52671 + Interview Questions in Chemistry in Class 11 Page 1013 InterviewSolution

50601.	A certain compound has the molecular formula X_(4)O_(6) having 57.2%X. Thus,
Answer» atomic mass of X is 32 X MAY contain six VALENCE electrons X is an electropositive metal X can be a non-metal Solution :`57.2=(4x)/(M)xx100,43.8=(96)/(M)xx100impliesx=32`

Discussion

50602.	A certain buffer solution contains equal concentration of X^(-) " and " HX.The K_(b) " for " X^(-) " is " 10^(-10) .Find the pH of the buffer .
Answer» Solution :`k_(a) . k_(b)^(-14) "" :. K_(a) = (10^(-14))/(10^(-10)) = 10^(-4)` `pH = pka + LOG.([x^(-)])/([HX]) ` ` :. pH = 4 + log. 1/1 = 4 "" :. pH = 4`

Discussion

50603.	A certain buffer solution contains equal concentration of X^(-) and HX. K_(b), for X^(-) is 10^(-10). Find the pH of buffer.
Answer» <P> Solution :` K_b=10 ^(-10 ), Ka = 10 ^(-4)[X^(-) ]=[HX]` ` PH= P^(Ka) +LOG ""([X^(-)])/([HX])=-log 10 ^(-4)=4`

Discussion

50604.	(A): Cellulose is used as sizing agent in place of starch in textile industry. (R) : By changing process and raw-materials polluted waste is reduced.
Answer» Both (A) and (R) are TRUE and (R) is the correct explanation'of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false (A) is false but (R) is true SOLUTION :Both (A) and (R) are true and (R) is not the correct explanation of (A)

Discussion

50605.	A cell is prepared by dipping copper rod in 1M copper suphate solutoin and zinc rod in 1M ZnSO_(4) solution The standard reduction potential s of copperand zinc are + 0.34 and -0.76 V respectively (i) what is the cell reaction ? (ii) what will be the standsard electromotive force (EMF) of the cell ? (iii) which electrode will be positive ? (iv) How will the cell be represented ?
Answer» Answer :(i) `Zn(s)+CuSO_(4)(aq)rarrZnSO_(4)(aq)+Cu(s);(ii)E_("cell")^(@)=1.1V`;(III)Cuelectrode (IV)`Zn\|ZnSO_(4)(1M)\|CuSO_(4)(1M)\|Cu`

Discussion

50606.	A cell is prepared by dipping a chromium rod in 1 MCr_(2)(sO_(4))_(3) solution and an iron rod in 1 M FeSO_(4) solution The standard reduction potentials of chromium and iron electrodes ae -0.75 V and -0.45 V respecitvely (a) what will be the cell reactoition ? (b) what will be the standard EMF of hte cell ? (C ) which electrode will act as anode ? (d) which electrode will act as cathods ?
Answer» SOLUTION :The two half cel reductio equation are : `2Fe^(2+)(aq) e^(-) rarr Fe(s),E^(@)=-0.45V` `Cr^(3+)(aq)+3 ^(-) rarr Cr(s),E^(@)=-0.75V` Since `Cr^(3+)//Cr` electrode has lower electrode POTENTIAL therfore it acts as the anode while `Fe^(2+)//Fe` electrode with higher electrode potential acts as the cathode to equalise the number of electrons MULTIPLY Eq (i) by 3 and Eq (II) by 2 but do not multiply their `E^(@)` values THUS To obtain equation for the cell reaction subtract Eq (iv) form Eq (iii) we have `2 Cr (s) +3Fe^(2+)(aq) rarr 2cr^(3+) (aq)+3 Fe(s) ,E_(cell)^(@) =-0.45 -(-0.75V)=+0.30 V` Thus the EmF of the cell =-+0.30 V

Discussion

50607.	A cell is prepared by dipping a chromium rod is 1 M Cr_2(SO_4)_3 solution and an iron and in 1M FeSO_4 solutions. The standard potentials of chromium and iron electrodes are -0.75 V and 0.45 respectively. (a) What will be the cell reaction? (b) What will the standard EMF of the cell? (c) Which electrode will act an anode? (d) WHich electrode will act as cathode?
Answer» Solution :The two half cell REDUCTION equations are: `Fe^(2+)(aq)+2e^(-) to Fe(s),E^@=-0.45V`....(i) `CR^(3+)(aq)+3e^(-)toCr(s),E^@=-0.75 V`.....(ii) Since `Cr^(3+)//Cr` electrode has a power reduction POTENTIAL it acts as the anode, while `Fe^(2+)//Fe` electrode with a higher electrode potential acts as the cathode. To equilize the number of electrons, multiply Eq (i) by 3 and Eq. (ii) by 2. But do not multiply their `E^@` values . Thus, `3Fe^(2+)(aq)+6e^(-)toFe(s),E^@=-0.45V`.....(III) `2Cr^(3r)(aq)+6e^(-)to2Cr(s),E^@=-0.75V`.....(iv) To obtain equations for the cell reactions, SUBTRACT Eq. (iv) from Eq. (iii) , we have, `2Cr(s)+3Fe^(2+)(aq)to2Cr^(3+)(aq)+3Fe(s)` `E_(cell)^o=-0.45-(-0.075V)=+0.30V` Thus, the EMF of the cell =+0.30 V

Discussion

50608.	A cell is constructed using Cu^(2+)//Cu and I^(3+)//AIelectrode what is the net cell reaction
Answer» SOLUTION :From we know that `E^(@)` of `AI^(3+)//AI(-1.66 V)` is LOWER than that of `Cu^(2+)//Cu(+0.34 V)` THEREFORE oxidation occurs at the `AI^(3+)//AI` electrode and reducation occurs at the `Cu^(2+) //Cu` electrode consequently the net cell reactin is `2 AI(s)+3 Cu^(2+)(aq)RARR 2 AI^(3+)(aq)+3Cu(s)`

Discussion

50609.	A catalyst will increase the rate of a chemical reaction by lowering the ……….
Answer» SOLUTION :ACTIVATION ENERGY

Discussion

50610.	A catalyst remains unchanged at the end of the reaction regarding
Answer» Mass Pysical STATE Physical state and CHEMICAL composition Mass and chemical composition Answer :D

Discussion

50611.	A catalyst increases the speed of an equilibrium reaction and hence the amount of products formed Increases.
Answer» SOLUTION :A CATALYST increases the speed of the forward as WELL as backward reactions to the same EXTENT.

Discussion

50612.	A catalyst is a substance which:-
Answer» Alters the EQUILIBRIUM in a REACTION Is always in the same phase as the reactants Participates in the reaction and provides easier PATHWAY for the same Does not PARTICIPATE in the reaction but speeds it up. Solution :A catalyst does not take part I the reaction but can speed it up. It can be RECOVERED after the reaction.

Discussion

50613.	A catalyst in the finely divided state is more efficient because in this state
Answer» it has LARGER ACTIVATION energy it can react with one of the REACTANT more efficiently it has large surface AREA All the above Solution :More the surface area , more is the number of ACTIVE centres and hence more is the efficiency of the catalyst.

Discussion

50614.	A catalyst can affect reversible reaction by:-
Answer» CHANGING EQUILIBRIUM Slowing forward reaction Attaining equilibrium in both direction None of these Answer :C

Discussion

50615.	A catalyst
Answer» ALTERS the equilibrium constant Increases the equilibrium CONCENTRATION of products helps establishing the equilibrium quickly Supplies energy to the reactants Solution :CATALYST PROPERTIES

Discussion

50616.	(A): CaSO_(4) is called dead burnt gypsum because it does not set with water (R) : Mixture of 1 part of slaked lime, 3 parts of sand and water is known asmortar
Answer» Both A and R are correct and R is the correct EXPLANATION of A. Both A and R are correct but R is not the correct explanation of A. A is True but R is False. R is False but A is True. ANSWER :A

Discussion

50617.	(A) : Carbon forms a large number of compounds (R) : Carbon has high catenation power
Answer» A and R are TRUE, R explains A A and R are true, R does not explain A A is true, but R is FALSE A is false, but R is true ANSWER :A

Discussion

50618.	(a ) Carbon dioxide is non -polar while water is polar. What conclusion do you draw about their structure from for this fact ? (b) Classify the following compounds into acidic basic and amphoteric oxides . AI_(2) O_(3),CI_(2)O_(7) .
Answer» SOLUTION :(a ) `Ooverset rarr =Coversetrarr = O` is LINEAR bond moments are equal and opposite net dipole moment is zero. Water is a BENT molecule it has a net dipole moment . ( b) `AI_(2)O_(3)` is AMPHOTERIC where as `CI_(2) O_(7)` is an acidic OXIDE.

Discussion

50619.	A carbon compound on analysis gave the following percentage composition, carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound.
Answer» SOLUTION :`C_(2)H_(3)Cl_(3)O_(2)`

Discussion

50620.	A carbon compound has many functional groups, then order of preference while naming it according to IUPAC nomenclature is
Answer» `-CHO GT - COOH gt -OH gt -NH_(2)` ` -COOH gt -CHO gt -NH_(2) gt -OH` `-COOH gt - OH gt - NH_(2) gt -CHO` `-COOH gt -CHO gt -OH gt - NH_(2)` ANSWER :D

Discussion

50621.	A carbon compound contains 12.8% Carbon, 2.1% Hydrogen, 85.1% Bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula.
Answer» Solution :Step 1: Percentage composition of the ELEMENTS present in the COMPOUND. `{:(C,H,Br),(12.8,2.1,85.1):}` Step 2: Dividing with the respective atomic WEIGHTS of the elements. `{:((12.8)/(12),(2.1)/(1),(85.1)/(80)),(1.067,2.1,1.067):}` Step 3: Dividing by the smallest number to get simple atomic RATIO. `{:((1.067)/(1.067),(2.1)/(1.067),(1.067)/(1.067)),(1,2,1):}` The empirical formula is `CH_(2)Br` Empricial formula weight `12 + (2 xx 1) + 80 = 94` The molecular weight = 187.9 (given) `therefore n=(187.9)/(94)=2` `"The molecular formula" = "(empirical formula)"_(2)` `(CH_(2)Br)_(2)=C_(2)H_(4)Br_(2)`

Discussion

50622.	A carbon compound containing only carbon and oxygen has an approximate molecular weight of 290. On analysis it is found to contain 50% by weight of each element. What is the molecular formula of the compound ?
Answer» Solution :Calculation of empirical formula : `therefore` the empirical formula `=C_(4)O_(3)` Calculations of MOLECULAR formula: Molecular MASS = 290 (given) Empirical formula mass `=(4 xx 12.01 + (3 xx 16.0) = 96.04` `therefore N=("Molecular mass")/("Empirical formula mass") = 290/(96.04) = 3.019 =3` `therefore` The molecular formula of the given compound `= 3 xx C_(4)O_(3) = C_(12)O_(9)`

Discussion

50623.	(A) : Carbon-carbon bond energy values are in the order C_(2)H_(2) gt C_(2)H_(4) gt C_(2)H_(6) (R ): Bond energy increases with increase in the bond order
Answer» A and R are true, R explains A A and R are true, R does not explain A A is true, but R is FALSE A is false, but R is true Answer :A

Discussion

50624.	A carbocation in which dispersal of charge does not take place
Answer» `R_(3)C^(oplus)` `R_(3)CH^(oplus)` `RCH_(2)^(oplus)` `""^(oplus)CH_(3)` Solution :AMONG the given cabo cataions there is no DISPERSAL of charge in `CH_(3)^(+)` (or) No-hyper conjugation

Discussion

50625.	(a) Calculate the energy needed to raise the temperature of10.0 g of iron from 25^(@) to 500^(@)C if specific heat capacity of iron is 0.45 J (^(@C) g^(-1). (b) What mass of gold of specific heat capacity 0.13 J ( .^(@)C)^(-1) g^(-1) can be heated through the same temperature difference when supplied with the same amount of energyas in (a) ?
Answer» SOLUTION :(a) Energy needed `(Q) =m XX C xx Delta T = 10.0xx 0.45xx ( 500 -25) J = 2137.5 J` (b) `q= m xx c xx Delta T` `2137.5 = m xx 0.13 xx ( 500 - 25) ` or ` m = 34.6 g`

Discussion

50626.	(A): C Cl_(2) does not undergo hydrolysis where as SiCl_(4) is readily hydrolysed. (R) : Carbon has no d-orbitals in its valence shell, but silicon has vacant d-orbitals in its valence shell.
Answer» a. Both A and R are true, and R is correct explanation of A b. Both A and R are true, and R is not correct explanation of A c. A is true, but R is false d. A is false, but R is true ANSWER :A

Discussion

50627.	Butyric acid contains only C, H and O. A 4.24 mg sample of butyric acid is completely burnt. It gives 8.45 mg of carbon dioxide and 3.46 mg of water. What is the mass percentage of each element in butyric acid?
Answer» Answer :( c) MOLECULAR FORMULA `=C_(4)H_(8)O_(2)`

Discussion

50628.	A burning strip of magnesium is introduced into a jar containing a gas. After sometime the walls of the container is coated with carbon. The gas in the container is
Answer» `O_(2)` `N_(2)` `CO_(2)` `H_(2)O` ANSWER :C

Discussion

50629.	A burning magnesium ribbon will continue to burn in
Answer» `CO_(2)` and `SO_(2)` `N_(2),CO_(2)` and `SO_(2)` `N_(2)` and steam `N_(2)`, `CO_(2)`, `SO_(2)` and steam Solution :A burning .Mg. ribbon will continue to burn in the ATMOSPHERES of `N_(2)`, `CO_(2)`, `SO_(2)` steam and EVEN in `N_(2)O`.

Discussion

50630.	A bulb of unknown volume 'V' Contains an ideal gas at 2 atm pressure. It was connected to another evacuated bulb of volume 0.5 litre through a stopcock. When the stopcock was opened the pressure in each bulb became 0.5 atm. Then V is
Answer» `21 C c ` `42c ,c` `105c c ` `63C c ` ANSWER :A

Discussion

50631.	A bulb emits light of wavelength 4500 Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second ?
Answer» Solution :ELECTRICAL energy in joules = Power in watts `xx` Time in seconds THUS 150 watt = 150 joules of energy emitted per second `:.` Energy emitted as light `= (8)/(100) xx 150 = 12J` `E = n HV = nh (c)/(lamda) " " :. n = (E xx lamda)/(h xx c) = ((12J) xx (4500 xx 10^(-10) m))/((6.626 xx 10^(-34) Js) xx (3 xx 10^(8) ms^(-1))) = 2.717 xx 10^(19)`

Discussion

50632.	A bulb emits light of wave length 4500Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second ?
Answer» ANSWER :`n=27.2 XX 10^(18)`

Discussion

50633.	A bulb containing N_(2)O_(4) is colourless in ice. Its colour inboiling water is …………. while in water at 298 K, it is………… .
Answer» ANSWER :REDDISH BROWN, PALE brown

Discussion

50634.	A buffer solution with pH 9 is to be prepared by mixing NH_(4)Cl that should be added to one litre of 1.0mNH_(4)OH kb1.8 xx 10^(-5)
Answer» ANSWER :`NH_(4)CL = 1.8 M`

Discussion

50635.	A buffer solution with pH 9 is to be prepared by mixing NH_(4) Cl and NH_(4)OH. Calculate the number of moles of NH_(4)Cl that should be added to one litre of 1.0 M NH_(4)OH (K_(b)=1.8xx10^(-5))
Answer» Solution :For a BASIC BUFFER, `pOH = pK_(b)+log.(["SALT"])/(["Base"])` Further, `pH + pOH = 14 ` so that `pOH = 14 - pH = 14 - 9 = 5` `pK_(b)=-log K_(b) = - log(1.8xx10^(-5))=4.7447` [Base] `=[NH_(4)OH]= 1 "MOL" L^(-1)` `:. 5 = 4. 7447 + log. (["Salt"])/(1) or log ["Salt"] = 0.2553 or ["Salt"]= 1.8 "mol" L^(-1)`

Discussion

50636.	A buffer solution of pH = 4.7 is prepared from CH_(3)COONa and CH_(3)COOH. Dissociation constant of acetic acid is 1.75xx10^(-5). Calculate the mole proportion of sodium acetate and acetic acid.
Answer» Solution :`pH=pK_(a)+"log"(["SALT"])/(["Acid"])rArrpH=-log_(10)K_(a)+"log"([CH_(3)COONa])/([CH_(3)COOH])` `4.7=-log_(10)(1.75xx10^(-5))+"log"([CH_(3)COONa])/([CH_(3)COOH])` `4.7=5-log1.75+"log"([CH_(3)COONa])/([CH_(3)COOH])=5-0.2430+"log"([CH_(3)COONa])/([CH_(3)COOH])` `"log"([CH_(3)COONa])/([CH_(3)COOH])="anti"log(-0.0557)="anti"log(-0.057+1-1)` `([CH_(3)COONa])/([CH_(3)COOH])=0.8770`.

Discussion

50637.	A buffer solution of pH 8.3 is prepared from ammonium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8xx10^(-5). What is the mole proportion of ammonium chloride and ammonium hydroxide?
Answer» Solution :pOH = 14 - pH = 14 - 8.3 = 5.7 `pH=pK_(a)+"log"(["SALT"])/(["Base"])rArrpH=-log_(10)K_(a)+"log"(["Salt"])/(["Base"])` `5.7=-log_(10)(1.8xx10^(-5))+"log"([NH_(4)CL])/([NH_(4)OH])rArr5.7=5-log1.8+"log"([NH_(4)Cl])/([NH_(4)OH])` `5.7=5-0.2553+"log"([NH_(4)Cl])/([NH_(4)OH])rArr5.7=4.7447+"log"([NH_(4)Cl])/([NH_(4)OH])` `therefore([NH_(4)Cl])/([NH_(4)OH])=5.7-4.7447=0.9553` `rArr([NH_(4)Cl])/([NH_(4)OH])="anti"log(0.9553)=9.022""` The ratio is 9 : 1.

Discussion

50638.	A buffer solution of acetic acid and sodium acetate is diluted 10 times. What is the effect on itspH ?
Answer» Solution :No effect because`pH = pK_(a) + LOG. (["Salt"])/(["ACID"])`. On dilution, ratio [Salt]/[Acid] remains same. Hence, pHdoes not change.

Discussion

50639.	A buffer solution is prepared in which the concentration of NH_(3) is 0.30 M and the concentrationof 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 ?
Answer» <P>8.73 9.08 9.43 11.72 Solution :`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`

Discussion

50640.	A buffer solution is prepared by mixing CH_(3) COONa and CH_(3) COOH. The pK_(a) of acetic acid is 4.74 . To maintain the pH of the buffer solution as 6.04 , the concentration ratio of CH_(3) COONa and CH_(3) COOH to be maintained is
Answer» `1 : 20` `1 : 10` `2 : 1` `20 : 1` ANSWER :D

Discussion

50641.	A buffer solution is formeed by mixing 100 mL of 0.1 M CH_(3)COOH with 200 mL of 0.02 M CH_(3)COONa. If this buffer solution is made to 1.0 L by adding 700 mL of water, pH will change by a factor of
Answer» Solution :No change in pH of buffer SOL. By DILUTION

Discussion

50642.	A buffer solutionis prepared by mixing 50 mL of 0.2 Mof acetic acid with .x. mL of 0.2 M of NaOHsolutions. If pH of the resulting buffersolutions is 4. 7 then value of .x.is (Ka= 2xx10 ^(-5))
Answer» 28 mL 22 m L 24 mL 25 mL Solution :` pH =pKa +log "" ([S])/([A]), 4.7 =4.7 +log "" ([S])/([A]) ` `RARR [S]=[A]` ` x XX 0.2 = 50 xx 0.2 - x xx 0.2 ` `rArr x = ( 50 xx 0.2)/(2 xx 0.2 )= 25 mL `

Discussion

50643.	A buffer solution contains a weak acid HA and A . When small quantity of NaOH is added, to keep p^(H)as constant. which of the following reaction takes place?
Answer» ` HA to H^(+) A^(-) ` ` H^(+) A^(-) toHA` ` HA+ OH^(-) to H_2O +A^(-) ` `A^(-) +H_2O to HA+ OH^(-) ` Solution :NaOHadded is a base hence acid COMPONENT of the buffter . `HA+ OH^(-)to H_2O+A^(-) `

Discussion

50644.	A buffer solution contains 0.40 mol of ammonium hydroxide and 0.50 mol of ammonium chloride to make a buffer solution of 1 L. Calculate the pH of the resulting buffer solution. Dissociation constant of ammonium hydroxide at 25^(@)C is 1.81xx10^(-5).
Answer» SOLUTION :`pOH = pK_(b) + log. (["Salt"])/(["Base"])=pK_(b) + log. ([NH_(4)Cl])/([NH_(4)OH])` `[NH_(4)OH]=0.40` mol `L^(-1)` `[NH_(4)Cl]=0.50 ` mol `L^(-1)` `pK_(b) = - log K_(b) = - log (1.81 xx 10^(-5))=5-0.2577 = 4.7423` `:. pOH = 4.742 + log. (0.5)/(0.4)= 4.742+ log 1.25 = 4.742 + 0.0969 = 4.8389 ~= 4.839` `:. PH = 14-pOH = 14 - 4.839 = 9.161`.

Discussion

50645.	A buffer solution contains 2 moles of ammonium hydroxide and 0.25 mole of ammonium chloride per dm^(3) of the solution. (K_(b) for ammonium hydroxide = 1.8xx10^(-5)). Calculate the pH of the buffer solution.
Answer» Solution :It is a BASIC BUFFER `pOH=pK_(B)+"LOG"(["Salt"])/(["Base"])` = `4.74+"log"0.25/2=4.74+"log"0.25/2` = 4.74 - 0.90 = 3.64 pH = 14 - 3.64 = 10.36

Discussion

50646.	A buffer solution contains 0.2mole of NH_4 OHand 0.2 mole of NH_4 Cl perlitre. The PK_bofNH_4 OHis 4.75 The pH of the buffer will be
Answer» `4.75` ` 5.75` `9.25` `2.25` Solution :` pOH =pK_b +LOG "" ([S])/([B])=4.75 +log ""(0.2)/(0.2) = 4.75` `rArr PH = 14 -4.75 =9.25`

Discussion

50647.	A buffer solution contains 0.1 mole of sodium acetate in 1000 cm^(3) of 0.1 M acetic acid. To the above buffer solution, 0.1 M aceticacid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The pH of the resultingbuffer is equal to ...........
Answer» `pK_(a) - log2` `pK_(a)` `pK_(a)+2` `pK_(a)+LOG 2` Solution :`pH = pK_(a) + log.(["Salt"])/(["Acid"])` [Salt]`=(0.1+0.1"mole")/(1000 ML) = (0.2"mole")/(1L) = 0.2M` [Acid]`=(0.1 "mole")/(1000mL) = (0.1"mole")/(1L)=0.1 M` `:. pH = pK_(a) + log 2`.

Discussion

50648.	A buffer solution consists of 1 mole each of HA and its conjugate base A^(-), Addition of which of the following decreases the pH of buffer ?
Answer» 0.1 MOLE of HA 0.1 mole of `A^(-)` 0.1 mole of `NAOH` WATER ANSWER :A

Discussion

50649.	A buffer solution can be prepared from a mixture of
Answer» SODIUM acetate and acetic acid in water sodium acetate and HYDROCHLORIC acid in water ammonia and ammonium chloride in water ammonia and sodium hydroxide in water SOLUTION :A buffer solution is a mixture of WEAK acid/base and its salt with conjugate base/acid (STRONG base /acid)

Discussion

50650.	A buffer solution 0.04 M in Na_2HPO_4 and 0.02M in Na_3 PO_4 is prepared. The electrolytic oxidation of 1.0 milli -mole of the organic compound RNHOH is carried out in 100 mLof the buffer. The reaction is RNHOH +H_2O to RNO_2 +4H ^(+) + 4e ^(-) The approximatepH of solution after the oxidation is complete is:[Given : forH_3O PO_4, pK_(a_1)=7.20 , pK_(a_2) =12]
Answer» ` 6.90 ` ` 7.20 ` ` 7. 5` None of these Solution :` {:(PO_4^(3-)+,H^(+)to, HPO_4^(-2)),( 0.02M , 0.04 M , 0.04 M),( -, 0.02 , 0.06M), (- , 0.02M, -):}` ` {:( HPO_4^(-2) ,H^(+) to, H_2PO_4^(-)),( 0.06, 0.02 , 0),( 0.02 , - , 0.02M),( 0.04M, , ):}` ` pH =7.2 +LOG ""(0.04)/(0.02 )=7.5`

Discussion

Explore topic-wise InterviewSolutions in .

A certain compound has the molecular formula X_(4)O_(6) having 57.2%X. Thus,

A certain buffer solution contains equal concentration of X^(-) " and " HX.The K_(b) " for " X^(-) " is " 10^(-10) .Find the pH of the buffer .

A certain buffer solution contains equal concentration of X^(-) and HX. K_(b), for X^(-) is 10^(-10). Find the pH of buffer.

(A): Cellulose is used as sizing agent in place of starch in textile industry. (R) : By changing process and raw-materials polluted waste is reduced.

A cell is constructed using Cu^(2+)//Cu and I^(3+)//AIelectrode what is the net cell reaction

A catalyst will increase the rate of a chemical reaction by lowering the ……….

A catalyst remains unchanged at the end of the reaction regarding

A catalyst increases the speed of an equilibrium reaction and hence the amount of products formed Increases.

A catalyst is a substance which:-

A catalyst in the finely divided state is more efficient because in this state

A catalyst can affect reversible reaction by:-

A catalyst

(A): CaSO_(4) is called dead burnt gypsum because it does not set with water (R) : Mixture of 1 part of slaked lime, 3 parts of sand and water is known asmortar

(A) : Carbon forms a large number of compounds (R) : Carbon has high catenation power

(a ) Carbon dioxide is non -polar while water is polar. What conclusion do you draw about their structure from for this fact ? (b) Classify the following compounds into acidic basic and amphoteric oxides . AI_(2) O_(3),CI_(2)O_(7) .

A carbon compound on analysis gave the following percentage composition, carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound.

A carbon compound has many functional groups, then order of preference while naming it according to IUPAC nomenclature is

A carbon compound contains 12.8% Carbon, 2.1% Hydrogen, 85.1% Bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula.

A carbon compound containing only carbon and oxygen has an approximate molecular weight of 290. On analysis it is found to contain 50% by weight of each element. What is the molecular formula of the compound ?

(A) : Carbon-carbon bond energy values are in the order C_(2)H_(2) gt C_(2)H_(4) gt C_(2)H_(6) (R ): Bond energy increases with increase in the bond order

A carbocation in which dispersal of charge does not take place

(A): C Cl_(2) does not undergo hydrolysis where as SiCl_(4) is readily hydrolysed. (R) : Carbon has no d-orbitals in its valence shell, but silicon has vacant d-orbitals in its valence shell.

Butyric acid contains only C, H and O. A 4.24 mg sample of butyric acid is completely burnt. It gives 8.45 mg of carbon dioxide and 3.46 mg of water. What is the mass percentage of each element in butyric acid?

A burning strip of magnesium is introduced into a jar containing a gas. After sometime the walls of the container is coated with carbon. The gas in the container is

A burning magnesium ribbon will continue to burn in

A bulb of unknown volume 'V' Contains an ideal gas at 2 atm pressure. It was connected to another evacuated bulb of volume 0.5 litre through a stopcock. When the stopcock was opened the pressure in each bulb became 0.5 atm. Then V is

A bulb emits light of wavelength 4500 Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second ?

A bulb emits light of wave length 4500Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second ?

A bulb containing N_(2)O_(4) is colourless in ice. Its colour inboiling water is …………. while in water at 298 K, it is………… .

A buffer solution with pH 9 is to be prepared by mixing NH_(4)Cl that should be added to one litre of 1.0mNH_(4)OH kb1.8 xx 10^(-5)

A buffer solution with pH 9 is to be prepared by mixing NH_(4) Cl and NH_(4)OH. Calculate the number of moles of NH_(4)Cl that should be added to one litre of 1.0 M NH_(4)OH (K_(b)=1.8xx10^(-5))

A buffer solution of pH = 4.7 is prepared from CH_(3)COONa and CH_(3)COOH. Dissociation constant of acetic acid is 1.75xx10^(-5). Calculate the mole proportion of sodium acetate and acetic acid.

A buffer solution of pH 8.3 is prepared from ammonium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8xx10^(-5). What is the mole proportion of ammonium chloride and ammonium hydroxide?

A buffer solution of acetic acid and sodium acetate is diluted 10 times. What is the effect on itspH ?

A buffer solution is prepared in which the concentration of NH_(3) is 0.30 M and the concentrationof 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 buffer solution is prepared by mixing CH_(3) COONa and CH_(3) COOH. The pK_(a) of acetic acid is 4.74 . To maintain the pH of the buffer solution as 6.04 , the concentration ratio of CH_(3) COONa and CH_(3) COOH to be maintained is

A buffer solution is formeed by mixing 100 mL of 0.1 M CH_(3)COOH with 200 mL of 0.02 M CH_(3)COONa. If this buffer solution is made to 1.0 L by adding 700 mL of water, pH will change by a factor of

A buffer solutionis prepared by mixing 50 mL of 0.2 Mof acetic acid with .x. mL of 0.2 M of NaOHsolutions. If pH of the resulting buffersolutions is 4. 7 then value of .x.is (Ka= 2xx10 ^(-5))

A buffer solution contains a weak acid HA and A . When small quantity of NaOH is added, to keep p^(H)as constant. which of the following reaction takes place?

A buffer solution contains 0.40 mol of ammonium hydroxide and 0.50 mol of ammonium chloride to make a buffer solution of 1 L. Calculate the pH of the resulting buffer solution. Dissociation constant of ammonium hydroxide at 25^(@)C is 1.81xx10^(-5).

A buffer solution contains 2 moles of ammonium hydroxide and 0.25 mole of ammonium chloride per dm^(3) of the solution. (K_(b) for ammonium hydroxide = 1.8xx10^(-5)). Calculate the pH of the buffer solution.

A buffer solution contains 0.2mole of NH_4 OHand 0.2 mole of NH_4 Cl perlitre. The PK_bofNH_4 OHis 4.75 The pH of the buffer will be

A buffer solution contains 0.1 mole of sodium acetate in 1000 cm^(3) of 0.1 M acetic acid. To the above buffer solution, 0.1 M aceticacid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The pH of the resultingbuffer is equal to ...........

A buffer solution consists of 1 mole each of HA and its conjugate base A^(-), Addition of which of the following decreases the pH of buffer ?

A buffer solution can be prepared from a mixture of