1869 + Interview Questions in IONIC-EQUILIBRIUM IN GENERAL AWARENESS Page 5 InterviewSolution

201.	The `pH` of a blood stream is maintained by a proper balance of `H_(2)CO_(3)` and `NaHCO_(3)` .What volume of `5MNaHCO_(3)` solution should be mixed with `10ml` of a sample of solution which is `2.5M` in `H_(2)CO_(3)` in order to maintain a `pH=7.4`(Take `pK_(a_(1))` for `H_(2)CO_(3)=6.7,log2=0.3)`
Answer» `pH=pK_(a_(1))+log""([HCO_(3)^(-)])/([H_(2)CO_(3)])` `rArr 7.4=6.7+log""([HCO_(3)^(-)])/([H_(2)CO_(3)]) rArr ([HCO_(3)^(-)])/([H_(2)CO_(3)])=5` `:.5xx"moles of" H2CO3`=Moles of `HCO_(3)^(-)` (since both are components of same solution,volume of solution would be same for both) `rArr :.5xx2.5xx10xx10^(-3)=5-VrArr "required volume"=0.025L=25mL`

Discussion

202.	The `ph` of blood stream is maintained by a proper balance of `H_(2)CO_(3)` and `NaHCO_(3)` concentrations. What volume of `5M NaHCO_(3)` solution, should be mixed with `10mL` sample of blood which is `2M` in `H_(2)CO_(3)` in order to maintain a `pH` of `7.4.K_(a)` for `H_(2)CO_(3)` in blood is `4.0 xx 10^(-7)`?
Answer» `[H_(2)CO_(3)]` in blood `=2M [NaHCO_(3)] = 5M` Volume of blood `= 10mL` Let volume of `NaHCO_(3)` used `= VmL` `[H_(2)CO_(3)]` in mixture =` (2xx10)/((V +10))` `[NaHCO_(3)]` in mixture `= ((5xxV))/((V +10))` `pH = pK_(a) + log .(["salt"])/(["acid"])` `7.4 = - log 4.0 xx 10^(-7) + "log"((5xxV)//(V+10))/((2xx10)//(V+10))` `V = 40mL`

Discussion

203.	The volume of water needed to dissolve `1mg` of `PbSO_(4) (K_(sp) = 1.44 xx 10^(-8), Mw of PbSO_(4) = 303g)` at `25^(@)C` isA. `80mL`B. `43mL`C. `27.5mL`D. 10mL`
Answer» Correct Answer - C `S = (K_(sp))^(1//2) = (1.44 xx 10^(-8))^(1//2)` `= 12 xx 10^(-5)` `=303 xx 12 xx 10^(-5)gL^(-1)` ` = 3636 xx 10^(-5)gL^(-1)` `= 0.03636 gL^(-1) = 0.03636 mg mL^(-1)` `V_(H_(2)O) = (1)/(0.03636) = (10^(5))/(3636) = 27.5mL`

Discussion

204.	The Bronsted acid which gives the weakest conjugated base isA. HFB. `H_(2)S`C. `HCl`D. `H_(2)O`
Answer» Correct Answer - C A strong acid gives a weak base. HCl being the strongest acid (out of these ) and hence, has weakest conjugate base.

Discussion

205.	pH of blood is maintained constant by mechanisms ofA. Common ion effectB. BufferC. SolubilityD. All of these
Answer» Correct Answer - B

Discussion

206.	Human blood has a narrow Ph range of `7.3-7.4`, which must be maintained for methabolic processes to function properly. To keep the Ph in this range requires a delicate balance between the concentration of the conjugate acid-base pairs making upto the buffer system. The main buffer is a carbonic acid/ hydrogencarbonate system, which involves the following three equilibria. `CO_(2)(g)hArrCO_(2)(aq)` `CO_(2)(aq)+H_(2)O(l)hArrH_(2)CO_(3)(aq)` `H_(2)CO(aq)+H_(2)O(l)hArrHCO_(3)^(-)(aq)+H_(3)O^(+)(aq)` Carbonic acid `(H_(2)CO_(3))` is a weak acid and `HCO_(3)^(-)`(aq) is its conjugate base. At the temperature of the human body, the `pK_(a)` for carbonic acid is 6.4 However, the normal concentration of `CO_(2)(g)` in the lungs maintanis a ratio of `HCO_(3)^(-)(aq)//H_(2)CO_(3)(aq)` in blood plasma of about `8:1`. The carbonic acid concentration in the bloos is largely controlled by breathing and respiration. Hydrogencarbonate ion concentration is largely controlled by excreation in urine. If blood pH rises above `7.4`, a potentially life-threatening conditon called alkalosis can result. This can happen in patients who are hyperventilating from severse anxiety, or in climbers suffereing from oxygen deficency at high altitude. (Given: log 2=0.3) Calculate the maximum permissible value of `([H_(2)CO_(3)])/([HCO_(3)^(-)])` in the human blood to just prevent alkalosis.A. `10`B. `8`C. `0.1`D. `0.35`
Answer» Correct Answer - c

Discussion

207.	Assertion (A): `pH` of `10^(8)M HCI` is not equal to `8`. Reason (R) : `HCI` does not dissociate properly in very dilute solution.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct.
Answer» Correct Answer - C Correct `(R)`: In such very dilute solutions both source of `H^(o+)` ions from `HCI` and `H_(2)O` mus be considered and due to common ion `(H^(o+))` supression of ionisation occurs.

Discussion

208.	The `K_(b)` of weak base is `10^(-4)`. The `["salt"]` to `["base"]` ratio to be maintained to keep the `P^(H)` of buffer solution as `9` is .A. `1 : 4`B. `4 : 2`C. `1 : 10`D. `10 : 1`
Answer» Correct Answer - D `P^(H) = 14 - P_(kb) - "log" (["Salt"])/(["Base"])` `([S])/([B]) = 10 : 1`

Discussion

209.	The volume of the water needed to dissolve `1 g` of `BaSO_(4) (K_(SP)=1.1xx10^(-10))` at `25^(@) C` is:A. 280 litreB. 410 litreC. 205 litreD. 500 litre
Answer» Correct Answer - b

Discussion

210.	The volume of the water needed to dissolve `1 g` of `BaSO_(4) (K_(SP)=1.1xx10^(-10))` at `25^(@) C` is:A. 820 litreB. 410 litreC. 205 litreD. none of these
Answer» Correct Answer - B Solubility of `BaSO_(4)=sqrt(K_(SP))` `=sqrt(1.1xx10^(-10))=1.05xx10^(-5)M` `:. wt.of BaSO_(4)=1.05xx10^(-5)xx233` or `w_(BaSO_(4))=244.37xx10^(-5) g//litre` `:.` Volume of water needed to dissolve 1 g `BaSO_(4)=(1)/(2.44.37xx10^(-5))=410 litre`

Discussion

211.	Human blood has a narrow Ph range of `7.3-7.4`, which must be maintained for methabolic processes to function properly. To keep the Ph in this range requires a delicate balance between the concentration of the conjugate acid-base pairs making upto the buffer system. The main buffer is a carbonic acid/ hydrogencarbonate system, which involves the following three equilibria. `CO_(2)(g)hArrCO_(2)(aq)` `CO_(2)(aq)+H_(2)O(l)hArrH_(2)CO_(3)(aq)` `H_(2)CO(aq)+H_(2)O(l)hArrHCO_(3)^(-)(aq)+H_(3)O^(+)(aq)` Carbonic acid `(H_(2)CO_(3))` is a weak acid and `HCO_(3)^(-)`(aq) is its conjugate base. At the temperature of the human body, the `pK_(a)` for carbonic acid is 6.4 However, the normal concentration of `CO_(2)(g)` in the lungs maintanis a ratio of `HCO_(3)^(-)(aq)//H_(2)CO_(3)(aq)` in blood plasma of about `8:1`. The carbonic acid concentration in the bloos is largely controlled by breathing and respiration. Hydrogencarbonate ion concentration is largely controlled by excreation in urine. If blood pH rises above `7.4`, a potentially life-threatening conditon called alkalosis can result. This can happen in patients who are hyperventilating from severse anxiety, or in climbers suffereing from oxygen deficency at high altitude. (Given: log 2=0.3) Calculate pH of blood at the temperature of the human body.A. `7.4`B. `7.3`C. `7.35`D. `6.7`
Answer» Correct Answer - b

Discussion

212.	The equilibrium equation and `K_(a)` values for the three acids are given at `25^(@)C`: `HA(aq)+H_(2)OhArrH_(3)O^(+)(aq)+A^(-)(aq),K_(a)=2xx10^(-5)` `HB(aq)+H_(2)OhArrH_(3)O^(+)(aq)+B^(-)(aq),K_(a)=4xx10^(-6)` `HC(aq)+H_(2)OhArrH_(3)O^(+)(aq)+C^(-)(aq),K_(a)=1xx10^(-4)` The pH of 0.1M aqueous NaC solution is : (NaC is sodium salt of acid HC)A. `2.5`B. `11.5`C. `9.5`D. `8.5`
Answer» Correct Answer - d

Discussion

213.	The following reaction is known to occur in the body `CO_(2) + H_(2)O hArr H_(2)CO_(3) hArr H^(+) + HCO_(3)^(-)`. If `CO_(2)` escapes from the systemA. pH will decreaseB. Hydrogen ion concentration will decreaseC. `H_(2)CO_(3)` concentration will be unalteredD. The forward reaction will be promoted
Answer» Correct Answer - B The equilibrium will shift in the backward direction.

Discussion

214.	The following reaction takes place in the body `CO_(2)+H_(2)OhArrH_(2)CO_(3)hArrH^(+)+HCO_(3)^(-)`. If `CO_(2)` escapes from the systemA. pH will decreaseB. `[H_(2)CO_(3)]`gets changedC. `[H_(3)O^(+)]` concentration will decreaseD. The forward reaction is favoured.
Answer» Correct Answer - C Backward reaction is favoured.

Discussion

215.	Choose the correct statement:A. pH of acidic buffer solution decreases if more salt is added.B. pH of acidic buffer solution increases if more salt is added.C. pH of acidic buffer decreases if more salt is added.D. pH of acidic buffer increases if more salt is added.
Answer» Correct Answer - b,c

Discussion

216.	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. `6.2 xx 10^(-5)g`B. `5.0 xx 10^(-8) g`C. `1.2 xx 10^(-10)g`D. `1.2 xx 10^(-9)g`
Answer» Correct Answer - D `\|Ag^(+)\|=0.05 M ,K_(sp)(AgBr)=[Ag^(+)][Br^(-)]` `:. Br^(-) =(K_(sp))/(\|Ag^(+)\|)=(5.0 xx 10^(-13))/(0.05)=10^(-11)M` `i.e., ` amount of KBr to be added to `=10^(-11)` mole `=10^(-11)xx120=1.2 xx 10^(-9) ` g

Discussion

217.	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. `1.2xx10^(-10)` gB. `1.2xx10^(-9)` gC. `6.2xx10^(-5)` gD. `5.0xx10^(-8)` g
Answer» Correct Answer - b

Discussion

218.	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. `1.2 xx 10^(-10) g`B. `1.2 xx 10^(-9) g`C. `6.2xx 10^(-5) g`D. `5.0 xx 10^(-8) g`
Answer» Correct Answer - B `k_(SP) = [Ag^(+)][Br^(-)]`

Discussion

219.	For the reaction `Ag(CN)_(2)^(ɵ)hArr Ag^(o+)+2CN^(ɵ)`, the `K_(c )` at `25^(@)C` is `4 xx10^(-19)` Calculate `[Ag^(o+)]` in solution which was originally `0.1 M` in `KCN` and `0.03 M` in `AgNO_(3)`.
Answer» Correct Answer - A

Discussion

220.	Which of the following is a set of amphiprotic species ?A. `H_(3)O^(+),HPO_(4)^(2-),HCO_(3)^(-)`B. `H_(2)O,HPO_(3)^(2-),H_(2)PO_(2)^(-)`C. `HSO_(4)^(-),H_(2)PO_(4)^(-),H_(2)PO_(3)^(-)`D. Both (B) and (C )
Answer» Correct Answer - C Amphiprotic species can undergo hydrolysis to form `OH^(-)` and can ionise to form `H_(3)O^(+)` ions. In set (A ) both `HPO_(3)^(2-)` and `HPO_(2)` cannot be hydrolysed to form `OH^(-)` ions. In set (B) both `HPO_(3)^(2-)` and `H_(2)PO_(2)^(-)` can not be ionised to form `H_(3)O^(+)` ions. As `H_(3)PO_(2)` is monoprotic and `H_(3)PO_(3)` is diprotic. All the species in set (C ) are amphiprotic.

Discussion

221.	Which of the following is an amphiprotic ionA. Chloirde ionB. Acetate ionC. SulphateD. Bicarbonate ion
Answer» Correct Answer - D The ions which is proton donar as well as proton acceptor is an amphiprotic ion

Discussion

222.	Liquid `NH_(3)`, like water, is an amphiprotic solvent. Write the equaiton for the auto-ionisation of `NH_(3)`.
Answer» `2NH_(3) hArr NH_(4)^(oplus)+overset(Theta)(N)H_(2)`

Discussion

223.	A certain weak acid has a dissocation constant of `1.0 xx 10^(-4)`. The equilibrium constant for its reaction with a strong base isA. `1.0 xx 10^(-4)`B. `1.0 xx 10^(-10)`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(14)`
Answer» Correct Answer - C `HA: K_(a) = 10^(-4)` `HA +NaOH hArr NaA +H_(2)O` Clearly, the reverse reaction is the hydrolysis reaction. `rArr K_("Required") = (1)/(K_(h)) = (K_(a))/(K_(w)) = (10^(-10))/(10^(-14)) = 10^(10)`

Discussion

224.	Auto-ionisation of liquid `NH_(3)` is `2NH_(3) hArr NH_(4)^(o+) +NH_(2)^(Theta)` with `K_(NH_(3)) = [NH_(4)^(o+)] [NH_(2)^(Theta)] = 10^(-30) at -50^(@)C` Number fo amide ions `(NH_(2)^(Theta))`, present per `mm^(3)` of pure liquied `NH_(3)` isA. `602`B. `301`C. `200`D. `100`
Answer» Correct Answer - A `2NH_(3) hArr NH_(4)^(o+) +NH_(2)^(Theta)` `[NH_(4)^(o+)] = [NH_(2)^(Theta)]= sqrt(K_(NH_(3))) = 10^(-15)M` `1dm^(3)` of solution contains `10^(-15)` moles `= 6.02 xx 10^(23) xx 10^(15)` molecules of `NH_(2)^(Theta)`. `1mm^(3)` of solution contains `6.02 xx 10^(23) xx 10^(-15) xx 10^(-6) = 602 NH_(2)^(Theta)` ions.

Discussion

225.	`Ca_(3)(PO_(4))_(2)` is insoluble in in water. On adding a few drops of HCI to solid `Ca_(3)(PO_(4))_(2)` in contant with water, the solid dissolves. The reason is:A. The solvent becomes more polar on adding HClB. `Ca_(3)(PO_(4))_(2)` combines with HCl to form soluble `CaCl_(2)` and `H_(3)PO_(4)`C. `Ca(H_(2)PO_(4))_(2)` is formed, which dissolvesD. `H_(3)PO_(4)` a weak acid is formed and the solubility product of `Ca_(3)(PO_(4))_(2)` decrease
Answer» Correct Answer - b

Discussion

226.	`Ca_(3)(PO_(4))_(2)` is insoluble in in water. On adding a few drops of HCI to solid `Ca_(3)(PO_(4))_(2)` in contant with water, the solid dissolves. The reason is:A. the solvent becomes more polar on adding HCIB. `Ca_(3)(PO_(4))_(2)` combines with HCI to form soluble `CaCI_(2)`C. `Ca(H_(2)PO_(4))_(2)` is formed, which dissolvesD. `H_(3)PO_(4)`, a weak acid is formed and the solubility product of `Ca_(3)(PO_(4))_(2)` decrease
Answer» Correct Answer - B `Ca_(3)(PO_(4))_(2)+6HCIrarr underset("soluble")(3CaCI_(2))+2H_(3)PO_(4)`

Discussion

227.	To a `200 ml` of `0.1 M` weak aicd `HA` solution `90 ml` of `0.1 M` solution of `NaOH` be added. Now, what volume of `0.1 M NaOH` be added into above solution so that `pH` of resulting solution be `5`. `[K_(a)(HA)=10^(-5)]`A. 2 mLB. 20 mLC. 10 mLD. 15 mL
Answer» Correct Answer - c

Discussion

228.	What are the units in which the solubility product of `Ca_(3)(PO_(4))_(2)` is expressed?A. `mol dm^(-3)`B. `mol^(2) dm^(-6)`C. `mol^(3) dm^(-9)`D. `mol^(5) dm^(-15)`
Answer» Correct Answer - D `{:(Ca_(3)(PO_(4))_(2)hArr,3Ca^(2+)+,2PO_(4)^(3-)),(,3x,2x):}` `K_(sp) = (3x)^(2) (2x)^(2) = 108 x^(5)` `:.` Units of `K_(sp)` is `("concentration")^(5) = ("mol dm"^(-3))^(5) = "mol"^(5) "dm"^(-15) = "mol"^(5) "dm"^(-15)`

Discussion

229.	The solubility of `Ca_(3)(PO_(4))_(2)` in water is y moles `//` litre. Its solubility product isA. `6 y^(2)`B. `36y^(4)`C. `64y^(5)`D. `108 y^(5)`
Answer» Correct Answer - D `underset(y " moles")(Ca_(3)(PO_(4))_(2))hArr underset(3y" moles")(3Ca^(2+))+underset(2y " moles")(2PO_(4)^(3-))` `K_(sp)=\|Ca^(2+)\|^(3)\|PO_(4)^(3-)\|^(2)=(3y)^(3)(2y)^(2)=108 y^(5)`

Discussion

230.	What is the degree of dissociation of weak acid HA (C=0.1 M) in presence of strong acid HB (C=0.1 M) ? Given :`K_a`=(weak acid)=`10^(-6)`A. `10^(-5)`B. `10^(-6)`C. `10^(-4)`D. `10^(-3)`
Answer» Correct Answer - a

Discussion

231.	20 mL of 0.1 M weak acid `HA(K_(a)=10^(-5))` is mixed with solution of 10 mL of 0.3 M HCl and 10 mL. of 0.1 M NaOH. Find the value of `[A^(-)]`//([HA]+[A^(-)])` in the resulting solution :A. `2 xx 10^(-4)`B. `2 xx 10^(-5)`C. `2 xx 10^(-3)`D. `5 xx 10^(-2)`
Answer» Correct Answer - A `[H^(+)] = (10 xx 0.3 - 10 xx 0.1)/(10 + 10 + 20) = 0.05` `[HA] = (20 xx 0.1)/(20 + 20) = 0.05` `{:underset(0.05-x)(HAhArr)underset(0.05+x x)(H^(+)+A^(-)):}` Due to common ion effect neglect x w.r.t `0.05` `K_(a) = ((0.5 + x)x)/((0.05-x))=x, :. x = 10^(-5)` `:. ([A^(-)])/([HA]+[A^(-)]) = (x)/(x+0.05)` `(x)/(0.05) = (10^(-5))/(5 xx 10^(-2)) = 2 xx 10^(-4)`

Discussion

232.	Calculate solubility of `Ca_(3)(PO_(4))_(2) (K_(sp) = 10^(-15))` in presence of `0.1 M CaCI_(2)` solution. If your answer is `x xx 10^(-y)` where x is single digit then fill `y -x` in OMR sheet, e.g., if answer is `2 xx 10^(-3)` then answer is 1.
Answer» Correct Answer - 2

Discussion

233.	pH when equal volume of 0.1 M HA `(K_a=10^(-5))` and 1 M HB `(K_a=10^(-6))` are mixed ?A. 3+ log 2B. 3-`1/2` log 2C. 3+`1/2` log 2D. 3
Answer» Correct Answer - d

Discussion

234.	`10ml` of `0.1M` weak acid `HA(K_(a) = 10^(-5))` is mixed with 10 ml `0.2 M HCI` and `10ml 0.1 M NaOH`. Find the concentration of `A^(-)` in the resulting solution. Write your answer as `(-log [A^(-)])`
Answer» Correct Answer - 5

Discussion

235.	A solution is a mixture of 0.05 M NaCl and 0.05 M AgI. The concentration of iodide in the solution when AgCl just starts precipitating is equal to: `(K_(sp)AgCl=1xx10^(-10)M^(2), K_(sp)AgI=4xx10^(-16)M^(2))`A. `4xx10^(-6)M`B. `2xx10^(-8)M`C. `2xx10^(-7)M`D. `8xx10^(-15)M`
Answer» Correct Answer - C `[Ag^(+)]=(K_(SP)AgCI)/([CI^(-)])` For AgCI precipitation `= (10^(-10))/(0.05)=2xx10^(-9)` `:. [Ag^(+)]=(K_(SP)AgI)/([I^(-)])` For AgI precipitation `= (4xx10^(-16))/(0.05)=8xx10^(-15)` Thus, AgI will precipitate first. AgCI will precipitate only, when `[Ag^(+)]=2xx10^(-9)`, Thus `[I^(-)]_(l eft)=(4xx10^(-16))/(2xx10^(-9))=2xx10^(-7)M`

Discussion

236.	A buffer solution can be prepared from a mixture ofA. Sodium acetate and acetic acid in water.B. Sodium acetat and `HCI` in waterC. Ammonia and ammonia chloride in water.D. Ammonia and sodium hydroxide in water.
Answer» Correct Answer - A::C a. `CH_(3)COONa+CH_(3)COOH` (Acidic buffer, mixture of salt of `W_(A)//S_(B)` and `W_(A))` b. It is not buffer. [Mixture of salt of `W_(A)//S_(B)`and `S_(A)]` c. `NH_(3)+NH_(4)CI` (Basic buffer, mixture of `W_(B)` and salt of `W_(A)//S_(B))` d. `NH_(3) + NH_(4)OH` (not buffer, mixture of `W_(B))`

Discussion

237.	Calculate the pH when equal volume of `0.01 M HA (K_(a) = 10^(-6))` and `10^(-3)M HB (K_(a) = 10^(-5))` are mixed.
Answer» Correct Answer - 4

Discussion

238.	Calculate the pH of a solution which contains 100 mL of `0.1M HCI` and `9.9 mL` of `1.0M NaOH`.
Answer» Correct Answer - `3.0409;`

Discussion

239.	A mixture of weak acid is `0.1M` in `HCOOH (K_(a) = 1.8 xx 10^(-4))` and `0.1M` in `HOCN (K_(a) = 3.1 xx 10^(-4))`. Hence, `[H_(3)O^(o+)]` isA. `7.0 xx 10^(-3)M`B. `4.1 xx 10^(-4)M`C. `0.20M`D. `4.1 xx 10^(-3)M`
Answer» Correct Answer - A Use: `[H^(o+)]_("mix") = sqrt(Sigma K_(ai)C_(i))` `=sqrt(1.8 xx 10^(-4)xx0.1+3.1xx10^(-4)xx0.1)` `=sqrt(4.9xx10^(-5)) M = 7 xx 10^(-3)M`

Discussion

240.	The `[H^(+)]` in a solution containing `0.1 M HCOOH` and `0.1 M HOCN` [Ka for `HCOOH` and `HOCN` are `1.8 xx 10^(-4)` and `3.3 xx 10^(-4)`] respectively will beA. `7.13 xx 10^(-2) M`B. `7.13 xx 10^(-3) M`C. `7.13 xx 10^(-5) M`D. `7.13 xx 10^(-6) M`
Answer» Correct Answer - B `underset((0.1-x))(HCOOH)hArr underset((x+y))(H^(o+))+underset(x)(HCOOO^(o+))` `underset((0.1-y))(HOCN)hArr underset((x+y))(H^(o+))+underset(y)(OCN^(o+))` `K_(HCOOH)= ([H^(+)][HCOO^(-)])/([HCOOH]) = 1.8 xx 10^(-4)` `K_(HOCN) = ([H^(+)][OCN^(-)])/([HOCN]) = 3.3 xx 10^(-4)`

Discussion

241.	A buffer solution can be prepared from a mixture ofA. Sodium acetate and acetic acid in waterB. Sodium acetate and hydrochloric acid in waterC. Ammonia and ammonium chloride in waterD. Ammonium and sodium hydroxide in water
Answer» Correct Answer - A::C A buffer solution is prepared by mixing a weak acid/base with salt of its conjugate base/acid.

Discussion

242.	Calculate `[H^(o+)]` in a soluton that is `0.1M HCOOH` and `0.1 M HOCN. K_(a)(HCOOH) = 1.8 xx 10^(-4), K_(a) (HoCN) = 3.3 xx 10^(-4)`.
Answer» Correct Answer - `7.13xx10^(-3)M;`

Discussion

243.	What are `[H^(o+)], [A^(Theta)`, and `[B^(Theta)]` in a solution that is `0.3M HA` and `0.1M HB? K_(a)` for `HA` and `HB` are `1.38 xx 10^(-4)` and `1.05 xx 10^(-10)`, respectively.
Answer» Correct Answer - `[H^(+)]=[A^(-)]=2.04xx10^(-3)M, [B^(-)]=5.15xx10^(-9)M;`

Discussion

244.	Which of the following will not function as a buffer solutionA. NaCl and NaOHB. NaOH and `NH_(4)OH`C. `CH_(3)COONH_(4)` and HClD. Borax and boric acid
Answer» Correct Answer - A::B::C Because buffer solution are mixture of weak acid or weak base and their salt.

Discussion

245.	Calculate the pH in a solution that is `0.1M` in acetic acid and `0.1M` in benozic acid. `K_(a) for C H_(3)COOH` and `C_(6)H_(5)COOH are 1.8xx10^(-5)` and `6.5xx10^(-5)` respectively.
Answer» Correct Answer - `2.2396;`

Discussion

246.	Which statement is/are correct?A. All Bronsted bases are laso Lewis basesB. All Bronsted acids are not Lewis acidsC. All cations are acids and anions are basesD. `H_(2)O` can behave as Lewis acid as well aas Lewis base
Answer» Correct Answer - a,b,c

Discussion

247.	Aqueous solutions of `HNO_(3), KOH, CH_(3)COOH` and `CH_(3)COONa` of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is (are)A. `HNO_(3)` and `CH_(3)COOH`B. KOH and `CH_(3)COONa`C. `HNO_(3)` and `CH_(3)COONa`D. `CH_(3)COOH` and `CH_(3)COONa`
Answer» Correct Answer - C::D `HNO_(3) + CH_(3)COONa` mixture can act as buffer solutions if volume of `HNO_(3)` solution taken is lesser than volume of `CH_(3)COONa` solution because of following reaction `CH_(3)COONa+HNO_(3) rarr CH_(3)COOH + NaNO_(3)` (d) `CH_(3)COOH + CH_(3)COONa`-mixture will act as buffer.

Discussion

248.	Aqueous solutions of `HNO_(3), KOH, CH_(3)COOH`,and `CH_(3)COONa` of identical concentrations are provided. The pair (s) of solution which form a buffer upon mixing is// areA. `HNO_(3)` and `CH_(3)COOH`B. `KOH` and `CH_(3)COONa`C. `HNO_(3)` and `CH_(3)COONa`D. `CH_(3)COOH` and `CH_(3)COONa`
Answer» Correct Answer - C::D a. Not buffer since it is a mixture of `S_(A)` and `W_(A)`. b. Not buffer sinxce it is a mixture of `S_(B)` and salt of `W_(A)//S_(B)`. c. It is an acidic buffer. Since some of `HNO_(3)` will react with `CH_(3)COONa` to produce weak acid and some `CH_(3)COONa` to produce weal acid and some `CH_(3)COONa` will be left. So buffer is formed. `{:(,HNO_(3)+,CH_(3)COONararr,NaNO_(3)+,CH_(3)COOH),("Initial",x,y,0,0),("Final",-,(y-x),(y-x),(y-x)):}` d. It is an acidic buffer (mixture of `W_(A)` and salt of `W_(A)//S_(B))`

Discussion

249.	The base imidazole has a `K_(b)` of `8.1 xx 10^(-8)`. a. In what amounts should `0.02 M HCI` and `0.02M` imidazole be mixted to make `100mL` of a buffer at `pH 7`? b. If the resulting solution is diluted to `1L`, what is the `pH` of the diluted solution?
Answer» a. `pOH = pK_(b) +log ("salt/base")` `7 = 8 - 0.90848 + log ("salt/base")` and `("base/salt") = 1.234458`. Let `xmL` of `HCI` and `ymL` of imisazole are mixed (since molarity is same) `{:(,,HCI+,Baserarr,"Salt"),("Initial",,x,(y),0),("Final",,0,(y-x),x):}` `x +y = 100mL, y = 100 - x` `("Base")/("Salt") = (y-x)/(x) = (100-2x)/(x) = 1.2358` `x = 30.9 mL ("volume of acid")` Volume of imidanzole `= 100 - 30.9 = 69.1` b. `pH remains same (since on dilution the concentration of salt and base changes in the same proportions).

Discussion

250.	For the reversible reaction, net rate is `2NO(g)+O_(2)(g) hArr 2NO_(2)(g)` `((dx)/(dt))_("net")=2.6 xx 10^(3)[NO]^(2)[O_(2)]-4.1 [NO_(2)]^(2)` If a reaction mixture contains 0.01 mol each of NO and `O_(2)` and 0.1 mol of `NO_(2)` in 1L closed flask, then above reaction isA. shifted in forward reactionB. shifted in backward reactionC. in equilibriumD. given values are incomplete
Answer» Correct Answer - B `K_(c )=(k_(f))/(k_(b))=(2.6 xx 10^(3))/(4.1)=6.34 xx 10^(2)` `Q_(c)=([NO_(2)]^(2))/([NO]^(2)xx[O_(2)])` `=([0.1 xx 1]^(2))/([0.01]^(2)[0.01])=10^(4)` i.e., `Q_(c )gt K` Hence the reaction shifts in backward direction so that reaction quotient `(Q_(c ))` approaches `K_(c )`

Discussion

Explore topic-wise InterviewSolutions in .

The `pH` of a blood stream is maintained by a proper balance of `H_(2)CO_(3)` and `NaHCO_(3)` .What volume of `5MNaHCO_(3)` solution should be mixed with `10ml` of a sample of solution which is `2.5M` in `H_(2)CO_(3)` in order to maintain a `pH=7.4`(Take `pK_(a_(1))` for `H_(2)CO_(3)=6.7,log2=0.3)`

The volume of water needed to dissolve `1mg` of `PbSO_(4) (K_(sp) = 1.44 xx 10^(-8), Mw of PbSO_(4) = 303g)` at `25^(@)C` isA. `80mL`B. `43mL`C. `27.5mL`D. 10mL`

The Bronsted acid which gives the weakest conjugated base isA. HFB. `H_(2)S`C. `HCl`D. `H_(2)O`

pH of blood is maintained constant by mechanisms ofA. Common ion effectB. BufferC. SolubilityD. All of these

The `K_(b)` of weak base is `10^(-4)`. The `["salt"]` to `["base"]` ratio to be maintained to keep the `P^(H)` of buffer solution as `9` is .A. `1 : 4`B. `4 : 2`C. `1 : 10`D. `10 : 1`

The volume of the water needed to dissolve `1 g` of `BaSO_(4) (K_(SP)=1.1xx10^(-10))` at `25^(@) C` is:A. 280 litreB. 410 litreC. 205 litreD. 500 litre

The volume of the water needed to dissolve `1 g` of `BaSO_(4) (K_(SP)=1.1xx10^(-10))` at `25^(@) C` is:A. 820 litreB. 410 litreC. 205 litreD. none of these

The following reaction is known to occur in the body `CO_(2) + H_(2)O hArr H_(2)CO_(3) hArr H^(+) + HCO_(3)^(-)`. If `CO_(2)` escapes from the systemA. pH will decreaseB. Hydrogen ion concentration will decreaseC. `H_(2)CO_(3)` concentration will be unalteredD. The forward reaction will be promoted

The following reaction takes place in the body `CO_(2)+H_(2)OhArrH_(2)CO_(3)hArrH^(+)+HCO_(3)^(-)`. If `CO_(2)` escapes from the systemA. pH will decreaseB. `[H_(2)CO_(3)]`gets changedC. `[H_(3)O^(+)]` concentration will decreaseD. The forward reaction is favoured.

Choose the correct statement:A. pH of acidic buffer solution decreases if more salt is added.B. pH of acidic buffer solution increases if more salt is added.C. pH of acidic buffer decreases if more salt is added.D. pH of acidic buffer increases if more salt is added.

For the reaction `Ag(CN)_(2)^(ɵ)hArr Ag^(o+)+2CN^(ɵ)`, the `K_(c )` at `25^(@)C` is `4 xx10^(-19)` Calculate `[Ag^(o+)]` in solution which was originally `0.1 M` in `KCN` and `0.03 M` in `AgNO_(3)`.

Which of the following is a set of amphiprotic species ?A. `H_(3)O^(+),HPO_(4)^(2-),HCO_(3)^(-)`B. `H_(2)O,HPO_(3)^(2-),H_(2)PO_(2)^(-)`C. `HSO_(4)^(-),H_(2)PO_(4)^(-),H_(2)PO_(3)^(-)`D. Both (B) and (C )

Which of the following is an amphiprotic ionA. Chloirde ionB. Acetate ionC. SulphateD. Bicarbonate ion

Liquid `NH_(3)`, like water, is an amphiprotic solvent. Write the equaiton for the auto-ionisation of `NH_(3)`.

A certain weak acid has a dissocation constant of `1.0 xx 10^(-4)`. The equilibrium constant for its reaction with a strong base isA. `1.0 xx 10^(-4)`B. `1.0 xx 10^(-10)`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(14)`

Auto-ionisation of liquid `NH_(3)` is `2NH_(3) hArr NH_(4)^(o+) +NH_(2)^(Theta)` with `K_(NH_(3)) = [NH_(4)^(o+)] [NH_(2)^(Theta)] = 10^(-30) at -50^(@)C` Number fo amide ions `(NH_(2)^(Theta))`, present per `mm^(3)` of pure liquied `NH_(3)` isA. `602`B. `301`C. `200`D. `100`

To a `200 ml` of `0.1 M` weak aicd `HA` solution `90 ml` of `0.1 M` solution of `NaOH` be added. Now, what volume of `0.1 M NaOH` be added into above solution so that `pH` of resulting solution be `5`. `[K_(a)(HA)=10^(-5)]`A. 2 mLB. 20 mLC. 10 mLD. 15 mL

What are the units in which the solubility product of `Ca_(3)(PO_(4))_(2)` is expressed?A. `mol dm^(-3)`B. `mol^(2) dm^(-6)`C. `mol^(3) dm^(-9)`D. `mol^(5) dm^(-15)`

The solubility of `Ca_(3)(PO_(4))_(2)` in water is y moles `//` litre. Its solubility product isA. `6 y^(2)`B. `36y^(4)`C. `64y^(5)`D. `108 y^(5)`

What is the degree of dissociation of weak acid HA (C=0.1 M) in presence of strong acid HB (C=0.1 M) ? Given :`K_a`=(weak acid)=`10^(-6)`A. `10^(-5)`B. `10^(-6)`C. `10^(-4)`D. `10^(-3)`

20 mL of 0.1 M weak acid `HA(K_(a)=10^(-5))` is mixed with solution of 10 mL of 0.3 M HCl and 10 mL. of 0.1 M NaOH. Find the value of `[A^(-)]`//([HA]+[A^(-)])` in the resulting solution :A. `2 xx 10^(-4)`B. `2 xx 10^(-5)`C. `2 xx 10^(-3)`D. `5 xx 10^(-2)`

Calculate solubility of `Ca_(3)(PO_(4))_(2) (K_(sp) = 10^(-15))` in presence of `0.1 M CaCI_(2)` solution. If your answer is `x xx 10^(-y)` where x is single digit then fill `y -x` in OMR sheet, e.g., if answer is `2 xx 10^(-3)` then answer is 1.

pH when equal volume of 0.1 M HA `(K_a=10^(-5))` and 1 M HB `(K_a=10^(-6))` are mixed ?A. 3+ log 2B. 3-`1/2` log 2C. 3+`1/2` log 2D. 3

`10ml` of `0.1M` weak acid `HA(K_(a) = 10^(-5))` is mixed with 10 ml `0.2 M HCI` and `10ml 0.1 M NaOH`. Find the concentration of `A^(-)` in the resulting solution. Write your answer as `(-log [A^(-)])`

A solution is a mixture of 0.05 M NaCl and 0.05 M AgI. The concentration of iodide in the solution when AgCl just starts precipitating is equal to: `(K_(sp)AgCl=1xx10^(-10)M^(2), K_(sp)AgI=4xx10^(-16)M^(2))`A. `4xx10^(-6)M`B. `2xx10^(-8)M`C. `2xx10^(-7)M`D. `8xx10^(-15)M`

A buffer solution can be prepared from a mixture ofA. Sodium acetate and acetic acid in water.B. Sodium acetat and `HCI` in waterC. Ammonia and ammonia chloride in water.D. Ammonia and sodium hydroxide in water.

Calculate the pH when equal volume of `0.01 M HA (K_(a) = 10^(-6))` and `10^(-3)M HB (K_(a) = 10^(-5))` are mixed.

Calculate the pH of a solution which contains 100 mL of `0.1M HCI` and `9.9 mL` of `1.0M NaOH`.

A mixture of weak acid is `0.1M` in `HCOOH (K_(a) = 1.8 xx 10^(-4))` and `0.1M` in `HOCN (K_(a) = 3.1 xx 10^(-4))`. Hence, `[H_(3)O^(o+)]` isA. `7.0 xx 10^(-3)M`B. `4.1 xx 10^(-4)M`C. `0.20M`D. `4.1 xx 10^(-3)M`

The `[H^(+)]` in a solution containing `0.1 M HCOOH` and `0.1 M HOCN` [Ka for `HCOOH` and `HOCN` are `1.8 xx 10^(-4)` and `3.3 xx 10^(-4)`] respectively will beA. `7.13 xx 10^(-2) M`B. `7.13 xx 10^(-3) M`C. `7.13 xx 10^(-5) M`D. `7.13 xx 10^(-6) M`

A buffer solution can be prepared from a mixture ofA. Sodium acetate and acetic acid in waterB. Sodium acetate and hydrochloric acid in waterC. Ammonia and ammonium chloride in waterD. Ammonium and sodium hydroxide in water

Calculate `[H^(o+)]` in a soluton that is `0.1M HCOOH` and `0.1 M HOCN. K_(a)(HCOOH) = 1.8 xx 10^(-4), K_(a) (HoCN) = 3.3 xx 10^(-4)`.

What are `[H^(o+)], [A^(Theta)`, and `[B^(Theta)]` in a solution that is `0.3M HA` and `0.1M HB? K_(a)` for `HA` and `HB` are `1.38 xx 10^(-4)` and `1.05 xx 10^(-10)`, respectively.

Which of the following will not function as a buffer solutionA. NaCl and NaOHB. NaOH and `NH_(4)OH`C. `CH_(3)COONH_(4)` and HClD. Borax and boric acid

Calculate the pH in a solution that is `0.1M` in acetic acid and `0.1M` in benozic acid. `K_(a) for C H_(3)COOH` and `C_(6)H_(5)COOH are 1.8xx10^(-5)` and `6.5xx10^(-5)` respectively.

Which statement is/are correct?A. All Bronsted bases are laso Lewis basesB. All Bronsted acids are not Lewis acidsC. All cations are acids and anions are basesD. `H_(2)O` can behave as Lewis acid as well aas Lewis base

Aqueous solutions of `HNO_(3), KOH, CH_(3)COOH` and `CH_(3)COONa` of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is (are)A. `HNO_(3)` and `CH_(3)COOH`B. KOH and `CH_(3)COONa`C. `HNO_(3)` and `CH_(3)COONa`D. `CH_(3)COOH` and `CH_(3)COONa`

Aqueous solutions of `HNO_(3), KOH, CH_(3)COOH`,and `CH_(3)COONa` of identical concentrations are provided. The pair (s) of solution which form a buffer upon mixing is// areA. `HNO_(3)` and `CH_(3)COOH`B. `KOH` and `CH_(3)COONa`C. `HNO_(3)` and `CH_(3)COONa`D. `CH_(3)COOH` and `CH_(3)COONa`

The base imidazole has a `K_(b)` of `8.1 xx 10^(-8)`. a. In what amounts should `0.02 M HCI` and `0.02M` imidazole be mixted to make `100mL` of a buffer at `pH 7`? b. If the resulting solution is diluted to `1L`, what is the `pH` of the diluted solution?