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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1201. |
A compound whose aqueous solution will have the highest pHA. NaClB. `Na_(2)CO_(3)`C. `NH_(4)Cl`D. `NaHCO_(3)` |
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Answer» Correct Answer - B `Na_(2)CO_(3)` when react with water form strong base and weak acid. So its aqueous solution is basic. |
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| 1202. |
The values of Ksp of `CaCO_(3)` and `CaC_(2)O_(4)` are `4.7 xx 10^(-9)` and `1.3 xx 10^(-9)` respectively at `25^(@)C`. If the mixture of these two is washed with water, what is the concentration of `Ca^(2+)` ions in waterA. `5.831 xx 10^(-5)M`B. `6.856 xx 10^(-5)M`C. `3.606 xx 10^(-5) M`D. `7.746 xx 10^(-5) M` |
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Answer» Correct Answer - D `CaCO_(3) rarr underset(x)(Ca^(2+)) + underset(x)(CO_(3))^(2-)` `CaC_(2)O_(4) rarr underset(y)(Ca^(2+)) + underset(y)(C_(2)O_(4))^(2-)` Now, `[Ca^(2+)] = x + y` and `x(x+y) = 4.7 xx 10^(-9)` `y(x+y) = 1.3 xx 10^(-9)` On solving we get `[Ca^(2+)] = 7.746 xx 10^(-5)M`. |
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| 1203. |
The `K_(a)` value of `CaCO_(3)` and `CaC_(2)O_(4)` in water are `4.7 xx 10^(-9)` and `1.3 xx 10^(-9)`, respectively, at `25^(@)C`. If a miaxture of two is washed with `H_(2)O`, what is `Ca^(2+)` ion concentration in water?A. `7.746 xx 10^(-5)`B. `5.831 xx 10^(-5)`C. `6.856 xx 10^(-5)`D. `3.606 xx 10^(-5)` |
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Answer» Correct Answer - A Since `K_(sp)` of `CaCO_(3)` and `CaC_(2)O_(4)` are very close, so concentration of any species cannot be neglected. Let the solubilities of `CaCO_(3)` and `caC_(2)CO_(4)` are `x` and `yM`. `{:(CaCO_(3)hArr,Ca^(2+)+,CO_(3)^(2-)),(,x,x):}` `{:(CaC_(2)O_(4)hArr,Ca^(2+)+,CO_(3)^(2-)),(,y,y):}` Total `[Ca^(2+)] = x +y` `K_(sp) "of" CaCO_(3) = [Ca^(2+)] [CO_(3)^(2-)] = (x +y)x` `K_(sp) "of" CaC_(2)O_(4) = [Ca^(2+)] [C_(2)O_(4)^(2-)] = (x+y)y` `:. x(x+y) = 4.7 xx 10^(-9) ...(i)` `y (x+y) = 1.3 xx 10^(-9) ...(ii)` `(x)/(y) =(4.7 xx 10^(-9))/(1.3 xx 10^(-9))` `:. x = 3.615y` Substituting the value of `x` in equation (i) or (ii) `3.615 y(3.15y+y) = 4.7 xx 10^(-9)` `3.615 xx 4.615y^(2) = 4.7 xx 10^(-9)` `:. y = ((4.7 xx 10^(-9))/(16.68))^(1//2) = 1.67 xx 10^(-5)M` `x = 6.03 xx 10^(-5)M` `[Ca^(2+)] = x +y = (6.03 xx 10^(-5) + 1.67 xx 10^(-5))` `= 7.707 xx 10^(-5)M` |
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| 1204. |
For pure water:A. pH increases and pOH decreases with rise in temperatureB. pH decreases and pOH increases with rise in temperatureC. Both pH and pOH increases with rise in temperatureD. Both pH and pOH decreases with rise in temperature |
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Answer» Correct Answer - d |
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| 1205. |
`pH` of water is `7.0` at `25^(@)C`. If water is heated to `70^(@)C`, the:A. pH will decrease and solution becomes acidicB. pH will increaseC. pH will remain constant as 7D. pH will decrease but solution will be neutral |
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Answer» Correct Answer - d |
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| 1206. |
Solubility product of a salt AB is `1 xx 10^(-8)` in a solution in which concentration of A is `10^(-3) M`. The salt will precipitate when the concentration of B becomes more thanA. `10^(-4) M`B. `10^(-7) M`C. `10^(-6) M`D. `10^(-5) M` |
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Answer» Correct Answer - D `[B] = (K_(sp)AB)/([A]) = (1 xx 10^(-8))/(10^(-3)) = 1 xx 10^(-5) M` Where ionic product `gt K_(sp)`, ppt formed `:.` B should be more then `10^(-5)M`. |
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| 1207. |
The solubility product of `BaSO_(4)` is `1.3 xx 10^(-9)`. The solubility of this salt in pure water will beA. `1.69 xx 10^(-9) "mol litre"^(-1)`B. `1.69 xx 10^(-18) "mol litre"^(-1)`C. `3.6 xx 10^(-18) "mol litre"^(-1)`D. `3.6 xx 10^(-5) "mol litre"^(-1)` |
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Answer» Correct Answer - D `{:(BaSO_(4)hArr,Ba^(++),+,SO_(4)^(--)),(,(S),,(S)):}` `K_(sp) = S^(2), S = sqrt(K_(sp)) = sqrt(1.3 xx 10^(-9))` `= 3.6 xx 10^(-5)` mol/litre |
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| 1208. |
Which of the following correctly explains the nature of boric acid in aqueous medium :A. `H_3BO_3 overset(H_2O)to H_3O^+ + H_2BO_3^-`B. `H_3BO_3 overset(2H_2O)to 2H_3O^+ + HBO_3^(2-)`C. `H_3BO_3 overset(3H_2O)to 3 H_3O^+ + BO_3^(3-)`D. `H_3BO_3 overset(H_2O)to B (OH)_4^(-)+H+` |
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Answer» Correct Answer - D |
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| 1209. |
The addition of `FeCl_(3)` to waterA. decreases the value of `K_(w)`B. increases the value of `K_(w)`C. has no effect on the value of `K_(w)`D. gives rise to a basic solution. |
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Answer» Correct Answer - C Ionic product of water will not change. |
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| 1210. |
Find the pH at which an acid indicator having concentration `1.0xx10^(-4)M`, having dissociation of `1%` shows a colour change. |
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Answer» Correct Answer - 8 `H InhArrH^(+)+In^(-)` `[H^(+)]=c.alpha=1.0xx10^(-4)xx0.01=1.0xx10^(-6)` `K_(a)=c.alpha.alpha=1.0xx10^(-6)xx0.01=1.0xx10^(-8)` Now, the indicator will show colour when `[In^(-)]=[H In]` `:. K_(a)=[H^(+)])` or `[H^(+)]=K_(a)=1.0xx10^(-8)` or `pH=8` |
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| 1211. |
The `pH` of a solution `7.00`. To this solution, sufficient base is added to increase the `pH` to `12.0`. The increase in `overset(Theta)OH` ion concentration isA. `5 times`B. `100 times`C. `10^(5)times`D. `4 times` |
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Answer» Correct Answer - C Initial `pH` or `pOH = 7, :. [overset(Theta)OH] = 10^(-7)M`. Final `pH = 12, :. pOH = 2, [overset(Theta)OH] = 10^(-2)M`. Increase in `[overset(Theta)OH] = (10^(-2))/(10^(-7)) = 10^(5)` times |
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| 1212. |
Basic lead carbonate is:A. PbSB. `PbCO_(3)`C. `PbSO_(4)`D. `2PbCO_(3).Pb(OH)_(2)` |
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Answer» Correct Answer - D It is basic lead carbonate. |
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| 1213. |
The indicator used in the titration of sodium carbonate with sulphuric acid isA. PhenolphthaleinB. Methyl organeC. Potassium ferrocynideD. Potassium ferricynide |
| Answer» Correct Answer - B | |
| 1214. |
What is the solubility of magnesion carbonate, `MgCO_(3)`, in water at `25^(@)`C? A. 0.22g.`L^(-1)`B. `2.6xx10^(-3)g.L^(-1)`C. `3.1xx10^(-5)g.L^(-1)`D. `8.1xx10^(-8)g.L^(-1)` |
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Answer» Correct Answer - a |
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| 1215. |
An example of a Lewis acid isA. Aluminium chlorideB. AmmoniaC. PyridingD. Amines |
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Answer» Correct Answer - A The central atoms (Al) in `AlCl_(3)` has incomplete octet. Hence, it acts as Lewis acid. |
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| 1216. |
0.1 mol HCl is dissolved in distilled water of volume V, then, at `underset(Vtooo)(lim) (pH)_("solution")` is equal to :A. zeroB. 1C. 7D. 14 |
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Answer» Correct Answer - c |
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| 1217. |
100 cc of HCl of pH value 1 is mixed with 100 cc of distilled water. The pH value of the resultant solution isA. 1.7B. 1.9C. 2.5D. 1.3 |
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Answer» Correct Answer - D `pH=1` means `[H^(+)]=10^(-1)M` On adding equal amount of water, `[H^(+)]=(10^(-1))/(2)` `pH=-log 5 xx 10^(-2) =2-0.7 =1.3` |
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| 1218. |
Classify each of the folowing as a strong acid, string base, weak acid, and weak base: i. `NaOH` ii. `HF` iii. `NH_(4)^(o+)` iv. `NH_(3)` v. `F^(Θ)` vi. `HI` |
| Answer» i. `S_(B)` ii. `W_(A)` ii. `W_(A)` iv. `W_(B)` v. `W_(B)` vi. `S_(A)` | |
| 1219. |
Which of the following solutions will have `pH = 10` at `298 K`?A. `1xx10^(-10)M HCl ` solutionB. `1xx10^(-4) M` NaOH solutionC. `1xx10^(-10) M ` NaOH solutionD. Both (A) and (B) |
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Answer» Correct Answer - B pH of `10^(-10) M` HCl will be `lt7` pH of `10^(-10)` NaOH will be slightly greater than 7. pH of `10^(-4)` NaOH `=- log [H^(+)]` `=- log 1 xx 10^(-10) =10`. |
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| 1220. |
pH of solution produced when aqueous solution of pH of `6` is mixed with an equalA. `4.3`B. `3.3`C. `4.0`D. `4.5` |
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Answer» Correct Answer - B `(10^(-6) xx 10^(-3))/(2) = (10^(-3))/(2)[10^(-3) + 1]` `[H^(+)] = 0.5 xx 10^(-3) = 5 xx 10^(-4)` `pH = -log 5 xx 10^(-4) = 4-0.6990 = 3.3010` |
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| 1221. |
An acid solution of `pH =6` is diluted `100` times. The `pH` of solution becomesA. `6.95`B. `6`C. `4`D. `8` |
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Answer» Correct Answer - A `pH = 6 :. [H^(o+)] = 10^(-6)M` Volume is made `100` times. `:. [H^(o+)] = (10^(-6))/(100) = 10^(-8)M` `pH = 6.95` |
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| 1222. |
Let the colour of the indicator Hin (coloueless) will be visible only when its ionised form (pink) is `25%` or more in a solution. Suppose Hin `(pK_(a)=9.0)` is added to a solution of pH=9.6 predict what will happen? (Take log `2=0.3`)A. Pink colour will be visibleB. Pink colour will not be visibleC. Percentage of ionised will be less than `25%`D. Precentage of ionised from will be more than `25%` |
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Answer» Correct Answer - a,d |
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| 1223. |
`Hin` is an acidic indicator `(K_(Ind) =10^(-7))` which dissociates into aqueous acidic solution of `30mL` of `0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13))` Calculate the `[(Ind^(Theta))/(Hin)]`A. `1.413 xx 10^(-5)`B. `1.413 xx 10^(-4)`C. `3.128xx10^(-5)`D. `3.128xx10^(-14)` |
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Answer» Correct Answer - A i. `{:(,H_(3)PO_(4)overset(K_(1)=10^(-3))hArr,H^(o+)+,H_(2)PO_(4)^(Theta),),("Initial",C,0,0,),("At equilibrium",C(1-alpha),Calpha,Calpha,):}` `1 xx 10^(-3) = (Calpha^(2))/(1-alpha) =(0.05xxalpha^(2))/(1-alpha)` (Neglecting `alpha` wrt 1) Using direct formula for `pH` of a weak acid `(pK_(a_(1)) = 3, C = 0.05)` `pH = (1)/(2) (pK_(a_(1)) - log C) = (1)/(2) (3-log 0.05)` `= (1)/(2) (3-0.7+2) = 2.15` ii. For an acidic indicator. `HIn overset(K_(Ind))hArr H^(o+) + Ind^(Theta)`. `:. pH = pK_(Ind) + "log" ([Ind^(Theta)])/([HIn])` `2.15 = 7 + "log" ([Ind^(Theta)])/([HIn]) (K_(Ind) = 10^(-7))` `"log"([Ind^(Theta)])/([HIn]) = 2.15 - 7 = - 4.85` `([Ind^(Theta)])/([HIn]) = "Antilog" (-4.85)` `= "Antilog" (-4-0.85 +1-1)` `= "Antilog" (bar(5).15) = 1.413 xx 10^(-5)` |
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| 1224. |
`Hin` is an acidic indicator `(K_(Ind) =10^(-7))` which dissociates into aqueous acidic solution of `30mL` of `0.05M H_(3)PO_(4) (K_(1) = 10^(-3), K_(2) = 10^(-7), K_(3) = 10^(-13))` If `Hin` and `Ind^(Theta)` posses colour `P` and `Q`, respectively, and concentration of `HIn` is `120` times than that of `Ind^(Theta)`. colour `Q` predominates over `P` when concnetration of `Ind^(Theta)` is `127` times of `HIn`. What is the `pH` range of the indicator.A. `4.896 to 9.0792`B. `4.896 to 8.0792`C. `4.896 to 7.0792`D. `4.896 to 6.0792` |
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Answer» Correct Answer - A When colour `P` redominates over `Q`, then, `pH = pK_(Ind) + log 120` `pH = 7+ log 120 = 9.0792` When colour `Q` predominates over `P`, then `pH = pK_(Ind) +"log"(1)/(127)` `pH = 7 - log 127 = 4.896` `pH` range of indicator is `4.896` to `9.0792` |
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| 1225. |
The indicator constant for an acidic indicator, `HIn` is `5xx10^(-6)M`. This indicator appears only in the colour of acidic form when `([In])/([HIn])le(1)/(20)` and it appears only in the colour of basic form when `([HIn])/([In])le(1)/(40)` . The pH range of indicator is `[` Given `: log 5=0.7]`A. `4.3-6.3`B. `4.0-6.6`C. `4.0-6.9`D. `3.7-6.6` |
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Answer» Correct Answer - 3 `pH=P_(ki n)+log_(10)[(ln^(-))/(Hln)]` `pH=[6-log_(10)5]+log_(10)[(1)/(20)]=4` `PH=P_(k i n)+log_(10)[(ln^(-))/(Hln)]` `=P_(ki n)+log_(10)[(40)/(1)]` `=6-log_(10)5+log_(10)40` `=6-0.7+1.6=6.9` |
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| 1226. |
A universal indicatorA. can be used in all acid-base titrationsB. is a mixture of several indicatorsC. is useful in the titration of a weak acid against weak baseD. has limited pH range like any other indicator. |
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Answer» Correct Answer - B A universal indicator is a mixture of several indicators. |
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| 1227. |
A solution of `HCI` has p`H = 5`. If `1mL` of it is diluted to `1L` what will be the `pH` of resulting solution? |
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Answer» [Refer illustration `8.7(b)(i)]` `[HCI] = 10^(-5)M` `N_(1)V_(1) = N_(2)V_(2)` `10^(-5) xx 1mL = N_(2) xx 100mL` `N_(2) = (10^(-5))/(1001mL)~~10^(-8)M` `pH = 6.98` |
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| 1228. |
If `K_(1)` and `K_(2)` are respective equilibrium constants for two reactions `:` `XeF_(6)(g) +H_(2)O hArr XeOF_(4)(g) +2HF_(g)` `XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g)` Then equilibrium constant for the reaction `XeO_(4)(g)+2HF(g) hArr XeO_(3)F_(2)(g)+H_(2)O(g)` will beA. `K_(1)K_(2)`B. `K_(1)//(K_(2))^(2)`C. `K_(1)(K_(2))^(-1)`D. `K_(2)//K_(1)` |
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Answer» Correct Answer - D `XeF_(6)(g)+H_(2)O(g) hArr XeOF_(4)(g)+2HF(g)` `K_(1)=([XeOF_(4)][HF]^(2))/([XeF_(6)][H_(2)O])` `XeO_(4)(g)+XeF_(6)(g) hArr XeOF_(4)(g)` `K_(2)=([XeOF_(4)][XeO_(3)])/([XeO_(4)][XeF_(6)])` `XeO_(4)(g)+2HF(g) hArr XeO_(3)(g)+H_(2)O(g)` Let its eqn. constant `=K` `:. K=([XeO_(3)F_(2)][H_(2)O])/([XeO_(4)][HF]^(2))=(K_(2))/(K_(1))` |
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| 1229. |
For which salt is the molar solubility, s, is approximately equal to `4xx10^(-6)` M?A. `AgC_(2)H_(3)O_(2) K_(sp)=2xx10^(-3)`B. `TIBr K_(sp)=4xx10^(-6)`C. `MnO_(3) K_(sp)=2xx10^(-11)`D. `Zn(OH)_(2) K_(sp)=3xx10^(-17)` |
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Answer» Correct Answer - c |
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| 1230. |
Which of the following when added to 35 mL of the 1.0 M solution of KOH would cause largest change in pH ?A. 25 mL of 1.0 M HClB. 25 mL of 0.2 M HClC. 25 mL of 0.5 M HClD. 25 mL distilled water. |
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Answer» Correct Answer - A The stronger acid will effect more change in the pH value. |
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| 1231. |
If the equilibrium concentration of the components in a reaction `A+B hArr C+D` are 3,5,10 and 15 mol `L^(-1)` respectively then what is `Delta G^(@)` for the reactiona at 300K ?A. `-1.381kcal`B. `-600cal`C. `-1140cal`D. `-300cal` |
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Answer» Correct Answer - A `K=(10 xx 15)/(3 xx 5)=10` Since `Delta G ^(@)=-2.303 RT log K` `=- 2.303 xx 2 xx 300 xx log 10 =1381.8 cal` |
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| 1232. |
`K_(p)//K_(c)` for the reaction `CO(g)+1/2 O_(2)(g) hArr CO_(2)(g)` isA. `1.0`B. `RT`C. `1//sqrt(RT)`D. `(RT)^(1//2)` |
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Answer» Correct Answer - C `K_(p)=K_(c )(RT)^(Deltan)`. For the given reaction `Deltan=-0.5 :. K_(p)//K_(c)=(1)/(sqrt(RT))` |
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| 1233. |
`CaF_(2)` has a `K_(sp)=3.9xx10^(-11)` at `25^(@)` C. What is the `[F^(-)]` in a saturated solution of `CaF_(2)` at `25^(@)` C?A. `2.1xx10^(-4)`B. `3.4xx10^(-4)`C. `4.3xx10^(-4)`D. `6.8xx10^(-4)` |
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Answer» Correct Answer - c |
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| 1234. |
Acid strength is proportional toA. `H^(+)`B. `pH`C. PohD. `K_(a)` |
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Answer» Correct Answer - C,D Factual question. |
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| 1235. |
The pH of `10^(-2)` M `Ca(OH)_(2)` isA. 112B. 6.96C. 7.2D. 6 |
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Answer» Correct Answer - B `Ca(OH)_(2)hArr Ca^(2+)+2OH` `|OH^(-)|=2 xx 10^(-2)M:` `|H^(+)|=(K_(w))/(|OH|)=(10^(-14))/(2 xx 10^(-2))=5 xx 10^(-13)M` `pH =log |H^(+)|=-log [5 xx 10^(-3)|` `=13-log5=13-0.69%=12.3010` |
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| 1236. |
For the given reaction identify the true (T) & false (F) statements `C_(2)H_(5)NH_(2)+HIhArr (C_(2)H_(5)NH_(3))^(+)Gamma` `S_(1):HI` is bronsted base `S_(2)HI` is bronsted acid `S_(3):HI` is arrhenius acid `S_(4):HI` is lewis acid `S_(5):HI` is arrhenius base `S_(6):HI` is lewis base.A. `TFFFT T`B. `FT T TFF`C. `FT TFFF`D. `TFFFTF` |
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Answer» Correct Answer - C `HI` has donated `H^(+)` in the above reaction.So it acts as an Arrhenius acid as well as a Bronished acid. |
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| 1237. |
Four species are listed below (i) `HCO_(3)^(-)` (ii) `H_(3)O^(+)` (iii) `HSO_(4)^(-)` (iv) `HSO_(3)F` Which one of the following is the correct sequence of their acid strengthA. `ii lt iii lt i lt iv`B. `i lt iii lt ii lt iv`C. `iii lt i lt iv lt ii`D. `iv lt ii lt iii lt i` |
| Answer» Correct Answer - B | |
| 1238. |
If 0.1 M of a weak monobasic acid is taken and its percentage degree of ionisation is `1.34 %`, then calculate its ionisation constant.A. `0.8 xx 10^(-5)`B. `1.79 xx 10^(-5)`C. `0.182 xx 10^(-5)`D. None of these |
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Answer» Correct Answer - B `alpha =1.34 %=(1.34)/(100)=1.34 xx 10^(-2)` `C=0.1 mol L^(-1)` `K_(a)=C alpha^(2) =0.1 xx (1.34 xx 10^(-2))^(2)` `=1.7956 xx 10^(-5)` |
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| 1239. |
At `90^(@)C`, pure water has `[H_(3)O^(o+)] = 10^(-6)M`. What is the value of `K_(w)` at `90^(@)C`A. `10^(-6)`B. `10^(-8)`C. `10^(-12)`D. `10^(14)` |
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Answer» Correct Answer - C In pure waer, `[OH^(-)]=[H_(3)O^(+)]=10^(-6) mol L^(-1)` `K_(w)=[OH^(-)][H_(3)O^(+)]=10^(-6)xx 10^(-6) =10^(-12)` |
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| 1240. |
At `90^(@)C` , pure water has `[H_(3)O^(+)]` as `10^(-6)` mol `L^(-1)`. What is the value of `K_(w)` at `90^(@)C` ?A. `10^(-6)`B. `10^(-14)`C. `10^(-12)`D. `10^(-8)` |
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Answer» Correct Answer - C `[H_3)O^(+)]=[OH^(-)]=10^(-6)M` `:. K_(w)=[H_(3)O^(+)][OH^(-)]=10^(-6)xx 10^(-6)=10^(-12)` |
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| 1241. |
At `90^(@)C`, pure water has `[H_(3)O^(o+)] = 10^(-6.7)mol L^(-1)`. What is the value of `K_(w)`at `90^(@)C`?A. `10^(-6)`B. `10^(-12)`C. `10^(13.4)`D. `10^(-6.7)` |
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Answer» Correct Answer - C Silver chromate is `Ag_(2)CrO_(4)` `:. K_(sp)=[Ag^(+)]^(2)[CrO_(4)^(2-)]`. If solubility is S mol `L^(-1), K_(sp) = 4 s^(3)`. |
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| 1242. |
At `90^(@)C` pure water has `[H_(3)O^(+)] = 10^(-6) M`, the value of `K_(w)` at this temperature will beA. `10^(-6)`B. `10^(-12)`C. `10^(-14)`D. `10^(-8)` |
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Answer» Correct Answer - B For pure water `[H^(+)] = [OH^(-)], :. K_(w) = 10^(-12)`. |
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| 1243. |
Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will beA. `3.98 xx 10^(8)`B. `3.88 xx 10^(6)`C. `3.68 xx 10^(-6)`D. `3.98 xx 10^(-6)` |
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Answer» Correct Answer - D `ph = -log[H^(+)]` `5.4 = -log[H^(+)], [H^(+)] = 3.98 xx 10^(-6)`. |
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| 1244. |
What is the pH value of `(N)/(100)` KOH solutionA. `10^(-11)`B. 3C. 2D. 11 |
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Answer» Correct Answer - D `10^(-3) N KOH` will give `[OH^(-)] = 10^(-3) M` pOH = 3 `because pH + pOH = 14, pH = 14 - 3 = 11`. |
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| 1245. |
a. What amount of `H_(2)SO_(4)` must be dissolved in `500mL` of solution to have a `pH` of `2.15`? b. What amount of `KOH` must be dissolved in `200mL` of solution to have a `pH` of `12.3`? c. What amount of `ca(OH)_(2)` must be dissolved in `100mL` of solution to have a `pH` of `13.85`? |
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Answer» a. `pH = 2.15` `-log [H^(o+)] = 2.15` `log [H^(o+)] =- 2.15 =- 2 - 0.15 +1 - 1 = bar(3).85` `[H^(o+)] = Antilog (bar(3).85) = 7 xx 10^(-3) N` strength of `H_(2)SO_(4) (gL^(-1)) = N xx Ew` (Ew of `H_(2)So_(4) = (98)/(2) = 49`) `= 7 xx 10^(-3) xx 49 = 0.343 g L^(-1)` `= (0.343 xx 500 mL)/(1000mL) = 0.1715g //500 mL` Alternatively: Strength of `H_(2)SO_(4) (g L^(-1))` `=M xx Mw, (M = (7 xx 10^(-3)N)/(2))` `= (7 xx 10^(-3))/(2) xx 98 = 7 xx 10^(-3) xx 49 gL^(-1)` `= (7 xx 10^(-3))/(2) g//500mL = 0.1715g//500mL` b. Since `pH gt gt`( basic solution), first calculate `pOH` and `[overset(Θ)OH]`. Then calculate amount of `KOH//200mL` `pH = 12.3, pOH = 14 - 12.3 = 1.7` `- log [overset(Θ)OH] = 1.7` `log[overset(Θ)OH] =- 1.7 =- 1 - 0.7 +1 - 1 = bar(2).3` `[overset(Θ)OH] = Antilog (bar(2).3) = 2 xx 10^(-2) N`. Strength of `KOH (gL^(-1))` `= N xx Eq` or M `xx` Mw = (Mw = Ew of `KOH = 56 g`) `= 2 xx 10^(-2) xx 56 gL^(-1)` `= (2 xx 10^(-2) xx 56 xx 200 mL)/(1000mL) g//200 mL` `= 0.244g//200 mL` c. [Mw of `Ca(OH)_(2) = 74g, Ew` of `Ca(OH)_(2) = (74)/(2) = 37g]` `pH = 13.85, pOH = 14 - 13.85 = 0.15` `log [overset(Θ)OH] = - 0.15 +1 = bar(1).85` `[overset(Θ)OH] = "Antilog" (bar(1).85) = 7 xx 10^(-1)N` or `(7 xx 10^(-1))/(2)M` Strength `(gL^(-1)) = (7 xx 10^(-1)N xx 37) gL^(-1)` or `((7 xx 10^(-1))/(2) xx 74)gL^(-1)` `= (7 xx 10^(-1) xx 37 xx 100mL)/(1000mL)g//100 mL` `=2.59g//100mL` |
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| 1246. |
The strongest conjugate base isA. `NO_(3)^(-)`B. `Cl^(-)`C. `SO_(4)^(2-)`D. `CH_(3)COO^(-)` |
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Answer» Correct Answer - D Factual question. |
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| 1247. |
pH of water isA. pressure dependentB. pressure independentC. temperature dependentD. temperature independent |
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Answer» Correct Answer - C Dissociation constant of water, `K_(w)` is temperature dependetn. As such, `H^(+)` as well as pH are tempeature dependent. |
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| 1248. |
Ammonia gas dissolves in water to give `NH_(4)OH`. In this reaction water acts asA. An acidB. A baseC. A saltD. A conjugate base |
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Answer» Correct Answer - A `H_(2)O` acts as acid as it provides `H^(+)` to `NH_(3)`. |
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| 1249. |
`CuSO_(4)` solution isA. acidicB. basicC. neutralD. none |
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Answer» Correct Answer - A A solution of `CuSO_(4)` [ salt of a strong acid, `H_(2) SO_(4)` and a weak base, `Cu(OH)_(2)` in water will be acidic in nature. |
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| 1250. |
A white substance was alkaline in solution. Which of the following substances could it beA. `Fe_(2)O_(3)`B. `Na_(2)CO_(3)`C. `NH_(4)Cl`D. `NaNO_(3)` |
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Answer» Correct Answer - B Aqueous solution of `Na_(2)CO_(3)` is alkaline due to hydrolysis of `Na_(2)CO_(3)` in `Na^(+)` and `CO_(3)^(-)`. |
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