1869 + Interview Questions in IONIC-EQUILIBRIUM IN GENERAL AWARENESS Page 34 InterviewSolution

1651.	Amino acid glycine `(NH_(2)-CH_(2)-COOH)` exists as a zwitter ion in aq. Solution. The `K_(a)` and `K_(b)` values of glycine are `1.6xx10^(-10)(Pk_(a)=9.8)`and `2.5xx10^(-12)(pk_(b)=11.6)` respectively. The `K_(a)` and `K_(b)` values are for zwitter ion of Amino acid with following strctures `[NH_(3)^(+)-CH_(2)-COO^(-)]` An aqueous solution of glycine has pH:A. nearly 7B. nearly 7.9C. nearly 6.1D. nearly 11.5
Answer» Correct Answer - c

Discussion

1652.	At `25^(@)C`, the `K_(sp)` value of `Fe(OH)_(2)` in aquoeus solution is `3.8 xx 10^(-38)`. The solubility of `Fe^(3+)` ions will increases whenA. `P^(H)` is increasedB. `P^(H)` is `7`C. `P^(H)` is decreasedD. `P^(H) = 14`
Answer» Correct Answer - C In acidic solution, `OH^(-)` ins are neutrarlized with `H^(+)` ions then equailibrium shift to forward direction.

Discussion

1653.	Boric acid `Br(OH)_(3)` is monobasic Lewis acid if gives `H^(+)` ions in aquoeus soluions as follows `B(OH)_(3) + H_(2)O hArr [B(OH)_(4)]^(-) + H^(+)` `K_(a) = 5.9 xx 10^(-10)` The `P^(H)` of `0.025M` aqueous solution is `[log 5.9 = 0.771]`A. `5.42`B. `4.52`C. `2.45`D. `4.675`
Answer» Correct Answer - A `P^(H) = - 1/2 [Log K_(a) + logC]` `= - 1/2["log"(5.9 xx 10^(-10)) + "log" (0.025)]` `= - 1/2[(-10 + "log"5.9) + (-2 + "log" 2.85)]` `= - 1/2[(-10 + 0.771 )+ (-2 + 0.3979)]` `= - 1/2[(-9229) + (-16021)]` `= - 1/2 [-9.229 - 1.6021]` `=- 1/2[-10.3811] = 5.4255`

Discussion

1654.	The pH of a solution obtained by mixing 100mL of a solution pH of `=3` with 400mL of a solution of pH `=4` isA. `3- log 2.8`B. `7- log 2.8`C. `4- log 2.8`D. `5-log 2.8 `
Answer» Correct Answer - C `N_(1)V_(1)+N_(2)V_(2)=N_(3)(100+400)` or `N_(3)=(0.1+0.04)/(500)=(0.14)/(500)=2.8 xx 10^(-4)M` `pH=-log (2.8 xx 10^(-4))=4=log 2.8`

Discussion

1655.	Calculate the `pH` of the resulting solution formed by mixing the following solutions: (a) `20ml" of "0.2M" "Ba(OH)_(2)+30mL" of "0.1M" "HCl` (b) `2 mL" of "0.1" M "HCl + 10 ml" of "0.01" M "KOH` (c ) `10mL" of "0.1" M "H_(2)SO_(4)+10 mL" of "0.1 M KOH`.
Answer» Milimoles of `H^(+)=10xx0.2+40xx0.1xx2=10` So `[H^(+)]=(10)/(50)M` `:.pH=0.7`.

Discussion

1656.	The concentration of `[H^(+)]` and concentration of `[OH^(-)]` of a `0.1` aqueous solution of `2%` ionised weak acid is [Ionic product of water `=1xx10^(-14)`]A. `0.02xx10^(-3)M` and `5xx10^(-11)M`B. `1xx10^(-3)M` and `3xx10^(-11)M`C. `2xx10^(-3)M` and `5xx10^(-12)M`D. `3xx10^(-2)M` and `4xx10^(-13)M`
Answer» Correct Answer - C `[H^(+)]` in monobasic acid = molarity `xx` degree of ionisation `=0.1xx(2)/(100)=2xx10^(-3)M` ionisation constant of water `K_(w)=(H^(+))(OH^(-))` `[OH^(-)]=(K_(w))/([H^(+)])=(1xx10^(-14))/(2xx10^(-3))=5xx10^(-12)M`

Discussion

1657.	The solubility product of CuS, CdS and HgS are `10^(-31),10^(-44),10^(-54)` respectively. The solubility of these sulphides are in the orderA. CdS gt HgS gt CuSB. HgS gt CdS gt CuSC. CdS gt CuS gt HgSD. CuS gt CdS gt HgS
Answer» Correct Answer - D All are binary salts, hence their solubility is equal to square root of their solubility products. So, order of solubility is CuS gt CdS gt HgS or greater the value of solubility product, greater will be the solubility.

Discussion

1658.	Calculate the pH of a given mixtures. (a) `(4g CH_(3)COOH+6g CH_(3)COONa)` in 100 mL of mixture, (`K_(a)` for `CH_(3)COOH=1.8xx10^(-5))` (b) 5 mL of `0.1MBOH+250` mL of `0.1 MBCI`, (`K_(a)` for `MOH=1.8xx10^(-5))` (c ) (`0.25` mole of `CH_(3)COOH+0.35` mole of `CH_(3COOH=3.6xx10^(-4))`
Answer» (a) We have pH= -log `K_(a)+log (["Salt"])/(["Acid"])` `:. ["Salt"]= (6xx1000)/(82xx100)M` and `["Acid"]=(4xx1000)/(60xx100)M` `:. pH= -log 1.8xx10^(5)+log(6xx1000//82xx100)/(4xx1000//60xx100)` `:. pH= 4.7851` (b) `pOH= -log K_(b)+log ((["Salt"])/(["Base"]))` `.: "Total volume after mixing"=250+5=255mL` Meq.of salt `= 250xx0.1=25` Meq.of base `=5xx0.1=0.5` `:. ["Salt"]= (25)/(255)` `:. pOH= -log 1.8xx10^(-5)+log((25//255)/(0.5//255))` pOH= 6.4437` `:. pOH=14-pOH=7.5563` (c ) `pH= -log K_(b)+log ((["Salt"])/(["Acid"]))` `= -llog 3.6xx10^(-4)+log ((0.35//500)/(0.25//500))` `pH= 3.5898`

Discussion

1659.	What is `[H^(+)]` in `mol//L` of a solution that is `0.20 M` in `CH_(3)COONa` and `0.1 M` in `CH_(3)COOH`? `K_(a)` for `CH_(3)COOH` is `1.8xx10^(-5)`?A. `9.0 xx 10^(-6)`B. `3.5 xx 10^(-4)`C. `1.1 xx 10^(-5)`D. `1.8 xx 10^(-15)`
Answer» Correct Answer - A The given solution is an acidi buffer. Hence `pH =pK_(a)+log.(["Salt"])/(["Acid"])` `=-log (1.8 xx 10^(-5))+log. (0.20)/(0.10)` or ` -log [H^(+)]=log .(2)/(1.8 xx 10^(-5))` or `log\|(1)/(H^(+))\|=log. (2)/(1.8 xx 10^(-5))` or `\|H^(+)\|=(1.8 xx 10^(-5))/(2)=0.9 xx 10^(-5)M`

Discussion

1660.	`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `5.0 xx 10^(-7) m^(3)`B. `0.6 xx 10^(-12) M^(3)`C. `4.0 xx 10^(-8)M^(3)`D. `5.0 xx 10^(-9) M^(3)`
Answer» Correct Answer - A Solubility expressed in terms of molarity `[OH^(-)] = 10^(-2)` `[Ba^(+2)] = 0.5 xx 10^(-2)` `K_(SP) = (0.5 xx 10^(-2)) (10^(-2))^(2)` `K_(SP) = 5 xx 10^(-7)`

Discussion

1661.	`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `4.00xx10^(-6)M^(3)`B. `4.00xx10^(-7)M^(3)`C. `5.00xx10^(-7)M^(3)`D. `5.00xx10^(-6)M^(3)`
Answer» Correct Answer - D Given, pH of `Ba(OH)_(2)=12` So, pOH = 2 `therefore [H^(+)]=[1xx10^(-12)]` `K_(w)=(H^(+))(OH^(-))` `K_(w)=1xx10^(-14)` `OH^(-)=(K_(omega))/(H^(+))` and `[OH^(-)]=(1xx10^(-14))/(1xx10^(-12))[because[H^(+)][OH^(-)]=1xx10^(-14)]` `=1xx10^(-2)mol//L` `Ba(OH)_(2)rarr underset(S)(Ba^(2+)+underset(2S)(2OH^(-))` `K_(sp)=[Ba^(2+)][OH^(-)]^(2)=[S][2S]^(2)` `=[(1xx10^(-2))/(2)](1xx10^(-2))^(2)` `=0.5xx10^(-6)=5.0xx10^(-6)M^(3)`

Discussion

1662.	`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `3.3xx10^(-7)`B. `5.0xx10^(-7)`C. `4.0xx10^(-6)`D. `5.0xx10^(-6)`
Answer» Correct Answer - B Given, pH of `Ba(OH)_(2)=12` `therefore pOH=14-pH` = 14 - 12 = 2 We know that, `pOH=-log[OH^(-)]` `2=-log[OH^(-)]` `[OH^(-)]` = antilog (-2) `[OH^(-)]=1xx10^(-2)` `Ba(OH)_(2)` dissolves in water as `underset("S mol L"^(-1))(Ba(OH)_(2))(S)hArr underset(S)(Ba^(2+))+underset(2S)(2OH^(-))` `therefore [OH^(-)]=2S=1xx10^(-2)` `S=([OH^(-)])/(2) " " [Ba^(2+)=S]` `[Ba^(2+)]=([OH^(-)])/(2)=(1xx10^(-2))/(2)` `K_(sp)=[Ba^(2+)][OH^(-)]^(2)` `=((1xx10^(-2))/(2))(1xx10^(-2))^(2)` `=0.5xx10^(-6)=5xx10^(-7)`

Discussion

1663.	`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `5.06 xx 10^(-7) M^(3)`B. `4.00xx10^(-6)M^(3)`C. `4.00xx 10^(-7) M^(3)`D. `5.00 xx 10^(-6) M^(3)`
Answer» Correct Answer - B `pH =12` means `\|H^(+)\|=10^(-12)M` `:. [OH^(-1)]=10^(-2)` `Ba(OH)_(2) hArr Ba^(2+)+2OH^(-) ` `pH =10^(-2) M :. Ba^(2+)=(10^(-2))/(2)M` `=5 xx 10^(-3)M` `K_(sp)=\|Ba^(2+)\|" "\|OH^(+)\|^(2)=(5 xx 10^(-3))(10^(-2))^(2)` `=5 xx 10^(-7) M^(3)`

Discussion

1664.	At `25^(@)C`, the solubility product of `Mg(OH)_(2)` is `1.0xx10^(-11)`. At which `pH`, will `Mg^(2+)` ions start precipitating in the form of `Mg(OH)_(2)` from a solution of `0.001 M Mg^(2+)` ions ?A. 9B. 10C. 11D. 8
Answer» Correct Answer - b

Discussion

1665.	The pH of a saturated solution of `Fe(OH)_(2)` is 8.67. What is the `K_(sp)` for `Fe(OH)_(2)`?A. `5xx10^(-6)`B. `2xx10^(-11)`C. `1xx10^(-16)`D. `5xx10^(-17)`
Answer» Correct Answer - d

Discussion

1666.	Equal volumes of three acid solution of pH 3,4 and 5 are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixtureA. `3.7 xx 10^(-3) M`B. `1.11 xx 10^(-3) M`C. `1.11 xx 10^(-4) M`D. `3.7 xx 10^(-4) M`
Answer» Correct Answer - D `[H^(+)]_(mix) = (10^(-3)x + 10^(-4)x + 10^(-5)x)/(3x)` `= (1)/(3)(10^(-3) + 10^(-4) + 10^(-5)) = 37 xx 10^(-5) = 3.7 xx 10^(-4)`.

Discussion

1667.	A base type indicator `B` differs in colour from its conjugate acid.`(BH^(+))` .Acidic form is red in colour while basic form is blue in colour. Human eye can sense blue colour distincity when ratio of blue form concentration to red form cocnentration is `(a)/(b)`. or more.However ,red colour can be sensed by human eye distinctly when ratio of red form concentration of blue form concentration is `(c)/(d)` or more. Determine the `pH` range of solution in which human eyes will be unable to observe distinct red or distinctly blue colour.Take ionisation of `B` as `K_(eq)`.
Answer» `:.pOH=pK_(eq)+log((["Acidic form"])/(["Baic form"]))=pK_(eq)+log((["Red"])/([Blue]))` `(pOH)_(blue)=pK_(eq)+log((b)/(a))` and `p(OH)_(red)=pK_(eq)+pK_(eq)+log((c)/(d))` so range `=pH_(blue)` to `pH_(red)=14-pK_(eq)-log((b)/(a))` to `14-pK_(eq)-log((c)/(d))`

Discussion

1668.	Equal volumes of three acid solutions of `pH 3, 4` and `5` are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixture?A. `1.11xx10^(-4)M`B. `3.7xx10^(-4)M`C. `3.7xx10^(-3)M`D. `1.11xx10^(-3)M`
Answer» Correct Answer - B Let the volume of each acid = V pH of first, second and third acids = 3,4 and 5 respectively `[H^(+)]` of first acid `(M_(1))=1xx10^(-3)[because H^(+)=1xx10^(-)pH]` `[H^(+)]` of second acid `(M_(2))=1xx10^(_4)` `[H^(+)]` of third acid `(M_(3))=1xx10^(-5)` Total `[H^(+)]` concentrated of mixture `(M)=(M_(1)V_(1)+M_(2)V_(2)+M_(3)V_(3))/(V_(1)+V_(2)+V_(3))` `=(1xx10^(-3)xxV+1xx10^(-4)xxV+1xx10^(-5)xxV)/(V+V+V)` `=(1xx10^(-3)xx V(1+0.1+0.01))/(3V)` `=(1.11xx10^(-3))/(3)=3.7xx10^(-4)M`

Discussion

1669.	An acid-base indicator has `K_(a) = 10^(-5)`. The acid form of the indicator is red and basic form is blue. Which of the following is//are correct?A. At `pH = 4.52`, solution is redB. At `pH - 5.47`, solution is blue.C. At `pH =6`, solution is `75%` redD. At `pH = 8`, solution is `75%` blue.
Answer» Correct Answer - A::B `{:(HIn,hArr,Ind^(Theta),+,H^(o+)(pK_(a)=5)),(Red,,Blue,,):}` a. When `pH lt pK_(a) (4.52 lt 5)`, colout of acid from i.e., red predominates. b. When `pH gt pK_(a) (5.47 gt 5)`, colour of basic from i.e., blue predominates. c. `75% red rArr [(In^(Theta))/(HIn)] = (0.75)/(0.25)` `rArr pH = 5+ log 3 = 5 + 0.48 = 5.48` (Hence (c) is wrong)

Discussion

1670.	The Ph of basic buffer mixtures is given by : `Ph=Pk_(a)+`log `(["Base"])/(["Salt"])` whereas Ph of acidic buffer mixtures is given by : Ph =`pK_(a)+"log"(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no appreciable change in Ph for all practical purposes, but sicne the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. A solution containing 0.2 mole of dichloroacetic acid `(K_(a)=5xx10^(-2))` and 0.1 mole sodium dichloracetate in one litre solution has `[H^(+)]`:A. 0.05 mB. 0.025 mC. 0.10 mD. 0.005 m
Answer» Correct Answer - a

Discussion

1671.	The Ph of basic buffer mixtures is given by : `Ph=Pk_(a)+`log `(["Base"])/(["Salt"])` whereas Ph of acidic buffer mixtures is given by : Ph =`pK_(a)+"log"(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no appreciable change in Ph for all practical purposes, but sicne the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. The ratio of pH of solution (I) containing 1 mole to pH of solution (II) containing 1 mole of `CH_(3)COONa` and 1 mole of acetic in one litre is :A. `1:2`B. `2:1`C. `1:3`D. `3:1`
Answer» `{:(CH_(3)COONa,+,HCl,rarr,CH_(3)COOH,+,NaCl),(1,,1,,0,,0),(0,,0,,1,,1):}` `:.[CH_(3)COOH]=(1)/(1)=1` `:.[H^(+)]=Calpha=Csqrt((K_(a))/(C))=sqrt(K_(a).C)=sqrt(K_(a))` or `pH_(1)=-(1)/(2)logK_(a)=(1)/(2)pK_(a)` `underset(1)(CH_(3)COOH)+underset(1)(CH_(3)COONa)` `:.pH=pK_(a)+(log)(1)/(1)` `pH_(2)=pK_(a)` `(pH_(1))/(pH_(2))=(1)/(2)`

Discussion

1672.	The Ph of basic buffer mixtures is given by : `Ph=Pk_(a)+`log `(["Base"])/(["Salt"])` whereas Ph of acidic buffer mixtures is given by : Ph =`pK_(a)+"log"(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no appreciable change in Ph for all practical purposes, but sicne the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. The amount of `(NH_(4))_(2)SO_(4)` to be added to 500mL of 0.01 M `NH_(4)OH` solution `(pH_(a)NH_(4)^(+)` is 9.26) to prepare a buffer of pH 8.26 is :A. 0.05 moleB. 0.025 moleC. 0.10 moleD. 0.005 mole
Answer» Correct Answer - b

Discussion

1673.	Equal volumes of `1xx10^(-4)` M solutions of `Cd^(2+)` and `CO_(3)^(2-)` ions are mixed in one flask and equal volumes of `1xx10^(-4)` M solutions of `Ag^(+) " and " CrO_(4)^(2-)` ions are mixed in a second. Which substances precipitate ? `{:("Formula",CdCO_(3), Ag_(2)CrO_(4)),(K_(sp), 5.2xx10^(-12), 1.1xx10^(-12)):}`A. `CdCO_(3)` onlyB. `Ag_(2)CrO_(4)` onlyC. BothD. Neither
Answer» Correct Answer - a

Discussion

1674.	Why the following compounds will produce acidic solution in water i. `H_(3)PO_(4)` ii. `CO_(2)` iii. `HNO_(2)` iv. `AICI_(3)`
Answer» a. All of them give `H^(o+)` ions in squeous solution. i. `H_(3)PO_(4) rarr H^(o+) +H_(2)PO_(4)^(Θ)` ii. `CO_(2) + H_(2)O rarr H_(2)CO_(3) rarr H^(o+) + HCO_(3)^(Θ)` iii. `HNO_(2) rarr H^(o+) + NO_(2)^(Θ)` iv `A1CI_(3) + H_(2)O rarr [A1(H_(2)O)_(6)]^(3+)CI_(3)` `[A1(H_(2)O)_(6)]^(3+) rarr [A1(H_(2)O)_(5)(OH)]^(2+) + H^(o+)`

Discussion

1675.	When different types of salts have nearly same solubility product constant `K_(SP)` but less than one the most soluble salt is that:A. which produces maximum number of ionsB. which produces minimum number of ionsC. which produces more charge on ionD. none of the above
Answer» Correct Answer - a

Discussion

1676.	The pH of basic buffer mixtures is given by : `pH=pK_(a)+log((["Base"])/(["Salt"]))` , whereas pH of acidic buffer mixtures is given by: `pH= pK_(a)+log((["Salt"])/(["Acid"]))`. Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` change, a slight decrease or increase in pH results in. A solution containing `0.2` mole of dichloroacetic acid `(K_(a)=5xx10^(-2))` and `0.1` mole sodium dichloroacetate in one litre solution has `[H^(+)]` :A. `0.05M`B. `0.025M`C. `0.10M`D. `0.005M`
Answer» Since `{:(CHCl_(2)COONa,rarr ,CHCl_(2)COO^(-),+,Na^(+)),(,,0.1,,0.1),(CHCl_(2).COOH,hArr ,CHCl_(2)COO^(-),+,H^(+)),(0.2,,0,,0),((0.2-x),,(x+0.1),,x):}` `K_(a)=([CHCl_(2)COO^(-)][H^(+)])/([CHCl_(2)COOH])` or `5xx10^(-2)=([0.1+x][x])/([0.2-x])` `:. x=0.05`

Discussion

1677.	A solution of a substances is titrated against a strong base ( or acid ) ,volume `V` of strong base (or acid) is plotted against `pH` of the solution (as shown in figure) .The substances could be : A. `Na_(2)CO_(3)`B. Ethylene diamineC. `H_(2)C_(2)O_(2)`D. `CH_(2)(COOH)_(2)`
Answer» Given graph is plotted for the titration of weak diprotic acid with base.

Discussion

1678.	When weak base solution `(50mL of 0.1 N NH_(4)OH)` is titrated with strong acid `(0.1N HCI)`, the `pH` of the solution initially decrease fast and then decreases slowely till near the equivalence point (as shown in figure). Which of the following is//are correct. A. The slow decrease of `pH` is due to the formation of an acidic buffer solution after the addition of some `HCI`.B. The slope of shown graph will be minimum when `25mL` of `0.1N HCI` is added.C. The slow decrease of `pH` is due to the formation of basic buffer solution.D. The initial fast decrease in `pH` is due to fast consumption of `overset(Theta)OH` ions by `HCI`.
Answer» Correct Answer - B::C::D (a) is wrong. For reaction of `S_(A)` with `S_(B)`, basic buffer is formed `beta` is maximum. When `pOH = pK_(b)`. At `50%` neutralisation `(25 mL "of" 0.1 N HCI)` Slope of the given graph will be least the buffer will have maximum buffer capacity

Discussion

1679.	A 40 mL solution of weak base BOH is tritrated with `0.1N HCI` solution. The pH of solution is found to be `10.04` and `9.14` after the addition of `5.0mL` and `20.0` mL of acid respecctively. Find out `K_(b)` for weak base.
Answer» Case I: `{:(,BOH+,HCI,rarrBCI+,H_(2)O),("Millimole before reaction", a,0.1xx5=0.5,0,0),("Millimole after reaction",(a-0.5),0,0.5,0.5):}` `{:( :.pH=10.04),( :. pOH=3.96):}:\|{:( :. pOH=-log((K_(b)+log([BC]))/([BOH]))....(1)),( :. 3.96=-logK_9b+log(0.5/((a-0.05)))....(2)):}` Case II: `{:(,BOH+,HCI,rarrBCI+,H_(2)O),("Millimole before reaction", alpha,0.1xx20=2,0,0),("Milliomole after reaction", (a-2),0,2,2):}` `{:( :.pH=9.14),( :. pOH=4.86):}:\|{:( :. pOH=-logK_(b)+log(([BC])/([BOH]))....(3)),( :. 4.86=-logK_b+log(2/((a-2)))....(4)):}` Solving eqs. (2) and (4), `K_(b)= 1.81xx10^(-5)`

Discussion

1680.	When weak base solution `(50mL of 0.1 N NH_(4)OH)` is titrated with strong acid `(0.1N HCI)`, the `pH` of the solution initially decrease fast and then decreases slowely till near the equivalence point (as shown in figure). Which of the following is//are correct. A. The initial fast decrement in `pH` is due to fast consumption of free `OH^(-)` ions by `HCl`B. The slow decreases of `pH` is due to formation of an acidic buffer solution after addition of some`HCl`C. The slope of shown `pH` graph (magnitude only) will be minimum when `25mL` of `0.1 M HCl` is added.D. The slow decreases of `pH` is due to formation of a basic buffer solution after addition of some `HCl`.
Answer» Initial decrement is due to consumptionof free `OH^(-)` ions, then slow decrement in `pH` is due basic buffer solution and minimum slope will be there whenthere is best buffer action `(["salt"]/["Base"]=1)`

Discussion

1681.	A 50 mL solution of weak base BOH is titrated with `0.1N HCI` solution. The pH of solution is found to be `10.04` and `9.14` after the addition of `5.0mL` and `20.0` mL of acid respectively. Find out `K_(b)` for weak base.
Answer» Correct Answer - A::B::D

Discussion

1682.	When a strong acid is titrated using a weak base, the pH at the equivalence point isA. 7B. `gt 7`C. `lt 7`D. `~~7`
Answer» Correct Answer - C

Discussion

1683.	A weak base `(BOH)` with `K_(b) = 10^(-5)` is titrated with a strong acid `(HCl)`, At `3//4` th of the equivalence point, pH of the solution is:A. `5 + "log"3`B. `14-5-"log"3`C. `14-5 + log3`D. `9.523`.
Answer» Correct Answer - B Let the initial no. of equivalents of `BOH` be x `{:(,BOH+HCl,rarr,BCl+,H_(2)O),("Initial equivalent",x,,0,0),("At"3//4th eqv. pt., x-(3x)/(4),,(3)/(4)x,(3)/(4)x):}` `= x/4` Now solution contains `x/4` moles of weak base and `(3x)/(4)` moles of its conjugated acid which is a basic buffer. `pOH = pK_(b) + "log" (["salt"])/(["Base"]) = 5+ "log" (3x xx 4)/(4 xx x)` `pH = 14 - 5 - log 3 = 8.523`

Discussion

1684.	A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:
Answer» Correct Answer - `8.219xx10^(-6);`

Discussion

1685.	A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:A. `1.8xx10^(-5)`B. `8.22xx10^(-6)`C. `1.8xx10^(-6)`D. `8.2xx10^(-5)`
Answer» Correct Answer - b

Discussion

1686.	A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:A. `1.8xx10^(-5)`B. `8xx10^(-6)`C. `1.8xx10^(-8)`D. `8xx10^(-5)`
Answer» m.moles of `HA` taken =`27xx0.1=2.7` `{:(,HA,+,OH^(-),rarr,A^(-),+,H_(2)O),(t=0,2.7,,1.2,,,,),(t_(eq),1.5,,-,,1.2,,):}` `pH=pK_(a)+log(([A^(-)])/([HA]))rArr 5=pK_(a)+log((1.2)/(1.5))=pK_(a)+log((4)/(5))` `:.pK_(a)=5.1 rArr K_(a)=8x10^(-6).`

Discussion

1687.	A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:A. `1.8xx10^(-5)`B. `8.2xx10^(-6)`C. `1.8xx10^(-6)`D. `8.2xx10^(-5)`
Answer» Correct Answer - B For neutralisation, Total Meq.of acid = Meq.of base `= 26.6xx0.1=2.66` Now for partial netralisation of acid, `{:(,HA+,BOHrarr,BA+,H_(2)O),("Meq. before reaction",2.66,1.2,0,0),("Meq. after reaction",1.46,0,1.2,1.2):}` The rersultant mixture acts as a buffer and, [HA] ans [BA] may be placed in terms of Meq. since volume of mixture is constant. `pH= -log K_(a)+log([Sal t])/([Acid])` or `5= -log K_(a)+log (1.2)/(1.46)` `K_(a)=8.219xx10^(-6)`

Discussion

1688.	The dihydrogen phosphate ion undergoes these reaction in water `H_2PO_4^(-)(aq)+H_2O(l)toHPO_4^(2-)(aq)+H_3O^(+)(aq) K=6.2xx10^(-8)` `H_2PO_4^(-)(aq)+H_2O(l)toH_3PO_4(aq)+OH^(-)(aq) K=1.6xx10^(-7)` What is the conjugate base of `H_2PO_4^(-)` ?A. `HPO_4^(2-)(aq)`B. `H_2O(l)`C. `OH^-` (aq)D. `H_3PO_4` (aq)
Answer» Correct Answer - A

Discussion

1689.	A weak base `BOH` is titrated with strong acid `HA`. When `10mL` of `HA` is added, the `pH` is `9.0` and when `25mL` is added, `pH` is `8.0`. The volume of acid required to reach the equivalence point isA. `50mL`B. `40mL`C. `35mL`D. `30mL`
Answer» Correct Answer - D `{:(BOH+,HArarr,BA+H_(2)O,,),(W_(B),S_(A),("Salt of" W_(B)//S_(A)),,):}` `[W_(B)` is in excess, so basic buffer is formed] `pOH = pK_(b) + "log" ([BA])/([BOH]) (pH = 9, pOH = 14 - 9 = 5)` `5 = pK_(b) + log ((10)/(V - 10)) …(i)` When `pH = 8, :. pOH = 14 - 8 = 6`. `6 = pK_(b) + log ((25)/(V - 25))` Solve equation (i) and (ii), `V = 30mL`.

Discussion

1690.	For ortho phosphoric acid, `H_3PO_4(aq)+H_2O(aq) to H_3O^(+)(aq)+H_2PO_4^(-)(aq),K_(a_1)` `H_2PO_4^-(aq)+H_2O(aq) to H_3O^(+)(aq)+H_2PO_4^(-)(aq),K_(a_2)` `HPO_4^(2-)(aq)+H_2O(aq) to H_3O^(+)(aq)+PO_3^(4)(aq),K_(a_3)` The correct order of `K_a` values is :A. `K_(a_1) gt K_(a_2) lt K_(a_3)`B. `K_(a_1) lt K_(a_2) lt K_(a_3)`C. `K_(a_1) gt K_(a_2) gt K_(a_3)`D. `K_(a_1) lt K_(a_2) gt K_(a_3)`
Answer» Correct Answer - c

Discussion

1691.	`H_2CO_3(aq)+H_2O(l)toHCO_3^-(aq) + H_3O^+(aq)` `HCO_3^(-)(aq) + H_2O(l)to CO_3^(2-)(aq)+H_3O^(+)(aq)` Accoding to the equations above, what is the conjugate base of `HCO_3^-`?A. `H_2CO_3(aq)`B. `H_2O(l)`C. `H_3O^+(aq)`D. `CO_3^(2-)(aq)`
Answer» Correct Answer - D

Discussion

1692.	0.2 mol each of AgCl and NaCl are heated with conc. `H_2SO_4` in presence of `K_2Cr_2O_7` giving deep red vapours which are dissolved in 30 litre of water.What is pH of the resulting solution ? [log 2=0.3, log 3=0.48] Given : `(K_(a_2))_(H_2CrO_4)=10^(-7)` Assume 100% first ionisation of `H_2CrO_4`A. 1.18B. 1.82C. 2.18D. 2
Answer» Correct Answer - d

Discussion

1693.	Statement: In an acid-basic titration involving a strong base and a weak acid, methyl orange can be used as an indicator. Explanation: Methyl orange changes its colour in the pH range 3 to 5.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-17B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 and Statement-2 is True
Answer» Correct Answer - d

Discussion

1694.	Statement: `CO+NaOH overset("High"P)underset("High"T)rarr HCOONa` Explanation: CO although being neutral can acts as acid in the given reaction.A. S is correct but E is wrong.B. S is wrong but E is correc.C. Both S and E are correct and E is correct explanation of S.D. Both S and E are correct but E is correct explanation of S.
Answer» Correct Answer - C CO is acid, NaOH is base and salt formed is HCOONa.

Discussion

1695.	The pH values 0.1 M solution of HCOONa (I), HCOOH (II),`CH_(3)COONH_(4)` (III), NaOH (IV) HCl (V), will be in the order :A. `IVgtIIIgtIgtIVgtV`B. `IVgtIgtIIIgtIIgtV`C. `IIgtIIIgtIgtIVgtV`D. `VgtIIgtIIIgtIgtIV`
Answer» Correct Answer - 2

Discussion

1696.	An acid buffer can be prepared by mixing solution ofA. sodiu chloride and hydrochloric acidB. sodium acetate and aceticeC. sodium hydroxide and boric acidD. sodium borate and boric acid
Answer» Correct Answer - B::D Acidic buffer is a mixture of weak acid and its salt with strong base.

Discussion

1697.	If the concentration of lead iodide in its saturated solution at `25^(@)C` be `2 xx 10^(-3)` moles per litre, then the solubility product isA. `4 xx 10^(-6)`B. `8 xx 10^(-12)`C. `6 xx 10^(-9)`D. `32 xx 10^(-9)`
Answer» Correct Answer - D `{:(PbI_(2) ,hArr,(Pb^(++)), + ,(2I^(-))),(,,S,,2S):}` `K_(sp) = 4S^(3) = 4 xx [2 xx 10^(-3)]^(3) = 32 xx 10^(-9)`.

Discussion

1698.	Buffer solution is prepared by mixingA. Strong acid + its salt of strong baseB. Weak acid + its salt of weak baseC. Strong acid + its salt of weak baseD. Weak acid + its salt of strong base
Answer» Correct Answer - D

Discussion

1699.	An acidic buffer solution can be prepared by mixing solution ofA. Sodium acetate and acetic acidB. Ammonium chloride and ammonium hydroxideC. Sulphuric acid and sodium sulphateD. Sodium chloride and sodium hydroxide.
Answer» Correct Answer - A `CH_93)COONa//CH_(3)COOH` combination constitutes acidic buffer among the given options.

Discussion

1700.	`50 cm^(3)` of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after adding `50 cm^(3)` of NaOH solution. The remaining titration is completed by adding 0.5 N KOH solution. What is the volume of KOH required for completing the titration ?A. `12 cm^(3)`B. `10 cm^(3)`C. `25 cm^(3)`D. `10.5 cm^(3)`
Answer» Correct Answer - B `{:("Acid",=,"Base"),(N_(1)V_(1),=,N_(2)V_(2)),(0.2 xx V_(1),=,0.1 xx 50):}` `:. V_(1)=(0.1 xx 50)/(0.2 ) =25cm^(3)` Thus , `25cm^(3)` is the volume of acid neutralized by `50 cm^(3)` of 0.1 N NaOH `:. ` volume of 0.2 N HCl left behind `=50-25=25 cm^(3)` This is neutralized by 0.5 KOH Acid= Base `N_(1)V_(1) = N_(2)V_(2)` `0.2 xx 25 =0.5 xx V_(2) :. V_(2)=(0.2 xx 25)/(0.5 ) =10cm^(3)` `:. `Volume of 0.5 N KOH required `=10cm^(3)`

Discussion

Explore topic-wise InterviewSolutions in .

At `25^(@)C`, the `K_(sp)` value of `Fe(OH)_(2)` in aquoeus solution is `3.8 xx 10^(-38)`. The solubility of `Fe^(3+)` ions will increases whenA. `P^(H)` is increasedB. `P^(H)` is `7`C. `P^(H)` is decreasedD. `P^(H) = 14`

Boric acid `Br(OH)_(3)` is monobasic Lewis acid if gives `H^(+)` ions in aquoeus soluions as follows `B(OH)_(3) + H_(2)O hArr [B(OH)_(4)]^(-) + H^(+)` `K_(a) = 5.9 xx 10^(-10)` The `P^(H)` of `0.025M` aqueous solution is `[log 5.9 = 0.771]`A. `5.42`B. `4.52`C. `2.45`D. `4.675`

The pH of a solution obtained by mixing 100mL of a solution pH of `=3` with 400mL of a solution of pH `=4` isA. `3- log 2.8`B. `7- log 2.8`C. `4- log 2.8`D. `5-log 2.8 `

Calculate the `pH` of the resulting solution formed by mixing the following solutions: (a) `20ml" of "0.2M" "Ba(OH)_(2)+30mL" of "0.1M" "HCl` (b) `2 mL" of "0.1" M "HCl + 10 ml" of "0.01" M "KOH` (c ) `10mL" of "0.1" M "H_(2)SO_(4)+10 mL" of "0.1 M KOH`.

The solubility product of CuS, CdS and HgS are `10^(-31),10^(-44),10^(-54)` respectively. The solubility of these sulphides are in the orderA. CdS gt HgS gt CuSB. HgS gt CdS gt CuSC. CdS gt CuS gt HgSD. CuS gt CdS gt HgS

Calculate the pH of a given mixtures. (a) `(4g CH_(3)COOH+6g CH_(3)COONa)` in 100 mL of mixture, (`K_(a)` for `CH_(3)COOH=1.8xx10^(-5))` (b) 5 mL of `0.1MBOH+250` mL of `0.1 MBCI`, (`K_(a)` for `MOH=1.8xx10^(-5))` (c ) (`0.25` mole of `CH_(3)COOH+0.35` mole of `CH_(3COOH=3.6xx10^(-4))`

What is `[H^(+)]` in `mol//L` of a solution that is `0.20 M` in `CH_(3)COONa` and `0.1 M` in `CH_(3)COOH`? `K_(a)` for `CH_(3)COOH` is `1.8xx10^(-5)`?A. `9.0 xx 10^(-6)`B. `3.5 xx 10^(-4)`C. `1.1 xx 10^(-5)`D. `1.8 xx 10^(-15)`

`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `5.0 xx 10^(-7) m^(3)`B. `0.6 xx 10^(-12) M^(3)`C. `4.0 xx 10^(-8)M^(3)`D. `5.0 xx 10^(-9) M^(3)`

`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `4.00xx10^(-6)M^(3)`B. `4.00xx10^(-7)M^(3)`C. `5.00xx10^(-7)M^(3)`D. `5.00xx10^(-6)M^(3)`

`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `3.3xx10^(-7)`B. `5.0xx10^(-7)`C. `4.0xx10^(-6)`D. `5.0xx10^(-6)`

`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `5.06 xx 10^(-7) M^(3)`B. `4.00xx10^(-6)M^(3)`C. `4.00xx 10^(-7) M^(3)`D. `5.00 xx 10^(-6) M^(3)`

At `25^(@)C`, the solubility product of `Mg(OH)_(2)` is `1.0xx10^(-11)`. At which `pH`, will `Mg^(2+)` ions start precipitating in the form of `Mg(OH)_(2)` from a solution of `0.001 M Mg^(2+)` ions ?A. 9B. 10C. 11D. 8

The pH of a saturated solution of `Fe(OH)_(2)` is 8.67. What is the `K_(sp)` for `Fe(OH)_(2)`?A. `5xx10^(-6)`B. `2xx10^(-11)`C. `1xx10^(-16)`D. `5xx10^(-17)`

Equal volumes of three acid solution of pH 3,4 and 5 are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixtureA. `3.7 xx 10^(-3) M`B. `1.11 xx 10^(-3) M`C. `1.11 xx 10^(-4) M`D. `3.7 xx 10^(-4) M`

Equal volumes of three acid solutions of `pH 3, 4` and `5` are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixture?A. `1.11xx10^(-4)M`B. `3.7xx10^(-4)M`C. `3.7xx10^(-3)M`D. `1.11xx10^(-3)M`

An acid-base indicator has `K_(a) = 10^(-5)`. The acid form of the indicator is red and basic form is blue. Which of the following is//are correct?A. At `pH = 4.52`, solution is redB. At `pH - 5.47`, solution is blue.C. At `pH =6`, solution is `75%` redD. At `pH = 8`, solution is `75%` blue.

Why the following compounds will produce acidic solution in water i. `H_(3)PO_(4)` ii. `CO_(2)` iii. `HNO_(2)` iv. `AICI_(3)`

When different types of salts have nearly same solubility product constant `K_(SP)` but less than one the most soluble salt is that:A. which produces maximum number of ionsB. which produces minimum number of ionsC. which produces more charge on ionD. none of the above

A solution of a substances is titrated against a strong base ( or acid ) ,volume `V` of strong base (or acid) is plotted against `pH` of the solution (as shown in figure) .The substances could be : A. `Na_(2)CO_(3)`B. Ethylene diamineC. `H_(2)C_(2)O_(2)`D. `CH_(2)(COOH)_(2)`

A 40 mL solution of weak base BOH is tritrated with `0.1N HCI` solution. The pH of solution is found to be `10.04` and `9.14` after the addition of `5.0mL` and `20.0` mL of acid respecctively. Find out `K_(b)` for weak base.

A 50 mL solution of weak base BOH is titrated with `0.1N HCI` solution. The pH of solution is found to be `10.04` and `9.14` after the addition of `5.0mL` and `20.0` mL of acid respectively. Find out `K_(b)` for weak base.

When a strong acid is titrated using a weak base, the pH at the equivalence point isA. 7B. `gt 7`C. `lt 7`D. `~~7`

A weak base `(BOH)` with `K_(b) = 10^(-5)` is titrated with a strong acid `(HCl)`, At `3//4` th of the equivalence point, pH of the solution is:A. `5 + "log"3`B. `14-5-"log"3`C. `14-5 + log3`D. `9.523`.

A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:

A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:A. `1.8xx10^(-5)`B. `8.22xx10^(-6)`C. `1.8xx10^(-6)`D. `8.2xx10^(-5)`

A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:A. `1.8xx10^(-5)`B. `8xx10^(-6)`C. `1.8xx10^(-8)`D. `8xx10^(-5)`

A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:A. `1.8xx10^(-5)`B. `8.2xx10^(-6)`C. `1.8xx10^(-6)`D. `8.2xx10^(-5)`

A weak base `BOH` is titrated with strong acid `HA`. When `10mL` of `HA` is added, the `pH` is `9.0` and when `25mL` is added, `pH` is `8.0`. The volume of acid required to reach the equivalence point isA. `50mL`B. `40mL`C. `35mL`D. `30mL`

`H_2CO_3(aq)+H_2O(l)toHCO_3^-(aq) + H_3O^+(aq)` `HCO_3^(-)(aq) + H_2O(l)to CO_3^(2-)(aq)+H_3O^(+)(aq)` Accoding to the equations above, what is the conjugate base of `HCO_3^-`?A. `H_2CO_3(aq)`B. `H_2O(l)`C. `H_3O^+(aq)`D. `CO_3^(2-)(aq)`

The pH values 0.1 M solution of HCOONa (I), HCOOH (II),`CH_(3)COONH_(4)` (III), NaOH (IV) HCl (V), will be in the order :A. `IVgtIIIgtIgtIVgtV`B. `IVgtIgtIIIgtIIgtV`C. `IIgtIIIgtIgtIVgtV`D. `VgtIIgtIIIgtIgtIV`

An acid buffer can be prepared by mixing solution ofA. sodiu chloride and hydrochloric acidB. sodium acetate and aceticeC. sodium hydroxide and boric acidD. sodium borate and boric acid

If the concentration of lead iodide in its saturated solution at `25^(@)C` be `2 xx 10^(-3)` moles per litre, then the solubility product isA. `4 xx 10^(-6)`B. `8 xx 10^(-12)`C. `6 xx 10^(-9)`D. `32 xx 10^(-9)`

Buffer solution is prepared by mixingA. Strong acid + its salt of strong baseB. Weak acid + its salt of weak baseC. Strong acid + its salt of weak baseD. Weak acid + its salt of strong base

An acidic buffer solution can be prepared by mixing solution ofA. Sodium acetate and acetic acidB. Ammonium chloride and ammonium hydroxideC. Sulphuric acid and sodium sulphateD. Sodium chloride and sodium hydroxide.