InterviewSolution
Saved Bookmarks
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.
| 1751. |
What is the pH of `10^(-3)` M ammonia cyanide solution, if `K_(HCN)=7.2 xx 10^(-11)` and `K_(NH_(3))=1.8 xx 10^(-5)mol L^(-1)` ?A. 14B. 9.7C. `12.0`D. `7.5` |
|
Answer» Correct Answer - B `[H^(+)]` for the salt of weak acid and weak base is given by the expression `[H^(+)]=sqrt((K_(w)xx K_(a))/(K_(b)))` `=sqrt((10^(-14)xx7.2 xx 10^(-11))/(1. 8 xx 10^(-5)))=2 xx 10^(-10)` `pH =-log 2 xx 10^(-10)=10-log2` `=10-0.3010=9.699` |
|
| 1752. |
Which of the following is the best conductor of electricity ?A. `1M HNO_(3)`B. `1M CH_(3)COOH`C. `1M NH_(4)OH`D. `1M H_(2)SO_(4)` |
|
Answer» Correct Answer - D Stong electrolyte with more number of ions Is the best conductor |
|
| 1753. |
At infinite dilution, the percentage ionisation of both strong and weak electrolytes isA. `25%`B. `10%`C. `75%`D. `100%` |
|
Answer» Correct Answer - D At infinite dultion weak electrolytes under goes `100%` Ionisation |
|
| 1754. |
The pH of a 0.1 M aqueous solution of a weak acid (HA) is 3. What is its degree of dissociation ?A. `10%`B. `25%`C. `50%`D. None of these |
|
Answer» Correct Answer - D `HA hArr H^(+)+A^(-)` `[H^(+)]=calpha=10^(-3)` As` c=0.1 M implies 0.1 alpha=10^(-3)` or `alpha=10^(=-2)` or `% alpha =1%` |
|
| 1755. |
HA is a weak acid. The pH of 0.1 M HA solution is 2. What is the degree of dissociation `(alpha)` of HAA. 0.5B. 0.2C. 0.1D. 0.301 |
|
Answer» Correct Answer - C `pH = 2, :. [H^(+)] = 10^(-2) = 0.01`. For weak acid, `[H^(+)] = alpha. C` or, `0.01 = alpha xx 0.1 rArr alpha = 0.1`. |
|
| 1756. |
The dissociation constant of a weak monobasic acid in `0.01 M` solution is `10^(-8)`. What is its `[OH^(-)]` concentration?A. `10^(-6)`B. `10^(-8)`C. `10^(-9)`D. `10^(-10)` |
|
Answer» Correct Answer - C `[H^(+)] = sqrt(KC)` `= sqrt(10^(-8) xx 10^(-2)) = 10^(-5)` `[OH^(-)] = (K_(w))/([H^(+)])` `= (10^(-14))/(10^(-5)) = 10^(-9) M` |
|
| 1757. |
For a weak acid (`alpha` is very small)A. `K_(a) = Calpha^(2)`B. `alpha = sqrt((K_(a))/(C))`C. `[H^(+)] = Calpha`D. All the above |
|
Answer» Correct Answer - D `k_(a) = calpha^(2), alpha = sqrt((k_(a))/(c)), [H^(+)] = calpha` |
|
| 1758. |
`0.01` mole of `AgNO_(3)` is added to 1 litre of a solution which is `0.1M` in `Na_(2)CrO_(4)` and `0.005M` in `NaIO_(3)`. Calculate the mole of precipitate formed at equilibrium and the concentrations of `Ag^(+), IO_(3)^(-)` and `CrO_(4)^(2-)`. `(K_(sP)` values of `Ag_(2)CrO_(4)` and `AgIO_(3)` are `10^(-8)` and `10^(-13)` erspectively) |
|
Answer» The `K_(SP)` values of `Ag_(2)CrO_(4)` and `AgIO_(3)` reveals that `CrO_(4)^(2-)` and `IO_(3)^(-)` will be precipitated on addition of `AgNO_(3)` as: `[Ag^(+)][IO_(3)^(-)]=10^(-13)` `[Ag^(+)]_("needed")=(10^(-13))/([0.005])=2xx10^(-11)` `[Ag^(+)]^(2)[CrO_(4)^(2-)]=10^(-8)` `[Ag^(+)]_("needed")=sqrt((10^(-8))/(0.1))=3.16xx10^(-4)` Thus, `AgIO_(3)` will be precipitated first. Now, in order to precipitate `AgIO_(3)`, one can show: `{:(AgNO_(3)+,NaIO_(3)rarr,AgIO_(3)+,NaNO_(3)),(0.01,0.05,0,0),(0.005,0,0.005,0.005):}` The left mole of `AgNO_(3)` are now used to precipitate `Ag_(2)CrO_(4)`. `{:(2AgNO_(3)+,NaCrO_(4)rarr,Ag_(2)CrO_(4)+,2NaNO_(3)),(0.005,0.1,0,0),(0,0.0975,0.0025,0.005):}` Thus, `[CrO_(4)^(2-)]` left solution `=0.0975` Now, solution has `{:(AgIO_(3(s))+,Ag_(2)CrO_(4(s))+,CrO_(4)^(2-)ions),(0.005,0.0025,0.0975):}` `:. [Ag^(+)]"left" = (K_(SP)ofAg_(2)CrO_(4))/([CrO_(4)^(2-)])` `=sqrt((10^(-8))/(0.00975))=3.2xx10^(-4)M` `:. [IO_(3)^(-)["left"= (K_(SP)of AgIO_(3))/([Ag^(+)])` `=(10^(-13))/(3.2xx10^(-4))=3.1xx10^(-10)M` |
|
| 1759. |
A weak acid is 0.1% ionised in 0.1 M solution. Its pH isA. 2B. 3C. 4D. 1 |
|
Answer» Correct Answer - C For a monobasic acid `[H^(+)] = C alpha = (1)/(10) xx 0.001 = 10^(-4) rArr pH =4`. |
|
| 1760. |
The ionisation constant of an acid base indicator (a weak acid) is `1.0 xx 10^(-6)`. The ionised form of the indicator is red and unionised form is blue. The p`H` change required to alter the colour of indicator form `80%` red isA. `0.80`B. `1.20`C. `1.40`D. `2.00` |
|
Answer» Correct Answer - B `HIn hArr H^(o+) + Ind^(Theta)`, `pH = pK_(Ind) + "log" ([Ind^(Theta)])/([HIn])` `pH_(1) = pK_(Ind) + "log" (20)/(80) = pK_(Ind) - 2 log 2` `pH_(2) = pK_(Ind) +"log" (80)/(20) = pK_(Ind) +2 log2` `Delta(pH) = pH_(2) - pH_(1) = 4 log 2 = 4 xx 0.3 = 1.2`. |
|
| 1761. |
Calculate the solubility of `AgCN` in a buffer solution of `pH = 3`, Given `K_(sp) of AgCN = 1.2 xx 10^(-16)` and `K_(a)` for `HCN = 4.8 xx 10^(-10)`. |
|
Answer» Solubility of salt of weak acid `(AgCN)` in a buffer is given by the formula: `S` in buffer `=[K_(sp)(1+(H^(o+))/(K_(a)))]^((1)/(2))` `=[1.2 xx 10^(-16) (1+(10^(-3))/(4.8xx10^(-10)))]^((1)/(2))` `= [1.2 xx 10^(-16) (1+2.08 xx 10^(6))]^((1)/(2))` `=[1.2 xx 10^(-16) (~~2.08 xx 10^(6))]^((1)/(2))` `= sqrt(2.496) xx 10^(-5) = 1.5 xx 10^(-5)M` |
|
| 1762. |
A weak acid is `0.1%` ionised in `0.1M` solution. Its ionisation constant is :A. `10^(-5)`B. `10^(-6)`C. `10^(-8)`D. `10^(-9)` |
|
Answer» Correct Answer - D `K_(a) = Calpha^(2) = 0.1 xx ((0.1)/(100))^(2)` `= 1 xx 10^(-9)` |
|
| 1763. |
A weak monobasic acid is half neutralized by a strong base. If the Ph of the solution is 5.4 its `pK_(a)` isA. 6.8B. 2.7C. 5.4D. 10.8 |
|
Answer» Correct Answer - C At the half neutralisation point. `pH=pK_(a)=5.4` |
|
| 1764. |
calculate `[H^(o+)]` and `[overset(Θ)OH]` in a `0.1M` solution of weak monoacitic base which is `2.0%` ionised. What is the `pH` of solution. |
|
Answer» `{:(,B +,H_(2)O,hArr,BH^(oplus) ,+overset(Theta)(O)H),("Initial",0.1,-,0,,0),("concentration",,,,,),("Equilibrium",0.1(1-alpha),-,C alpha,,C alpha),("concentration",,,,,):}` Given `alpha = (2.0)/(100) = 2xx10^(-2)` `[overset(Θ)OH] = C alpha = 0.1 xx2xx10^(-2) = 2xx10^(-3)M` `pOH =- log (2xx10^(-3)) =- 0.3 +3 = 2.7` `pH = 14 - 2.7 = 11.3`. |
|
| 1765. |
Calculate `[H^(o+)]` and `[overset(Θ)OH]` in `10^(-3)M` solution of monobasic acid which is `4.0%` ionised. What is the `pH,K_(a)` and `pK_(b)` of the acid. |
|
Answer» `{:(,CH_(3)COOH+,H_(2)O,hArr,CH_(3)COO^(Theta) +,H_(3)O^(oplus)),("initial",10^(-3)M,-,,-,-),("concentration",,,,,),("Equilibrium",10^(-3)(1-alpha),-,,C alpha,C alpha),("concentration",~~ 10^(-3),,,,):}` Given `alpha = (4.0)/(100) = 4 xx 10^(-2)` a. `[H_(3)O^(o+)] = Calpha = 10^(-3) xx 4xx 10^(-2) = 4 xx 10^(-5)` `pH =- log (4 xx 10^(-5)) =- 2log 2+5` `=- 2xx 0.3 +5 = 4.4` `:. pH = 4.4` b. `pOH = 14 - 4.4 = 9.6` `- log [overset(Θ)OH] = 9.6, log [overset(Θ)OH]` `=- 9.6` `=- 9-0.6 +1-1= bar(10).4` `:. [overset(Θ)OH] = 2.5 xx 10^(-10)M` c. `K_(a) = Calpha^(2) = 10^(-3) xx (4xx10^(-2))^(2) = 1.6 xx 10^(-6)` `pK_(a) =- log (1.6 xx 10^(-6)) =- log (16 xx 10^(-7))` `=- 4log 2+7` `=- 1.2 +7 = 5.8` `:.pK_(a) = 5.8` |
|
| 1766. |
The pH of decinormal solution of KOH isA. 1B. 4C. 10D. 13 |
|
Answer» Correct Answer - D pH of `10^(-1)` N KOH `=1.0` |
|
| 1767. |
How many gram equivalents of NaOH are required to neutralize `25 cm^(3)` of decinormal HCl solution ?A. 0.00125B. 0.0025C. 0.005D. 0.025 |
|
Answer» Correct Answer - B 1000`cm^(3)` of 0.1 N Cl neutralize 0.1 g. eq. of NaOH `25cm^(3)` of 0.1 N HCl neutralized of NaOH `=(0.1 )/(1000)xx25=0.0025`g eq. |
|
| 1768. |
When the following is added to `20 ml` of `0.1 M CH_(3)COOH + 20 ml 0.1 M` of `CH_(3)COONa`. The `pH` of the solution does not changeA. `10mL` of `0.1 M CH_(3)COOH`B. `10 mL` of `0.1 M CH_(3)COONa`C. `10 mL` of waterD. `20 mL` of `0.2 M CH_(3)COOH` |
|
Answer» Correct Answer - C `pH = pka + "log" ([S])/([A])` pH of buffe depends upon the ka, `["Salt"]` and `["Acid"]` |
|
| 1769. |
For which of the following shall the ratio of `K_(p)//K_(c)` is unity at 300K and 1 atm. PressureA. `N_(2)(g)+3H_(2)(g ) hArr 2NH_(3)(g)`B. `N_(2)(g)+O_(2)(g) hArr 2NO(g)`C. `NH_(4)(g) hArr HCl(g)+NH_(3)(g)`D. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` |
|
Answer» Correct Answer - B We know that `K_(p)=K_(c )(RT)^(Deltan)` For reaction where `Deltan=0` `K_(p)=K_(c )(RT)^(0)` or `(K_(p))/(K_(c ))=1` |
|
| 1770. |
In a reversible process, `Delta S_(sys) + Delta S_(surr)` isA. `gt0`B. `lt0`C. `=0`D. `ge0` |
|
Answer» Correct Answer - C Factual question. |
|
| 1771. |
For the reaction `S_(2)O_(3)^(2-)(aq) +2H_(3)O^(+)(aq) hArr S(s)+H_(2)SO_(3)(aq)+H_(2)O(l)` Rate `=k[H_(3)O^(+)][S_(2)O_(3)^(2-)]` Reaction is fastest inA. 0.1 M `CH_(3)COOH`B. 0.1 M `H_(2)SO_(4)`C. 0.1 M HClD. 0.1 M NaOH |
|
Answer» Correct Answer - B The reaction will be fastest when `[H_(3)O^(+)]` is maximum `[H^(+)]` from 0.1 M `CH_(3)COOH lt0.1 M` `[H^(+)]` from 0.1 M `H_(2)SO_(4) =0.2 M` `[H^(+)]` from 0.1 M HCl`= 0.1 M` `[H^(+)] ` from 0.1 M NaOH` = 10^(-13)M` |
|
| 1772. |
In the reaction `2H_(2)O hArr H_(3)O^(+) + OH^(-)`, water isA. A weak baseB. A weak acidC. Both a weak acid and a weak baseD. Neither an acid nor a base |
|
Answer» Correct Answer - C Because their conjugate base and conjugate acids are strong. |
|
| 1773. |
The conjugate acid of `H_(2)PO_(4)^(-)` isA. `H_(3)PO_(4)`B. `H_(2)PO_(4)^(-)`C. `PO_(4)^(3-)`D. `H_(3)O^(+)` |
|
Answer» Correct Answer - A `underset("Conjugate acid")(H_(3)PO_(4)) hArr H^(+) + H_(2)PO_(4)^(-)`. |
|
| 1774. |
Which of the following is not used as a Lewis acidA. `SnCl_(4)`B. `FeCl_(3)`C. KClD. `BF_(3)` |
|
Answer» Correct Answer - C KCl is a ionic compound. |
|
| 1775. |
An aqueous solution of ammonia consists ofA. `H^(+)`B. `OH^(-)`C. `NH_(4)^(+)`D. `NH_(4)^(+)` and `OH^(-)` |
|
Answer» Correct Answer - D `H_(2)O + NH_(3) rarr NH_(4)^(+) + OH^(-)`. |
|
| 1776. |
`M(OH)_(x)` has `K_(SP)=4xx10^(-12)` and its solubility in water is `10^(-4)` M. Calculate the value of x. |
|
Answer» `M(OH)_(x)hArr underset(S)(M^(x+))+underset(XS)(XOH^(-))` `K_(SP)=S.(XS)^(x)` , `("Give" S= 10^(-4))` or `4xx10^(-12)=10^(-14). (X.10^(-4))^(x)` `10^(-4)xx10^(-4x).X^(x)=4xx10^(-12)` `X^(x).(10^(-4))^(1+x)=4xx10^(-12)` `:. X=2` |
|
| 1777. |
Solubility of `BaF_(2)` in a solution of `Ba(NO_(3))_(2)`, will be represented by the concentration term:A. `[Ba^(2+)]`B. `[F^(-)]`C. `(1)/(2)[F^(-)]`D. `2[NO_(3)^(-)]` |
|
Answer» Correct Answer - c |
|
| 1778. |
On adding `0.04g` solid `NaOH` to a `100mL,(M)/(200)Ba(OH)_(2)` solution, determine change in `pH`:A. `0`B. `+0.3`C. `-0.3`D. `+0.7` |
|
Answer» initial `rArr[OH^(-)]=(2)/(200)=10^(-2)M:.pH=12` final `rArr [OH^(-)]=10^(-2)+(0.04)/(40xx0.1)=2xx10^(-2):.pH=12.3` So change =`12.3-12=+0.3` |
|
| 1779. |
Which of the following species is more soluble in water ?A. `M(OH)_(3)` , `(K_(SP)=1xx10^(-35))`B. `M(OH)_(2)` , `(K_(SP)=1xx10^(-30))`C. MOH , `(K_(SP)=1xx10^(-28))`D. MOH , `(K_(SP)= 1xx10^(-26))` |
|
Answer» Correct Answer - A S of `M(OH)_(3)=4sqrt((K_(SP))/(27))=4sqrt((10^(-35))/(27))` `= 7.8xx10^(-10)` S of `M(OH)_(2) = 3sqrt((K_(SP))/(4))` `S of MOH= sqrt(K_(CP))` |
|
| 1780. |
The number of millimoles present per litre of solution of `CH_3COONa` so that 150 ppm (150 moles of its gets hydrolysed out of one million moles added) of if `K_a=1.8xx10^(-5)` is :A. 25B. 35C. 45D. 55 |
|
Answer» Correct Answer - a |
|
| 1781. |
pH of `Ca(OH)_(2)` solution is 12. Millimoles of `Ca(OH)_(2)` present in 100mL of solution will beA. 1B. 0.5C. 0.05D. 5 |
|
Answer» Correct Answer - B `pH =12` `pOH =14-Ph =14-12=2` `:. [OH^(-)]=10^(-2) M` `:. [Ca(OH)_(2)]=0.5 xx 10^(-2) M` ( Assuming complete dissociation at this conc. ) Mili moles of `Ca(OH)_(2)` in solution `=`MV (in mL ) `=0.5 xx 10^(-2) xx 100 =0.5 ` |
|
| 1782. |
Which of the following species is more soluble in water ?A. `M(OH)_(3)` (`k_(sp)=1xx10^(-35)`)B. `M(OH)_(2)` `(K_(sp)=1xx10^(-30)`)C. MOH (`K_(sp)=1xx10^(-28)`)D. MOH (`K_(sp)=1xx10^(-26)`) |
|
Answer» Correct Answer - a |
|
| 1783. |
In qualitative analysis, the metals of Group I can be separated from other ion by precipitating them as chloride salts. A solution initially contains `Ag^(+)` and `Pb^(2+)` at a concentration of 0.10 M. Aqueous HCl is added to this solution until the `Cl^(-)` concentration is 0.10 M. What will the concentrations of `Ag^(+)` and `Pb^(2+)` be at equilibrium (`K_(sp)` for `AgCl = 1.8 xx 10^(-10), K_(sp)` fr `PbCl_(2) = 1.7 xx 10^(-5)`)A. `[Ag^(+)] = 1.8 xx 10^(-9) M, [Pb^(2+)] = 1.7 xx 10^(-3)M`B. `[Ag^(+)] = 1.8 xx 10^(-11) M, [Pb^(2+)] = 1.7 xx 10^(-4)M`C. `[Ag^(+)] = 1.8 xx 10^(-7) M, [Pb^(2+)] = 1.7 xx 10^(-6)M`D. `[Ag^(+)] = 1.8 xx 10^(-11) M, [Pb^(2+)] = 1.7 xx 10^(-5)M` |
| Answer» Correct Answer - A | |
| 1784. |
`H_(2)S` behaves as a weak diprotic acid in aqueous solution. Which of the following is the correct explanantion for `pH` of a solution of `H_(2)S` in terms of its `pK_(1), pK_(2), [H_(2)S]` and `[S^(2-)]`A. `pH = (1)/(2) (pK_(1) + pK_(2))`B. `pH = (1)/(2) (pK_(1) + pK_(2) - "log"([S^(2-)])/([H_(2)S]))`C. `pH = (1)/(2)(pK_(1) + pK_(2) + "log"([S^(2-)])/([H_(2)S]))`D. `pH = (1)/(2) (pK_(1) - pK_(2) + "log"([H_(2-)S])/([S^(2-)]))` |
|
Answer» Correct Answer - C Refer to illustration `H_(2)S hArr H^(o+) + hS^(Theta) (K_(1))` `HS^(Theta) hArr H^(o+) + S^(2-) (K_(2))` `K_(1) = ([H^(o+)][HS^(Theta)])/([H_(2)S])` `K_(2) = ([H^(o+)][S^(Theta)])/([HS^(Theta)])` `K_(1) xx K_(2) = ([H^(o+)]^(2)[S^(2-)])/([H_(2)S])` `[H^(o+)]^(2) = (K_(1)xxK_(2)xx[H_(2)S])/([S^(2-)])` Taking negative logarithms to both sides. `-2log [H^(o+)] =- log K_(1) - log K_(2)` `=- log [H_(2)S] + log [S^(2-)]` `:. pH = (1)/(2) (pK_(1) + pK_(2) + "log"([S^(2-)])/([H_(2)S]))` |
|
| 1785. |
Aniline behaves as a weak base. When 0.1 M, 50 mL solution of aniline was mixed with 0.1 M , 25mL solution of HCl the pH of resulting solution was 8. The the pH of 0.01 M solution of anilinium chloride will be `(K_w =10^(-14))`A. 6B. 6.5C. 5D. 5.5 |
|
Answer» Correct Answer - c |
|
| 1786. |
Aniline `(C_(6)H_(5)NH_(2))` is a organic base in aqueous solution. Suggest a solvent in which aniline would become a weak base. |
| Answer» A solvent is needed which has appreciably stornger acid proporeties than `H_(2)O` , one such solvent is liquid acetic acid `(CH_(3)COOH)`. | |
| 1787. |
`NH_(4)OH` is weak base but it becomes still weaker in the aqueous solutions ofA. 0.1 M HClB. 0.1 M `NH_(4)Cl`C. 0.1 M `H_(2)SO_(4)`D. 0.1 M `CH_(3)COOH`. |
|
Answer» Correct Answer - B Degree of ionisation of `NH_(4)OH` is supressed by `NH_(4)Cl` due to common ion effect. |
|
| 1788. |
Soluility product of `Ba(OH)_(2)` and `Al(OH)_(3)` are `1.8 xx 10^(10)` and `2.4 xx 10^(-20)` respectively. If both `Al^(3+)` and `Ba^(+2)` ions are present in a solution, which one will be ppt on addition of Ammonium hydroxide solutionA. `Ba(OH)_(2)`B. `Al(OH)_(3)`C. Both are precipitated at same timeD. Both are not precipitated |
|
Answer» Correct Answer - B Compound having low ksp precipitated first `CaF_(2) rarr Ca^(+2) + 2F^(-)` |
|
| 1789. |
The solution which consumes `[H^(+)]` or `[OH^(-)]` or both simultaneously from externally added base in order to give negligible change in `pH`, is known as buffer solution. In general, the solution resists the change in pH. Buffer solution does not mean that there does not occur a pH change in pH. Buffer solution des not mean that there does not occur a pH change at all. It implies the pH change occurs but in neglibible amount. There are two types of buffer (i) Acidic buffer: it is a mixture of weak acid and its salt acid strong base. (ii) Basic buffer : It is a mixture of weak base and its salt with strong acid. To prepare a buffer of `pH 8.26`, amount of `(NH_(4))_(2)OH` solution `[pK_(a)(NH_(4^(+)) = 9.26]`A. `0.05`molB. `0.025` molC. `0.01` molD. `0.005` mol |
|
Answer» Correct Answer - B `5.74 = 4.74 + "log" (x)/(5 xx 10^(-3))`, calculating `x = 0.02`, hence `[(NH_(4))_(2)SO_(4)]` required `= 0.01` |
|
| 1790. |
The expression for the solubility product of `Ag_(2)CO_(3)` will be `:`A. `K_(sp)=s^(2)`B. `K_(sp)=4s^(3)`C. `K_(sp)=27s^(4)`D. `K_(sp) = s` |
|
Answer» Correct Answer - B `Ag_(2)CO_(3) hArr underset(2s)(2Ag)+underset(s)(CO_(3)^(2-))` `K_(sp)=(2s)^(2)xx=4s^(3)` |
|
| 1791. |
If equal volumes of `BaCI_(2)` and `NaF` solutions are mixed, which of these combination will not give a precipitate? `(K_(sp) of BaF_(2) =1.7 xx 10^(-7))`.A. `10^(-3)BaCI_(2)` and `2xx10^(-2)M NaF`B. `10^(-3)M BaCI` and `1.5 xx 10^(-2)M NaF`C. `1.5 xx 10^(-2)M BaCI_(2)` and `10^(-2)M NaF`D. `2xx10^(-2)M BaCI_(2)` and `2xx10^(-2)M NaF` |
|
Answer» Correct Answer - C When equal volume of `BaCI_(2)` and `NaF` solutions are mixed. (volume becomes double and concentration is halved). a. `[Ba^(2+)] = (10^(-3))/(2), [F^(Theta)] = (2xx10^(-2))/(2) = 10^(-2)` `Q_(sp) of BaF_(2) = [Ba^(2+)] [F^(Theta)]^(2)` `=(0.5 xx 10^(-3)) (10^(-2))^(2) = 5 xx 10^(-6)` `Q_(sp) gt K_(sp) (6 xx 10^(-6) gt 1.7 xx 10^(-7))` will be precipitated. b. `[Ba^(2+)] = (10^(-3))/(2), [F^(Theta)] = (1.5 xx 10^(-2))/(2)` `Q_(sp) of BaF_(2) = [Ba^(2+)] [F^(Theta)]^(2) = (0.5 xx 10^(-2)) (0.75 xx 10^(-2))^(2) = 0.28 xx 10^(-6)` `Q_(sp) gt K_(sp)`. Hence precipitation occurs. c. `[Ba^(2+)] = (1.5 xx 10^(-3))/(2), [F^(Theta)] = 10^(-2))/(2)`. `Q_(sp) pf BaF_(2) = [Ba^(2+)] [F^(Theta)]^(2) = ((1.5 xx 10^(-3))/(2)) ((10^(-2))/(2))^(2)` `= 0.187 xx 10^(-7)` `Q_(sp) lt K_(sp) (0.187 xx 10^(-7) lt 1.7 xx 10^(-7))` So solution in (c) will not precipitate out. d. `[Ba^(2+)] = (2xx10^(-2))/(2) = 10^(-2)M`, `[F^(Theta)] = (2xx10^(-2))/(2) = 10^(-2)M` `Q_(sp) BaF_(2) = (10^(-2)) (10^(-2))^(2) = 10^(-6)` `:. Q_(sp) gt K_(sp)`. Hence precipitation occurs. |
|
| 1792. |
A solution containing `NH_(4)CI` and `NH_(4)OH` has `[overset(Theta)OH] = 10^(-6) molL^(-1)`, which of the following hydroxides would be precipitated when this solution in added in equal volume to a solution containing `0.1M` of metal ions?A. `Mg(OH)_(2),(K_(sp)=3xx10^(-11))`B. `Fe(OH)_(2)(K_(sp)=8xx10^(-16))`C. `Cd(OH)_(2)(K_(sp)=8xx10^(-6))`D. `AgOH(K_(sp)=5xx10^(-3))` |
|
Answer» Correct Answer - B When equal volumes of `(NH_(4)CI + NH_(4)OH)` and metal ions are mixed (volume becomes double and concentration is halved) `:. [overset(Theta)OH] = (10^(-6))/(2), [M^(+n)] = (0.1)/(2)` `Q_(sp) (or IP)` of metal hydroxides of `M(OH)_(2)`type `=[M^(+n)] [overset(Theta)OH]^(2)` `= ((0.1)/(2)) ((10^(-6))/(2))^(2) = (1)/(8) xx 10^(-13)` `= 0.125 xx 10^(-13) = 12.5 xx 10^(-11)` `:. Q_(sp) gt K_(sp) of Mg(OH)_(2) (12.5 xx 10^(-11) gt3 xx 10^(-11))` and `Q_(sp) gt K_(sp)` of `Fe(OH)_(2) (12.5 xx 10^(-11) gt 8 xx 10^(-16))` So both can be precipitated. Since the `k_(sp)` of `Fe(OH)_(2)` is less than `Ksp` of `Mg(OH)_(2)`, so `Fe(OH)_(2)` will be precipiated first. Similarly, `Q_(sp)` (or IP) of metal hydroxides of `M(OH)` type `=[M^(+1)] [overset(Theta)OH] = ((0.1)/(2)) ((10^(-6))/(2))` `= 0.25 xx 10^(-7)` `:.Q_(sp) lt K_(sp) of AgOH (0.25 xx 10^(-7) lt 5 xx10^(-3))` IT can not be precipitated out. Hence `Fe(OH)_(2)` will be precipitated. |
|
| 1793. |
Given: `Ag(NH_(3))_(2)^(+)hArrAg^(+)2NH_(3), K_(C)=6.2xx10^(-8)` and `K_(SP)` of `AgCI=1.8xx10^(-10)` at 298 K. Calculate the concentration of the complex in `1.0M` aqueous ammonia. |
|
Answer» Correct Answer - C `[Ag(NH_(3))_(2)]^(o+) (aQ) hArr Ag^(o+) (aq) +2NH_(3)(aq)` `K_(c) = ([Ag^(o+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)]^(o+))= ([Ag^(o+)][CI^(Theta)][NH_(3)]^(2))/([Ag(NH_(3))_(2)][CI^(Theta)])` `K_(sp) of AgCI = [Ag^(o+)] [CI^(Theta)] = 1.8 xx 10^(-10)` `6.2 xx 10^(-8) = (K_(sp) "of" AgCI[NH_(3)]^(2))/([Ag(NH_(3))_(2)]^(o+)[CI^(Theta)])` `= (1.8 xx 10^(-10) xx [NH_(3)]^(2))/([Ag(NH_(3))_(2)]^(o+)[CI^(Theta)])` `{:(AgCI,+2NH_(3)hArr,[Ag(NH_(3))_(2)]^(o+)+,CI^(Theta)),(,1,0,0),(,(1-a),a,a):}` `6.2 xx 10^(-8) = (1.8 xx 10^(-10)xx[NH_(3))]^(2)/([Ag(NH_(3))_(2)]^(o+)[CI^(Theta)])` `=(1.8 xx 10^(-10)(1-a)^(2))/(axxa)` Since `a` is very small, it can be neglected. Therefore, `6.2 xx 10^(-8) = (1.8 xx 10^(-10))/(a^(2))` or `a = sqrt((1.8xx10^(-10))/(6.2xx10^(-8))) ~~ 0.0538` ` =[Ag(NH_(3))]^(o+)` |
|
| 1794. |
The acidic strength of two monobasic acids may be compared by comparing the hydronium ion concentrations of their aqueous solution of same molar concentration. The dissociation constants of acids HA and HB are `4 xx 10^(-6)` and `2.5 xx 10^(-9)`, respectively. How many times HA is stronger than HB ? Assume that for both the acids, the degree of dissociation at the concentration take, is negligible. |
|
Answer» Correct Answer - `0040` |
|
| 1795. |
An aqueous solution of ammonium acetate isA. Faintly acidicB. Faintly basicC. Fairly acidicD. Almost neutral |
|
Answer» Correct Answer - D It is a salt of weak acid and weak base. |
|
| 1796. |
`K_a` for `CH_3COOH` is `1.8xx10^(-5)` and `K_b` for `NH_4OH` is `1.8xx10^(-5)` The pH of ammonium acetate will be :A. 7.005B. 4.75C. `7.0`D. Between 6 and 7 |
|
Answer» Correct Answer - c |
|
| 1797. |
When 0.10 N solutions of ammonium acetate, barium acetate, and sodium acetate are ranked from basic to most basic, what is the correct order ?A. `NH_4C_2H_3O_2 lt NaC_2H_3O_2 lt Ba(C_2H_3O_2)_2`B. `Ba(C_2H_3O_2)_2 lt NH_4C_2H_3O_2 lt NaC_2H_3O_2`C. `NaC_2H_3O_2 lt Ba(C_2H_3O_2)_2 lt NH_4C_2H_3O_2`D. `NaC_2H_3O_2 lt NH_4C_2H_3O_2 lt Ba(C_2H_3O_2)_2` |
|
Answer» Correct Answer - a |
|
| 1798. |
Ammonium acetate which is 0.01 M, is hydrolysed to 0.001 M concentration .Calculate the change in Ph in 0.001 M solution, if initially `Ph =pK_(a)`.A. 5B. 10C. 100D. 1 |
|
Answer» Correct Answer - D `CH_(3)COO^(-)+NH_(4)^(+)+H_(2)O hArr CH_(3)COOH+NH_(4)OH` `{:("Initial",0.01M,0.01M,(0.01-0.001)M,(0.01-0.001)M),("At eqm.",0.001M,0.001M,=0.009M,=0.009M):}` `K_(h)=([CH_(3)COOH][NH_(4)OH])/([CH_(3)COO^(-)][NH_(4)^(+)])=((0.009)^(2))/((0.001)^(2))=10^(2)` `K_(h)=(K_(w))/(K_(a) xx K_(b))` `:. K_(b)=(K_(w))/(K_(a)xxK_(h))=(10^(-14))/(K_(a)xx 10^(2))=(10^(-16))/(K_(a))` `[H^(+)]=sqrt((K_(a)xxK_(w))/(K_(b)))=sqrt((K_(a)xx10^(-14))/(10^(-16)//K_(a)))=10K_(a)` `(pH)_("Initial")=pK_(a)` `(pH)_("Final")=-log[H^(+)]=-log(10K_(a))=pK_(a)-1` Change in pH `=(pH)_("Final")-(pH)_("Inital")` `=pK_(a)-1-pK_(a)= -1` |
|
| 1799. |
Aqueous solution of `NaCl` is neutral becauseA. `Na^(+)` undergoes hydrolysisB. `Cl^(-)` undergoes hydrolysisC. Both `Na^(+)` and `Cl^(-)` undergo hydrolysisD. Does not undergo hydrolysis |
|
Answer» Correct Answer - D `NaCl` does not under go Hydrolysis |
|
| 1800. |
The hydrolysis constant of ammonium acetate is given byA. `(K_(w))/(K_(a))`B. `(K_(w))/(K_(b))`C. `(K_(w))/(K_(a).K_(b))`D. `(K_(a),K_(b))/(K_(w))` |
|
Answer» Correct Answer - C `CH_(3)COONH_(4)` is a salt of `W.A` & `W.B` `kh = (kw)/(ka xx kb)` |
|