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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The pH of a solution containing 0.20M CH_(3)COOH and 0.30M CH_(3)COONa is |
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Answer» `2.89` `pH=pK_(a)+"log"(C_("salt"))/(C_("acid"))` In this CASE, we need to calculate `pK_(a)` and the acid FIRST: `pK_(a)= -logK_(a)` `= -log(1.8xx10^(-5))` `=4.74` Now, we substitute the value of `pK_(a)` and concentrations of the acid and its salt: `pH=pK_(a)+"log"(C_("salt"))/(C_("acid"))` `=4.74+"log"(0.30)/(0.20)` `=4.74+log(1.5)` `=4.92` |
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| 2. |
The pH of a soft drink is 2.42. What is the hydronium ion concentration in the drink? |
| Answer» SOLUTION :`3.8 XX 10^(-3) M` | |
| 3. |
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. |
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Answer» Solution :pH=-LOG `[H^+]` = -3.76 `THEREFORE` log `[H^+]`=-3.76 = `bar4.24` `therefore [H^+]`= Antilog `(bar4.24)` `=1.7378xx10^(-4) = 1.74xx10^(-4)` M |
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| 4. |
The pH of a sample of H_(2) SO_(4) is 1.3979 . The percentage of the solution is 73.5 % (w/w) , the density of the solution is |
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Answer» `2.66 XX 10^(-3) g//cc` |
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| 5. |
The pH of a sample KOH solution is 12.3979 . The weight of solid KOH of 70% pure required to prepare 2.5 lit of this solution is |
| Answer» Answer :B | |
| 6. |
The P^(H) of a dibasic acid is 3.699. Its molarity is |
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Answer» ` 2xx10^(-4)M` ` log [H^(+) ] = - pH=- 3.7 =-4 + 0.3 = overset(-) 4.3` ` [H^(+)] ="ANTI long " (overset( -)4.3) =2 xx10 ^(-4)N ` ` ["acid"] = 2xx 10 ^(-4) N RARR M = (N)/( n- F)= 1xx 10 ^(-4) M ` |
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| 7. |
The pH of a buffer solution is 4.745 . When 0.044 mole of Ba(OH)_(2) is added to 1 lit. of the buffer , the pH changes to 4.756 . Then the buffer capacityis |
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Answer» 4 |
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| 8. |
The pH of a base solution is 12.699 . The amount of crystalline oxalic acid required to react with 2 lit of this base solution is (assume oxalic acid dissociates completely) |
| Answer» Answer :B | |
| 9. |
The pH of a10^(-8)M solution of HCl is water is |
| Answer» Answer :D | |
| 10. |
The P^(H) of a I lit solution is 2. It is diluted with water till its p^(H) becomes 4, How many litres of water is added? |
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Answer» 99 ` V_t = 1 " lit" , V_2 =?rArr N_1V_1 = N_2V_2` ` 10 ^(-2)xx 1 =V_2xx 10 ^(-4)rArr V_2 = 100 "lit " ` VOL of WATER added `= V_2 -V_1 = 99 ` lit |
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| 11. |
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, K_a of this acid is |
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Answer» `1XX10^(-3)` MOLARITY = 0.1 M `[H^+]=sqrt(K_a C)` `H^(+) =10^(-pH)=10^(-3)` `10^(-3) = sqrt(K_axx0.1)` or `10^(-6) =K_a xx0.1` `therefore K_a=10^(-5)` |
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| 12. |
The pH of a 0.1 M aqueous solution of a weak acid (HA) is 3. What is its degree of dissociation ? |
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Answer» 0.01 `[H^(+)]=c ALPHA = 10^(-3) " " (. :' pH=3)` i.e., `0.1xx alpha = 10^(-3) "or" alpha=10^(-2)=1%` |
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| 13. |
The pH of a 0.005 M H_2SO_4solution is |
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Answer» `5.0` |
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| 14. |
The P^(H)of 40 ml of 0.02M HClwill notbe changed by adding |
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Answer» `1 ml of 1M HCL` |
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| 15. |
The pH of 1.2 % (w/v) aqueous solution of CH_(3) COOH at 25^(@) C is (pK_(a) = 4.7) |
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Answer» `5.0` |
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| 16. |
The pH of 10^(-8) M acid solution lies between .........and ............... |
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Answer» |
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| 17. |
The P^(H) of 10^(-8) M HCI is |
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Answer» SOLUTION :`[H^(+) ]_("total ")= 10 ^(-7)M+10 ^(8)M= 10 ^(-7)[1+ 0.1 ]` ` "" =1.1 xx 10 ^(-7)M ` `pH "" = - log _10 (1.1 xx 10 ^(-7)) ` ` ""= 7- log ^(1.1)= 7 -0. 0 414 = 6. 96` |
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| 18. |
The pH of 10^(-2) M NaOH solution is .............. Times the pH of 10^(-2) M HCl solution |
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Answer» pH of `10^(-2)` M HCL = 2. |
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| 19. |
The P^(H)of 10^(-3) M mono acidic base, if it is 1% ionised is |
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Answer» 5 ` = 10 ^(-5) rArr pOH =5, pH =9` |
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| 20. |
The pH of 10^(-10) M NaOH solution lies between........and ............ |
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Answer» |
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| 21. |
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionizationin the solution. |
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Answer» Solution :Calculation of pH `[H^+]` on base of pH : `pH=-log [H^+]=-log (2.34)=-log bar3.76` `=4.157xx10^(-3)` `therefore [H^+] = 10^(-2.34) = 4.57xx10^(-3)` M Calculation of ionization constant `(K_a)`: SUPPOSE ionized HCNO=X M At equili. [HCNO] = (0.1 -x) `approx` 0.1 and `[CNO^-]=[H_3O^+]=xM` `{:(,HCNO_((aq))+ ,H_2O_((l)) HARR, CNO_((aq))^(-)+ , H_3O_((aq))^(+)),("Initially M:",0.1,-,0.0,0.0),("At equilibrium M",(0.1-x),-,x M, x M),(,approx 0.1 M, ,=4.57xx10^(-3)M, =4.57xx10^(-3)M):}` `K_a=([CNO^-][H_3O^+])/([HCNO])` `=((4.57xx10^(-3))(4.57xx10^(-3)))/0.1= 2.088xx10^(-4)` Calculation of degree of ionization of HCNO : Suppose degree of ionization = `alpha` So, `(C xx alpha) =0.1 alpha `, HCNO is ionized , `[H^+]=[CNO^-]`= (ionized HCNO) `therefore [H^+] = 0.1 alpha` According to above calculation `[H^+]=4.57xx10^(-3)` M `therefore 0.1 alpha = 4.57xx10^(-3)` `therefore alpha = 4.57xx10^(-2)` =0.0457 Thus degree of ionization of HCNO = 0.0457 `=4.57xx10^(-2)` |
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| 22. |
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution . |
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Answer» Solution :`HCNO HARR H^(+)+CNO^(-)` `pH = 2.34 "means" - LOG [H^(+)]= 2.34 or log [H^(+)]=2.34 or log [H^(+)]=-2.34 = bar(3).66` or `[H^(+)] ="Antilog" bar(3).66 = 4.57 xx 10^(-3) M` `[CNO^(-)]=[H^(+)]=4.57 xx 10^(-3) M` `K_(a)=((4.57xx10^(-3))(4.57xx10^(-3)))/(0.1)=2.09xx10^(-4)` `alpha = sqrt(K_(a)//C)=sqrt(2.09xx10^(-4)//0.1)=0.0457`. |
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| 23. |
The pH of 0.1 M solution of an organic acid is 3.0. Calculate the dissociation constant of the acid. |
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Answer» `[H^(+)]=[RCO O^(-)]=10^(-3)M(.:' pH =3)` `[RCO OH] = 0.1 - 10^(-3) ~~ 0.1 M` `K_(a) = ([RCO O^(-)][H^(+)])/([RCO OH])=(10^(-3)xx10^(-3))/(0.1)=10^(-5)`. |
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| 24. |
The p^(H) of 0.1 M NaCI solution is |
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Answer» 1 |
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| 25. |
The pH of 0.1 M monobasic acid is 4.50. Calculate the concentration of the species H^(+), A^(-) and HA at equilibrium. Also determine the value of K_(a) and pK_(a) of the monobasic acid. |
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Answer» `:. [H^(+)]=3.16xx10^(-5)M` `[H^(+)]=[A^(-)]=3.16xx10^(-5)M` `[HA]=(0.1-3.16xx10^(-5))~= 0.1M` `K_(a) = (3.16xx10^(-5))^(2//0.1)=1.0 xx 10^(-8), pK_(a)=8`. |
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| 26. |
The pH of 0.1 M monobasic acid is 4.50 . Calculate the concentration of species H^(+), A^(-)and HA at equilibrium . Also , Determine the value of K_a and pK_a of themonobasic acid. |
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Answer» Solution :pH=4.5 =-log `[H^+]` `therefore log [H^+]=-4.5 = bar5.5` `therefore [H^+]` =Antilog `bar5.5=3.1623xx10^(-5)` M `approx 3.16xx10^(-5)` M The ionic EQUILIBRIUM of weak acid HA is under : `{:(,HA_((aq)) + ,H_2O_((l)) hArr , H_3O_((aq))^(+)+, A_((aq))^(-)),("Initial", 0.1M, -,0,0),("Equilibrium" , (0.1-alpha)M, ,alphaM, alphaM):}` `approx` 0.1 M So, `[H^+]=[A^-]=3.16xx10^(-5)` M At equilibrium [HA]=`(0.1 - alpha)` =`(0.1-3.16xx10^(-5))` `approx` 0.1 M Value of `alpha` is very small so, neglect. `K_a=([H^+][A^-])/([HA])=((3.16xx10^(-5))(3.16xx10^(-5)))/0.1` `=9.9856xx10^(-9)` `approx 10xx10^(-9) approx 1.0xx10^(-8)` `pK_a =-log K_a=-log (1.0xx10^(-8))` =-(0.0-8) = + 8 |
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| 27. |
The pH of 0.1 M HCN solution is 5.2 calculate K_a of this solution. |
| Answer» SOLUTION :`3.98xx10^(-10)` | |
| 28. |
The pH of 0.08 M HOCI is 2.85. Calculate the degree of dissociation. |
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Answer» Solution :Given value of `pH=2.85, [H^+] = 1.4 XX 10^(-3) mol L^(-1)` `[H^(+)] =C alpha, 1.4 xx 10^(-3)= 0.008 xxalpha,alpha= 1.75xx10^(-2)` The DEGREE of dissociation, `alpha =1.77 xx 10^(-2)`. |
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| 29. |
The pH of 0.05M aqueous dimethyl amine is 12. Calculate the K_b value of the base in aqueous solution. |
| Answer» SOLUTION :`2 XX 10^(-3)` | |
| 30. |
The pH of 0.05 M aqueous solution of diethylamine is 12.0 . Calculate its K_(b). |
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Answer» Solution :`(C_(2)H_(5))_(2)NH+H_(2)O hArr(C_(2)H_(5))_(2)NH_(2)^(+)+OH^(-)` As `PH = 12, :. [H^(+)]=10^(-12)M or [OH^(-)]=10^(-2)M, [(C_(2)H_(5))_(2)NH]=0.05-0.01 = 0.04 M ` `K_(b) = ([(C_(2)H_(5))_(2)NH_(2)^(+)][OH^(-)])/([(C_(2)H_(5))_(2)NH]) = (10^(-2)xx10^(-2))/(0.04)=2.5xx10^(-3)" "{[C_(2)H_(5))_(2)NH_(2)^(+)]=[OH^(-)]}` |
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| 31. |
The pH of 0.05 M aqueous solution of diethyl amine is 12.0. The K_(b) = x xx 10^(-3) then what is x value? |
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Answer» ` pOH= (1)/(2) (pK_b -log C) ` ` 2= (1)/(2) (pK_b-log ( 5 xx10 ^(_2))) ` `pK_b -log 5 +2=4 rArr pK_b =2+ 0.7=2.7` ` -log K_b =2.7 rArr K_b = 10 ^(-2.7)=10 ^(-3+0.3) ` Anti log ` (0.3 ) xx 10 ^(-3)= 2XX 10 ^(-3) ` |
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| 32. |
The pH of 0.04 M hydrazine solution is 9.7 . Calculate its ionization constant K_(b) and pK_(b). |
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Answer» `PH = 9.7:. Log [H^(+)]=-9.7 = bar(10).3 or [H^(+)]= 1.67 xx 10^(-10)` `:. [OH^(-)]=(K_(w))/([H^(+)])=(10^(-14))/(1.67xx10^(-10))=5.98xx10^(-5)` `K_(b)=([NH_(2)NH_(3)^(+)][OH^(-)])/([NH_(2)NH_(2)])=((5.98xx10^(-5))^(2))/(0.004) = 8.96xx10^(-7)` `pK_(b) = - log K_(b) = - log (8.96 xx 10^(-7))=6.04` |
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| 33. |
The pH of 0.02M solution of HF is 2.62. Calculate K_4 of HF. What is the pH of 0.0002M HF solution ? |
| Answer» SOLUTION :`K_a= 3.2 XX 10^(-4), PH= 3.62` | |
| 34. |
The pH of 0.005 M codeine (C_18H_21NO_3) solution is 9.95 . Calculate its ionization constant and pK_b. |
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Answer» Solution :Calculation of concentrationof `OH^-` : pH=-log `[H^+]`= 9.95 So, pOH=(14.0-9.95)=4.05 pOH=-log `[OH^-]` =4.05 = `bar5.95` `therefore [OH^-]`=-Antilog `(bar5.95)` `[OH^-]=10^(-4.05) = 8.9125xx10^(-5)` pH = 9.95 `gt` 7.0, So codeine is weak BASE. So, ionic EQUILIBRIUM is as under : `{:(,C_18H_21NO_(3(aq))+H_2O_((L)) hArr, C_18H_21NO_3H_((aq))^(+) + , OH_((aq))^(-)),("INITIAL concen.", 0.005 M,0.0 M, 0.0M),("Change in equilibrium:",0.005 M ,0.0 M ,0.0 M),("Molarity at equili.",underset(approx0.005 M)(0.005-x),underset(=8.91xx10^(-5))(xM),underset(=8.91xx10^(-5))(x M)):}` Calculation of `K_a` : `K_b=([C_18 H_21 NO_3H^+][OH^-])/([C_18H_21NO_3])` `K_b=((8.9125)^2(10^(-5))^2)/0.005=1.5886xx10^(-6)` `pK_b=-log (K_b)=-log (1.5886xx10^(-6))` =-(0.2010-6.0)=-(-5.799)=+ 5.799 `APPROX` 5.8 |
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| 35. |
The pH of 0.005 M codeine (C_(18)H_(21)NO_(3)) solution is 9.95. Calculate its ionization constant and pK_(b). |
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Answer» Form GIVEN pH, `[H^(+)] = 1.12xx10^(-10) M` `:. [OH^(-)] = (K_(w))/([H^(+)]) = (10^(-14))/(1.12xx10^(-10))=8.93xx 10^(-5)M` `K_(B) = ([B^(+)] [OH^(-)])/([BOH]) = ((8.93xx10^(-5))^(2))/(0.005)=1.6xx10^(-6)`. |
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| 36. |
The P^(H) of 0.005 M Ba (OH)_(2) is |
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Answer» `2.301` ` [OH^(-) ] = 10 ^(-2)RARR [H^(+) ] = 10 ^(-12)rArr pH = 12` |
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| 37. |
The pH of 0.004 M hydrazine solution is 9.7. its ionisation constant (K_(b)) is |
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Answer» `7.79xx10^(-8)` |
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| 38. |
The pH of 0.004 M hydrazine solution is 9.7 . Calculate its ionization constant K_b and pK_b. |
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Answer» SOLUTION :WEAK base Hydrazine , `NH_2NH_2` the CALCULATION of `[OH^-]` is , pH=9.7 and pH+POH=14 `therefore` pOH=14.0-9.7 = 4.3 `[pOH^-] = -log [OH^-]`=4.3 `therefore log [OH^-]=-4.3 = bar5.7` `therefore [OH^-]`=Antilog `(bar5.7)=5.012xx10^(-5)` =xM Calculation of ionization constant `K_b`: Ionic equilibrium of `(NH_2NH_2)` hydrazine solution is as under. `{:(,NH_2NH_(2(aq))+,H_2O_((l)) hArr, NH_2NH_(3(aq))^(+) + , OH_((aq))^(-)),("At equili." , (0.004-x),M approx x, x M, x M):}` `therefore K_b=([NH_2NH_3^+][OH^-])/([NH_2NH_2])` `=((5.012xx10^(-5))(5.012xx10^(-5)))/0.004` `=6.028xx10^(-7)` Calculation of `pK_b` : `pK_b=-log K_b=-log (6.28xx10^(-7))` =-(0.7980- 7.000) = 6.020 |
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| 40. |
The P^(H)of 0. 1Msolution of the following compoundsincreases in the order |
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Answer» ` NACL LT NH_4 Cl lt NaCN lt HCl` pH of `NH_4Cl lt 7 , pH` of NaCN` GT 7` |
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| 41. |
ThepH 0.1 M solution of the following salts increases in the order : |
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Answer» `NACL lt NH_(4) CL lt NaCN lt HCl` `(pH lt 7) , NaCN = "basic" (pH gt 7) , HCl = "Strongly acidic"(pH lt lt 7)`. `:.` ORDER will be `HCl lt NH_(4)Cl lt NaCl lt NaCN`. |
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| 42. |
The peroxideeffect in antimarkovnikoff additioninvolves a ………………….mechanism |
| Answer» SOLUTION :freeradical | |
| 43. |
The peroxide effect is observed only in addition of HBr, and not with HCl and HI. |
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Answer» |
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| 44. |
The peroxide effect occurs by : |
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Answer» IONIC mechanism |
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| 45. |
The periodictime of radiationis 4 xx10^(-10)s findwavelengthwave numberandfrequency |
| Answer» SOLUTION :`v= 2.5 xx 10^(9) Hzlambda = 0.12 m VEC( v)=-8.33 m^(-1)` | |
| 46. |
The period that includes all blocks of elements is |
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Answer» 1 |
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| 47. |
The period that contains only gaseous elements is |
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Answer» 1 |
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| 48. |
The period number in the long form of the periodic table is equal to... |
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Answer» magnetic quantum number any element of the PERIOD. The period number in the form of periodic table refers to the maximum principal quantum number of any element, which is the period. `:.` Period number = maximum n of any element (n = principal quantum number). |
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| 49. |
The period numberin the longformof theperiodictableis equalto |
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Answer» magneticquantum number of anyelement OFTHE period. |
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