[S^(2-)])/([H_(2)S])` `10^(-7)xx1.3xx10^(-13)=([H^(+)]^(2)[1.2xx10^(-19)])/((0.1)) or [H^(2)]^(2)=(1.3xx10^(-20)xx10^(-1))/(1.2xx10^(-19))=1.083xx10^(-2)` or `[H^(+)]=1.041xx10^(-1)=0.1041 M` `:. pH = - log (0.1041)=0.9826` |
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| 139. |
Calcium lactate is a salt of weak organic acid and represented as `Ca(Lac)_(2)`. A saturated solution of `Ca(Lac)_(2)` contains 0.13 mole of this salt in 0.50 litre solution. The pOH of this solution is 5.60 . Assuming complete dissociation of the salt , calculate `K_(a)` of lactic acid. |
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Answer» In solution, `Ca(Lac)_(2)` is hydrolysed as follows : `Ca(Lac)_(2)+2H_(2)O hArr Ca(OH)_(2) + underset(underset("(weak)")("lactic acid"))(2HLac)` or `Ca^(2)+2 Lac^(-1) + 2 H_(2)O hArr Ca^(2+) + 2 OH^(-) + 2 Hlac` or `2 Lac^(-) + 2H_(2)O hArr 2 OH^(-) + 2 HLac` or `Lac^(-) + H_(2)O hArr OH^(-) + HLac` Hydrolysis constant, `K_(h) = ([OH^(-)][HLac])/([Lac^(-)])` But `[Ca(Lac)_(2)]=0.26 ` mol `L^-1)` so that `[Lac^(-)]=0.52 ` mol `L^(-)` and `pOH = 5.60 ` so that `- log [OH^(-) ] = 5.60` or `[OH^(-)] = ` antilog `(-5.6)=2.51xx10^(-6)M` `:. K_(h) = ((2.51xx10^(-6))^(2))/(0.52)=1.21xx10^(-11)` Further, `HLac hArr H^(+) + Lac^(-)` `K_(a) = ([H^(+)][Lac^(-)])/([HLac]) " " ...(ii)` Also `K_(w) = [H^(+)][OH^(-)]` ...(iii) From (i), (ii) and (iii), `K_(a)=(K_(w))/(K_(h))=(10^(-14))/(1.21xx10^(-11))=8.26xx10^(-4)` |
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| 140. |
Why `PO_(4)^(3-)` ion is not amphiprotic ? |
| Answer» an amphiprotic ion is one which can donate proton as well as accept proton . `PO_(4)^(3-)` ion can accept proton(s) but cannot donate any proton. Hence, `PO_(4)^(3-)` is not amphiprotic. | |
| 141. |
How are KP and Kc related? Mention the condition under which KP = KC. |
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Answer» Kp = Kc (RT)Δn when Δn = 0 i.e. no. of moles of gaseous products = no. of moles of gaseous reactants, e.g., H2(g) + I2(g) ⇌ 2HI(g)Kp = Kc |
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| 142. |
Which one of the following is not the Lewis acid: BF3 ,NH3, PCl5 and SO2 |
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Answer» NH3 ................. |
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| 143. |
Which of the following are Lewis acids?H2O, BF3, H+ and NH4+. |
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Answer» BF3, H+ are Lewis acids. |
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| 144. |
For the following reaction in equlibrium `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)` Vapour density is found to be 100 when 1 mole of `PCl_(5)` is taken in a 10 litre flask at `27^(@)C` Calculate the equlibrium pressure. Also calculate percentage dissociation of `PCl_(3)` |
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Answer» `PCl_(5) hArr PCl_(3)(g)+Cl_(2)(g)` `{:("Initial moles",1,0,0),("At equlibrium",1-alpha,alpha, alpha),("Concentration",(1-alpha)/(10),(alpha)/(10),(alpha)/(10)):}` `D=104.25` `D=100` `y=2" "therefore alpha=0.0425or 4.25%` Total moles at equlibrium `=1+alpha=1+0.0425=1.0425 ` Let total pressure at equlibrium be P atm `PV=nRT` `P=(1.0425xx0.282xx300)/(10)=2.57atm` |
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| 145. |
Calculate the pH of 0.10 M solution of `Nh_(4)Cl.` The dissociation constant `(K_(b))` of `NH_(3)` is `1.8xx10^(-5)` |
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Answer» `NH_(4)Cl` is a salt of strong acid and weak base. `pH=7-1/2[pK_(b)+logc]` `=7-1/2[-log(1.8xx10^(-5))+log0.10]` `=7-1/2(5-0.26-1)=5.12` |
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| 146. |
Calculate `[OH^(-)]and %` dissociation of 0.01 M solution of ammonium hydroxide solution. The ionization constant for `NH_(4)OH(K_(b))=1.8xx10^(-5)` |
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Answer» `{:(NH_(4)OH,hArrNH_(4)^(+)+,OH^(-)),(C-Calpha,C alpha,Calpha):}` `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(4)OH])` `K_(b)=(Calpha^(2))/(1-alpha)` `1.8xx10^(-5)=(0.01xxalpha^(2))/(1-alpha)` `alpha=sqrt(1.8xx10^(-3))=4.24xx10^(-2)[1-alpha=1]` `%` dissociation `=4.24` `[OH^(-)]=Calpha=0.01xx4.24xx10^(-2)` `[OH^(-)]=4.24xx10^(-4)M` |
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| 147. |
1 M solution of `CH_(3)COOH` is diluted to x times so that pH of solution is doubled. Calculate x Given `K_(a)=1.8xx10^(-5)` |
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Answer» `pH=1/2[pK_(a)-logC]=1/2[pK_(a)-log1]=(pK_(a))/(2)` `pH=2xxpH=2xx(pK_(a))/(2)=pK_(a)` `pH=pK_(a)=1/2[pK_(a)-logC]` `pK_(a)=-logC=-logK_(a)` `C=K_(a)=1.8xx10^(-5)` Dilution `x=1/C=0.55xx10^(4)` times |
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| 148. |
`K_(a) ` for `CH_(3)CO OH` is `1.8xx10^(-5) and K_(b)` for `NH_(4)OH ` is `1.8xx10^(-5)`. The pH of ammonium acetate will beA. 7.005B. 4.75C. `7.0`D. between 6 and 7. |
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Answer» Correct Answer - C `CH_(3)CO ONH_(4)` is a salt of weak acid and weak base. Hence, `pH = 7 + (1)/(2) (pK_(a) - pK_(b))=7+0=7` |
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| 149. |
Calculate the pH of a solution containing 0.1 M acetic acid and 0.1 M benzoic acid. `K_(a)` for `CH_(3)CO OH` and `C_(6)H_(5)COOH` are `1.8xx10^(-5) and 6.5xx10^(-5)` respectively . |
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Answer» `[H_(3)O^(+)]=sqrt((1.8xx10^(-5))(0.1)+(6.5xx10^(-5))(0.1))` `=sqrt(1.8xx10^(-6)+6.5xx10^(-6))=sqrt(8.3xx10^(-6))=2.88xx10^(-3)M` pH `=-log [H_(3)O^(+)]=- log (2.88xx10^(-3))=3-0.4594=2.5406`. |
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| 150. |
If 1 mole of acetic acid and 1 mole of ethyl alchol are mixed and reaction proceeds to equilibrium , the concentrations of acetic acid and water are found to be `1//3 and 2//3` mole respectively . If 1 mole of ethyl acetate and 3 moles of water are mixed , how much ester is present when equilibrium is reached ? |
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Answer» Correct Answer - `0.465 " mole ". ` {:(,CH_(3)COOH ,+,C_(2)H_(5)OH,+,hArr,CH_(3)COOC_(2)H_(5),+,H_(2)O),("Intial amounts ",1 " mole ",,1 " mole ",,,,,),("Amounts at eqm.",1//3 " mole",,1//3 " mole",,,2//3 " mole",,2//3 " mole"),(" Molar concs.(mol " L^(-1),1//3 V,,1//3 V,,,2//3 V,,2//3 V):}` `K_(c) = ((2//3 V) (2//3 V))/((1//3 V) (1//3 V))=4` For the reverse reaction ` {:(,CH_(3)COOC_(2)H_(5),+,H_(2)O,hArr,CH_(3)COOH,+,C_(2)H_(5)OH),("Intial",1 " mole",,3 " moles",,,,),("At. eqm.",(1-x),,(3-x),,x,,x):}` ` :. (x^(2))/((1-x) (3-x))= 1/4 or 4 x^(2)= x^(2) + 3- 4 x or 3 x^(2) + 4 x - 3 = 0 ` `or x= (-4 pm sqrt(16+ 4 xx 3 xx 3))/(2 xx 3)=0*535 " mole"` `:. " Ester present at eqm."= 0*535 = 0*465 " mole " ` |
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