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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.
| 1. |
Which of the following salts will give highest `pH` in water?A. `KCl`B. `Nacl`C. `Na_(2)CO_(3)`D. `CuSO_(4)` |
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Answer» Correct Answer - C `Na_(2)CO_(3)` is a salt of strong base `(NaOH)` and weak acid `(H_(2)CO_(3))`. On hydrolysis this salt will produce strongly basic solution. i.e., `pH` will be highest `(pHgt7)` for this solution. Others are combination of `KCl=` strong acid `+` Strong base `rarr` neutral solution `(pH=7)` `NaCl=` Strong aicd `+` Strong base `rarr` neutral solution `(pH=7)` `CuSO_(4)=` Strong acid `+` weak base `rarr` Acidic solution `(pHlt7)` |
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| 2. |
The `pH` of `0.1 M` solution of the following salts increases in the orderA. `NaClltNH_(4)ClltNaCNltHCl`B. `HClltNH_(4)ClltNaClltNaCN`C. `NaCNltNH_(4)ClltNaClltHCl`D. `HClltNaClltNaCNltNH_(4)Cl` |
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Answer» Correct Answer - D Undergoes cationic hydrolysis, hence `pH` is `gt7` `NaCN` undergoes anionic hydrolysis hence `pH` is `gt7` `HCl` is strong acid and `NaCl` is neutral solution. Hence the `pH` of given solutions will increases. `HClltNaClltNaCNltNH_(4)Cl` |
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| 3. |
Lemon juice normally has a `pH` of `2`. If all the acid the lemon juice is citric acid and there are no citrate salts present, then what will be the citric acid concentration `[Hcit]` in the lemon juice? (Assume that only the first hydrogen of citric acid is important) `HCithArrH^(+)+Cit^(-)`, `K_(a)=8.4xx10^(-4) mol L^(-1)`A. `8.4xx10^(-4) M`B. `4.2xx10^(-4) M`C. `16.8xx10^(-4) M`D. `11.9xx10^(-2) M` |
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Answer» Correct Answer - D `[H^(+)]=sqrt(K_(a)C), pH=2, :. [H^(+)]=10^(-2) M` `10^(-2)=sqrt(K_(a)C)` `10^(-4)=8.4xx10^(-4)xxC` `:. C=11.9xx10^(-2) M` |
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| 4. |
If an aqueous solution at `25^(@)C` has twice as many `OH^(-)` as pure water its `pOH` will beA. `6.699`B. `7.307`C. `7`D. `6.98` |
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Answer» Correct Answer - A `[OH^(-)]=2xx10^(-7)` `pOH=14-pH or-log[OH^(-)]` |
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| 5. |
`pOH` of `H_(2)O` is `7.0` at `298K`. If water is heated at `350 K`, which of the following statement should be true?A. `pOH` will decrease.B. `pOH` will increase.C. `pOH` will remain `7.0`.D. concentration of `H^(+)` ions will increase but that of `OH^(-)` will decrease. |
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Answer» Correct Answer - A `[OH^(-)]=sqrt(K_(w))` in pure water. So as temperature increases `K_(w)` decreases `implies [OH^(-)]` decreases. |
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| 6. |
`10 mL` of `10^(-6) M HCl` solution is mixed with `90mL H_(2)O`. `pH` will change approximately:A. By one unitB. By `0.3` unitC. By `0.7` unitD. By `0.1` unit |
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Answer» Correct Answer - C `pH` of `10^(-6) M HCl=6` on dilution `[HCl]=(10^(-6)xx10)/(100)=10^(-7)` Thus new `pH` of `HCl` is not `7` because it is acid. `pH` new solution: `[H^(+)]=10^(-7)+10^(-7)` (from`H_(2)O`) `=2xx10^(-7)` `:. pH=6.7` (approximately) |
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| 7. |
`10ml` of `1 M H_(2)SO_(4)` will completely neutraliseA. `10 ml` of `1M NaOH` solutionB. `10 ml` of `2 M NaOH` solutionC. `5 ml` of `2 M KOH` solutionD. `5 ml` of `1 M Na_(2)CO_(3)` solution |
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Answer» Correct Answer - B `H_(2)SO_(4)+2H_(2)OhArr2H_(3)O^(+)+SO_(4)^(--)` `NaOHhArrNa^(+)+OH^(-)` `1` mole of `H_(2)SO_(4)` acid gives `2` moles of `H_(3)O^(+) ions`. So `2` moles of `OH^(-)` are required for complete neutralization. |
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| 8. |
The `pH` of an aqueous solution of `0.1 M` solution of a weak monoprotic acid which is `1%` ionised isA. `1`B. `2`C. `3`D. `11` |
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Answer» Correct Answer - C `{:(,HA,hArr,H^(+),+,A^(-)),(,0.1,,,,),(,0.1(1-(1)/(1100)),,(1xx0.1)/(100),,(1xx0.1)/(100)):}` `[H^(+)]=10^(-3)` ` :. pH=3` |
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| 9. |
Three reactions involving `H_(2)PO_(4)^(-)` are given below `I. H_(3)PO_(4)+H_(2)OrarrH_(3)O^(+)+H_(2)PO_(4)^(-)` `II. H_(2)PO_(4)^(-)+H_(2)OrarrHPO_(4)^(2-)+H_(3)O^(+)` `III. H_(2)PO_(4)^(-)+OH^(-)rarrH_(3)PO_(4)+O^(2+)` In which of the above does `H_(2)PO_(4)^(-)` act as an acid?A. `II` onlyB. `I` and `II`C. `III` onlyD. `I` only |
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Answer» Correct Answer - A Only in reaction (`II`), `H_(2)PO_(4)^(-)`, gives `H^(+)` to `H_(2)O`. Thus, behaves as an acid. |
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| 10. |
The conjugate base of `H_(2)PO_(4)^(-)` is :A. `HPO_(4)^(2-)`B. `P_(2)O_(5)`C. `H_(3)PO_(4)`D. `PO_(4)^(3-)` |
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Answer» Correct Answer - A `underset(Acid)(H_(2)PO_(4)^(-)) rarr underset(Base)(H^(+)+HPO_(4)^(2-))` |
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| 11. |
For `10^(-2) (M) H_(3)PO_(3)` solution which of the following relations is correct?A. `[H_(3)PO_(3)]+[H_(2)PO_(3)^(-)]+[HPO_(3)^(2-)]+[PO_(3)^(2-)]=10^(-2)`B. `[H_(3)PO_(3)]+[H_(2)PO_(3)^(-)]+[HPO_(3)^(2-)]=10^(-2)`C. `[H_(2)PO_(3)^(-)]+[HPO_(3)^(2-)]+[PO_(3)^(3-)]=10^(-2)`D. `[H_(3)PO_(3)]+[H_(2)PO_(3)^(-)]+2[HPO_(3)^(2-)]=10^(-2)` |
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Answer» Correct Answer - B `H_(3)PO_(3) oversetlarrtoundersetH_(2)PO_(3)^(-)+H^(+)` `H_(2)PO_(3) oversetlarrtounderset HPO_(3_^(2-))+H^(+)` `H_(3)PO_(3)` is a dibasic acid. |
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| 12. |
A monoprotic acid in `0.1 M` solution has `K_(a)=1.0xx10^(-5)`. The degree of dissociation for acid isA. `1.0%`B. `99.9%`C. `0.1%`D. `99%` |
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Answer» Correct Answer - A `K_(a)=c alpha^(2)` or `1.0xx10^(-5)=0.1xxalpha^(2)` `:. alpha=10^(-2)` or `1%` |
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| 13. |
The solubility product of a sparingly soluble salt `AB` at room temperature is `1.21xx10^(-6)`. Its molar solubility isA. `1.21xx10^(-6)`B. `1.21xx10^(-3)`C. `1.1xx10^(-4)`D. `1.1xx10^(-3)` |
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Answer» Correct Answer - D `AB` is a binary electrolyte. `S=sqrt(K_(sp)=sqrt(1.21xx10^(-6))=1.1xx10^(-3) M` |
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| 14. |
`K_(sp)` of a sparingly soluble salt `AB_(2)` is `4xx10^(-12) mol^(3) L^(-3)`. The solubility of the salt isA. `2xx10^(-6) M`B. `4xx10^(-4) M`C. `1xx10^(-12) M`D. `1xx10^(-4) M` |
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Answer» Correct Answer - D `AB_(2)hArrunderset(x)(A^(2+))+underset(2x)(2B)` `K_(sp)=[A^(2+)][B]^(2)=(x)(2x)^(2)=4x^(3)` `x^(3)=(K_(sp))/(4)=(4xx10^(-12))/(4)=1xx10^(-12)` `x=1xx10^(-4) M` |
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| 15. |
The solubility of a springly soluble salt `AB_(2)` in water is `1.0xx10^(-5) mol L^(-1)`. Its solubility product is:A. `10^(-15)`B. `10^(-10)`C. `4xx10^(-15)`D. `4xx10^(-10)` |
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Answer» Correct Answer - C `K_(SP)=4S^(3)=4xx(10^(-5))^(3)` `=4xx10^(-15)M^(3)` |
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| 16. |
Which of the following is most soluble in water ?A. `MnS (K_(SP)=8xx-37)`B. `ZnS (K_(SP)=7xx10^(-16))`C. `Bi_(2)S_(3) (K_(SP)= xx10^(-70))`D. `Ag_(2)S (K_(SP)=6xx10^(-5))` |
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Answer» Correct Answer - B Find solubility for each separtely by `S^(2)=K_(SP)` for `MnS` and `ZnS`. `108S^(5)=K_(SP)` for `Bi_(2)S_(3)` and `4S^(3)=K_(SP)` for `Ag_(2)S`. |
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| 17. |
The dissociation constant of a monobasic acid which is `3.5%` dissociated in `(N)/(20)` solution at `20^(@)C` isA. `3.5xx10^(-2)`B. `5xx10^(-3)`C. `6.34xx10^(-5)`D. `6.75xx10^(-2)` |
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Answer» Correct Answer - C Concentration of acid `=(N)/(20)=0.05N` Out of `100` molecules, `3.5` molecules has been dissociated `:.` Out of `1` molecule the no. of disscoiated molecules `=(35)/(100)=0.035=alpha` `K_(a)=(c alpha^(2))/((1-alpha))=(0.05xx(0.035)^(2))/(1-0.035)` `=6.34xx10^(-5)` |
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| 18. |
The volume of the water needed to dissolve `1 g` of `BaSO_(4) (K_(SP)=1.1xx10^(-10))` at `25^(@) C` is:A. `280 litre`B. `410 litre`C. `205 litre`D. None of these |
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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` `=244.37xx10^(-5)g//L` `:.` Volume of water needed to dissolve `1 g BaSO_(4)` is equal to `(1)/(244.37xx10^(-5))=410 L` |
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| 19. |
The number of `S^(2-)` ions present in `1 L` of `0.1 (M) H_(2)S` solution having `[H^(+)]=0.1 (M)` is ( Given `H_(2)ShArr2H^(+)+S^(2-) K_(a)=1.1xx10^(-21)`)A. `6.625xx10^(3)`B. `6.625xx10^(4)`C. `6.625xx10^(5)`D. `6.625xx10^(6)` |
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Answer» Correct Answer - A `H_(2)S underset(larr)(rarr) 2H^(+)+S^(2-)` `K_(a)=1.1xx10^(-21)=([H^(+)]^(2)[S^(2-)])/([H_(2)S])=(10^(-2)xx[S^(2-)])/(0.1)` or `[S^(2-)]=1.1xx10^(-20)(m)` ` :. ` no. of `S^(2-) ions =1.1xx10^(-20)xx6.0^(23)xx10^(23)` `=1.1xx6.023xx10^(3)=6.625xx10^(3)` |
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| 20. |
The `pH` of a `0.01 M` solution of a monobasic acid is four. Which one of the following statement about the acid is incorrectA. When a little `NaOH` is added, it will form a buffer solutionB. It is a weak acidC. Its sodium salt will be acidicD. Its sodium salt will be basic |
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Answer» Correct Answer - C Concentration of monobasic acid `= 0.01 M` `pH=4` If the acid is completely ionised, the `pH` of the acid would be `pH= -log0.01= -log10^(-2)=2` So it is a weak acid. When the sodium salt of a weak acid is dissolved , the anions will be hydrolysed giving rise to `OH^(-)ion` concentration. The solution will be basic . So statement (`c`) is incorrect. |
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| 21. |
At `25^(@)C pH` range of phenolphthalein is `8-10`. At `100^(@)C pH` range of phenophthalein would be |
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Answer» Correct Answer - B The `pH` range of the indicator is `pK_("in")+1`. On increase in temperature, `pK_("in")` decreases therefore `pH` range of indicator decreases. |
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| 22. |
`2H_(2)OhArrH_(3)O^(+)+OH^(-)` `K_(w)=1xx10^(-14)` at `25^(@)C`. Hence, `K_(a)` is:A. `1xx10^(-14)`B. `5.55xx10^(-13)`C. `1.8xx10^(-16)`D. `1.00xx10^(-7)` |
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Answer» Correct Answer - C `K_(w)=[H^(+)][OH^(-)]=1xx10^(-14)` `:. [H^(+)]=1xx10^(-7) M`, `C= 1g//mL=1000 g//L=55.55 M` Also in weak acid `[H^(+)]=sqrt(K_(a)C)` `:. K_(a)=([H^(+)]^(2)]/(C )=(10^(-14))/(55.55)=1.8xx10^(-16)` |
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| 23. |
`pH` of `0.01 M HS^(-)` will be:A. `pH=7+(pK_(a))/(2)+(log C)/(2)`B. `pH=7-(pK_(a))/(2)+(log C)/(2)`C. `pH=7+(pK_(1)+pK_(2))/(2)`D. `pH=7+((pK_(a)+pK_(b)))/(2)` |
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Answer» Correct Answer - A `HS^(-)+H_(2)OhArrH_(2)S+OH^(-)` `:. [OH^(-)]=Ch=sqrt((K_(w)C)/(K_(a))` `:. [H^(+)]=(K_(w))/(sqrt((K_(w).C)/(K_(a))))=sqrt((K_(w)K_(a))/(C))` or `pH=1//2[pK_(w)+pK_(a)+log C]` |
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| 24. |
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` decreaseB. `[H^(+)]` will decreaseC. `[H_(2)CO_(3)]`remains the sameD. forward reaction will be promoted |
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Answer» Correct Answer - B If `CO_(2)` escapes from the system, the following equilibrium will not be attained `H_(2)O+CO_(2)hArrH_(2)CO_(3)hArrH^(+)+HCO_(3)^(-)` `:. [H^(+)]` will decrease |
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| 25. |
Assertion : A substance that can either act as an acid a base is called ampholyte. Reason : Bisulphide ion `(HS^(-))` and biscarbonate ion `(HCO_(3)^(-))` are ampholytes.A. If both assertion and reason are true and the reason is the correct explantion of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
| Answer» Correct Answer - B | |
| 26. |
The salt that forms neutral solution in water isA. `NH_(4)Cl`B. `Nacl`C. `Na_(2)CO_(3)`D. `K_(3)BO_(3)` |
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Answer» Correct Answer - B Because it is a salt of strong acid and strong base. `H^(+)(aq)+Cl^(-)(aq)+Na^(+)(aq)+OH^(-)(aq)hArrH_(2)O_((1))+Na^(+)(aq)+Cl^(-)` |
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| 27. |
An indicator is a weak acid and `pH` range of its colour is `3` to `5`. If the neutral point of the indicator lies in the centre of the `[H^(+)]` corresponding to given `pH` range, then `pH` at the equivalence point isA. `7.0`B. `4.0`C. `3.3`D. `5.0` |
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Answer» Correct Answer - C The neutral point is at the centre of the `H^(+)` ion concentration. `[H^(+]` at `(pH=3)` is `10^(-3)=100xx10^(-5)` `[H^(+)]` at `(pH=5)` is `10^(-5)` Thus, `[H^(+)]` at the centre `=(100xx10^(-5)+10^(-5))/(2)=(101xx10^(-5))/(2)` `:. pH= -log``((101xx10^(-5))/(2))=3.3` |
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| 28. |
What `%` of the carbon in the `H_(2)CO_(3)-HCO_(3)^(-)` buffer should be in the form of `HCO_(3)^(-)` so as to have a neutral solution? `(K_(a)=4xx10^(-7))`A. `20%`B. `40%`C. `60%`D. `80%` |
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Answer» Correct Answer - D `pH=pK_(a)+log``([HCO_(3)^(-)])/([H_(2)CO_(3)])` `implies 7=7-log4+log``([HCO_(3)^(-)])/([H_(2)CO_(3)])implies ([HCO_(3)^(-)])/([H_(2)CO_(3)])=4` `%` of carbon in the form of `HCO_(3)^(-)` `=([HCO_(3)^(-)])/([HCO_(3)^(-)]+[H_(2)CO_(3)])xx100=(4)/(1+4)xx100=80%` |
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| 29. |
For an aqueous solution to be neutral it must haveA. `pH=7`B. `[H^(+)]=[OH^(-)]`C. `[H^(+)]=sqrt(K_(w))`D. `[H^(+)]lt[OH^(-)]` |
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Answer» Correct Answer - B For neutral solution `[H^(+)]` always must be equal to `[OH^(-)]` |
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| 30. |
`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` isA. `4.0xx10^(-6)`B. `5.0xx10^(-6)`C. `3.3xx10^(-7)`D. `5.0xx10^(-7)` |
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Answer» Correct Answer - D `pH=12 :. pOH=2` `:. [OH^(-)]=10^(-2)` Now `Ba(OH)_(2)=Ba^(2+)+2OH^(-)` `:. K_(sp)=[Ba^(2+)][OH^(-)]^(2)` `=[10^(-2)/(2)][10^(-2)]^(2)` `:.[(Ba^(2+))=(1)/(2)xx(OH^(-))]=5xx10^(-7)` |
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| 31. |
The `K_(SP)` for `Cr(OH)_(3)` is `1.6xx10^(-30)`. The molar solubility of this compound in water isA. `2sqrt(1.6xx10^(-30))`B. `4sqrt(1.6xx10^(-30))`C. `4sqrt((1.6xx10^(-30))/(27))`D. `1.6xx10^(-30)//27` |
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Answer» Correct Answer - C Let molar solubility of `Cr(OH)_(3)=s mol L^(-1)` `Cr(OH)_(3)(s)hArrunderset(s)(Cr^(3+)(aq)+underset(3s)(3OH^(-)(aq))` `K_(SP)=1.6xx10^(-30)` `=[Cr^(3+)][OH^(-)]^(3)=(s)(3s)^(3)=27s^(4)` `s^(4)=(1.6xx10^(-30))/(27)` `s=4sqrt((1.6xx10^(-30))/(27))` |
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| 32. |
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)g`B. `1.2xx10^(-9)g`C. `6.2xx10^(-5)g`D. `5.0xx10^(-8)g` |
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Answer» Correct Answer - B `[AgBr]=[Ag^(+)]=0.05 M` `K_(sp)[AgBr]=[Ag^(+)][Br^(-)]` `implies [Br^(-)]=(K_(sp)(AgBr))/([Ag^(+)])` `=(5.0xx10^(-13))/(0.05)=10^(-11) M [mol L^(-1)]` Moles of `KBr` needed to precipitate `AgBr` `=[Br^(-)]xxV=10^(-11) mol L^(-1)xx1L=10^(-11) mol` Therefore, amount of `KBr` needed to precipitate `AgBr` `10^(-11) molxx120 g mol^(-1)=1.2xx10^(-9) g` |
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| 33. |
In a buffer solution containing equal concentration of `B^(-)` and `HB`, the `K_(b)` for `B^(-)` is `10^(-10)`. The `pH` of buffer solution isA. `10`B. `7`C. `6`D. `4` |
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Answer» Correct Answer - D `pH=pK_(a)+log``([CB])/([Acid])` `pK_(w)-pK_(b)+log``([CB])/([Acid])` `=14-10+log 1~~4` |
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| 34. |
The buffer solution of `100 ml` having a `pH` value `4` when added to `1 ml` dilute `HCl`, then the `pH` of buffer solutionA. Converts to `7`B. Does not changeC. Converts to `2`D. Changes to `10` |
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Answer» Correct Answer - B `pH` does not change on addition of some concentration of `HCl`. |
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| 35. |
Calculate the `pH` of a buffer solution prepared by dissolving `10.6 g` of `Na_(2)CO_(3)` in `500 ml` of an aqueous solution containing `80 ml` of `1 M HCl`. `K_(a)` for `HCO_(3)^(-)=6xx10^(-11)`A. `8.6`B. `9.6`C. `11.6`D. `12.6` |
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Answer» Correct Answer - B `{:(,Na_(2)CO_(3),+,HCl,to,NaCl,+,NaHCO_(3)),("Meq. before",(10.6)/(106)xx1000,,2,,80xx1,,4),("Reaction",=100,,80,,0,,0),("Meq. After",20,,0,,80,,80),("Reaction",1,,2,,2,,2):}` The reaction has `Na_(2)CO_(3)` and `HCO_(3)^(-)` and thus acts as buffer `pH= -logK_(a)+log``([CO_(3)^(-2)])/([HCO_(3)^(-)])` `= -log 6xx10^(-11)+log``(20)/(80)=9.6` |
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| 36. |
Assertion : The dissociation constant of water at `60^(@)C` is `10^(-13)`. Reason : The `pH` of water is `6.5` and that it behaves as acid at `60^(@)C`.A. If both assertion and reason are true and the reason is the correct explantion of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C `K_(w)=10^(-13)` at `60^(@)C` `:. pH=6.5`, but water is neutral because `pH` scale contracts to `0` to `13`. |
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| 37. |
Solubility of `MX_(2)` type electrolytes is `0.5xx10^(-4) mol//L`, then find out `K_(sp)` of electrolytes.A. `5xx10^(-12)`B. `25xx10^(-10)`C. `1xx10^(-13)`D. `5xx10^(-13)` |
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Answer» Correct Answer - D `underset(Solubility 0.5xx10^(-4)M)(MX_(2)) rarr underset(0.5xx10^(-4)M)(M^(2+))+underset(2xx0.5xx10^(-4)M)(2X^(-))` (On `100%` ionisation) ` :. K_(sp)` of `MX_(2)=[M^(2+)][X^(-)]^(2)` `=(0.5xx10^(-4))(1.0xx10^(-4))^(2)` `=0.5xx10^(-12)=5xx10^(-13)` |
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| 38. |
Solubility if `M_(2)S` type salt is `3.5xx10^(-6)`, then find out its solubility productA. `1.7xx10^(-6)`B. `1.7xx10^(-16)`C. `1.7xx10^(-18)`D. `1.7xx10^(-12)` |
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Answer» Correct Answer - B Solubility of `M_(2)S` salt is `3.5xx10^(-6)M` `underset(Solubility 3.5xx10^(-6)M)(M_(2)S)hArrunderset(2xx3.5xx10^(-6))(2M^(+))+underset(M3.5xx10^(-6)M)(S^(2-))` `:. K_(sp)` (Solubility product of `M_(2)S`)=`[M^(+)]^(2)[S^(2-)]` `= (7.0xx10^(-6))^(2)(3.5xx10^(-6))=171.5xx10^(-16)` |
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| 39. |
Assertion : Salting out action of sodium soap in oresence of `NaCl` is based on common ion effect. Reason : Salting out action of soap is based on the fact that as the concentration of `Na^(+)` increases, the `RCOONa` shows precipitation because `[RCOO^(-)][Na^(+)]gtK_(sp)`.A. If both assertion and reason are true and the reason is the correct explantion of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D `RCOONa rarr RCOO^(-)+Na^(+)`, In presence of `NaCl`, `[Na^(+)]` increases and `[RCOO^(-)][Na^(+)]` exceeds thatn `K_(sp)` of `RCOONa. |
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| 40. |
The hydrogen ion concentration of a `10^(-8) M HCl` aqueous soultion at `298 K(K_(w)=10^(-14))` isA. `1.0xx10^(-6) M`B. `1.0525xx10^(-7) M`C. `9.525xx10^(-8) M`D. `1.0xx10^(-8) M` |
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Answer» Correct Answer - B In aqueous solution of `10^(-8) M HCl`, `[H^(+)]` is based upon the concentration of `H^(+) ion` of `10^(-8) M HCl` and concentration of `H^(+) ion` of water. `K_(w)` of `H_(2)O=10^(-14)=[H^(+)][OH^(-)]` or `[H^(+)]=10^(-7) M` (due to its neutral behaviour) So, in aqueous solution of `10^(-8) M HCl`, `[H^(+)]=[H^(+)]` of `HCl+[H^(+)]` of water `=10^(-8)+10^(-7)=11xx10^(-8) M` `1.10xx10^(-7) M` |
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| 41. |
The hydrogen ion concentration of a `10^(-8) M HCl` aqueous soultion at `298 K(K_(w)=10^(-14))` isA. `9.525xx10^(-8) M`B. `1.0xx10^(-8) M`C. `1.0xx10^(-6) M`D. `1.0525xx10^(-7) M` |
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Answer» Correct Answer - D `HClrarrunderset(10^(-8) M)(H^(+))+Cl^(-)` `H_(2)OhArrunderset(10^(-8)+a)(H^(+))+underset(a)(OH^(-))` `K_(w)=[H^(+)][OH^(-)]` `:. K_(w)=(10^(-8)+a)xxa` or `10^(-14)=10^(-8)a+a^(2)` or `a=0.95xx10^(-7)` `:. [H^(+)]=10^(-8)+0.95xx10^(-7)` `=1.05xx10^(-7)` |
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| 42. |
An acid-base indicator which is a weak acid has a `pK_(a)` value `=5.5`. At what concentration ratio of sodium acetate to acetic acid would the indicator show a colour half-way between those of its acid and conjugate base forms?A. `4.93 : 1`B. `6.3 : 1`C. `5.62 : 1`D. `2.37 : 1` |
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Answer» Correct Answer - C At Half way `[HIn]=In` `pH=5.5+(["salt"])/([acid])5.5` `=pK_(a)+log``(["salt"])/([acid])` `log``(["salt"])/([acid])=0.75implies (["salt"])/([acid])=5.62` |
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| 43. |
An indicator has `pK_(In)=5.3`. In a certain titration, this indicator is found to be `80%` ionized in its acid form. Thus, `pH` of the solution isA. `4.7`B. `5.3`C. `5.9`D. `6.2` |
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Answer» Correct Answer - A `underset(Acid form)(HIn)hArrunderset(Basic form)(H^(+))+In^(-)` Thus, acid form `[HIn] =0.80 M` `[In^(-)]=0.20 M` `pH=pK_(In)+log``([In^(-)])/([HIn])=5.3+log``(0.2)/(0.8)` `=5.3+log(1)/(4)=5.3-0.60=4.7` |
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| 44. |
A certain indicator (an organic dye) has `pK_(a)=5`. For which of the following titrations may it be suitableA. acetic acid against `NaOH`B. aniline hydrochloride against `NaOH`C. sodium carbonate against `HCl`D. barium hydroxide against oxalic acid |
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Answer» Correct Answer - C `pK_(HIn)=5` (`a`) `CH_(3)COOH+NaOH`, end point `pHgt7` (`b`) Aniline hydrochloride `+ NaOH`, end point `pHgt7` (`c`) `NaHCO_(3)+HCl`, end point `pHlt7` `Ba(OH)_(2)+H_(2)C_(2)O_(4)`, end point `pHgt7` |
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| 45. |
In a saturated solution of the spatingly soluble strong electrolyte `AgIO_(3)` (molecular mass `=283`) the equilibrium which sets in is `AgIO_(3) (s)hArrAg^(+)(aq)+IO_(3)^(-)(aq)` If the solubility product constant `K_(SP)` of `AgIO_(3)` at a given temperature is `1.0xx10^(-8)`, what is the mass of `AgIO_(3)` cotained in `100 mL` of its saturated solution?A. `28.3xx10^(-2) g`B. `28.3xx10^(-3) g`C. `1.0xx10^(-7) g`D. `1.0xx10^(-4) g` |
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Answer» Correct Answer - B `AgIO_(3)(S)hArrAg_((aq))^(+)+IO_(3(aq))^(-)` Let the solubility of `AgIO_(3)` be `S` `K_(SP)=[Ag^(+)][IO_(3)^(-)]` `1.0xx10^(-8)=S^(2)` `S=10^(-4)mol//litre` `= (10^(-4)xx283)/(1000)xx100=283xx10^(-5)` `=2.83xx10^(-3) g//100mL` |
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| 46. |
A certain mixture of `HCl` and `CH_(3)-COOH` is `0.1 M` in each of the acids. `20 ml` of this solution is titrated against `0.1 M NaOH`. By how many units does the `pH` change from the start to the stage when the `HCl` is almost completely neutralised and acidic acid remains unreacted? `K_(a)` for acetic acid `=2xx10^(-5)`.A. `1.5`B. `3`C. `2`D. `3.25` |
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Answer» Correct Answer - B mmoles of `HCl=0.1xx20=2` mmoles of `CH_(3)COOH=0.1xx20=2` After titration of `HCl` by `NaOH` `[CH_(3)COOH]=(2)/(40)=(1)/(20)M` `pH=(1)/(2)(pK_(a)-log C)` `=(1)/(2)[5-log2-log``((1)/(20))]=3` |
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| 47. |
A volume of `50.00 mL` of a weak acid of unknown concentration is titrated with `0.10 M` solution of `NaOH`. The equivalence point is reached after `39.30 mL` of `NaOH` solution has been added. At the half-equivalence point `(19.65 mL)`, the `pH` is `4.85`. Thus, initial concentration of the acid and its `pK_(a)` values areA. `[HA]` initial `=0.1 M`, `pK_(a)=4.85`B. `[HA]` initial `=0.079 M`, `pK_(a)=2.93`C. `[HA]` initial `=0.1 M`, `pK_(a)=3.70`D. `[HA]` initial `=0.079 M`, `pK_(a)=4.85` |
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Answer» Correct Answer - D `underset(Acid)(M_(1)V_(1))=underset(Base)(M_(2)V_(2)` `M_(1)xx50.0=0.10xx39.30` `:. M_(1)=(0.10xx39.30)/(50)=0.0786 M` When half-equivalence point is reached `[HA]=[A^(-)]` `pH=pK_(a)+log``([A^(-)])/([HA])` `:. pH=pK_(a)=4.85` |
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| 48. |
When `CH_(3)COOCH_(3)+HCl` is titrated with `NaOH` then at neutral point the colour of phenopthalein becomes colourless form pink due to:A. due to formation of `CH_(3)OH`B. due to formation of `CH_(3)COOH` which act as a weak acid.C. Phenolpthalein vaporizes.D. due to presence of `HCl` |
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Answer» Correct Answer - B `CH_(3)COOCH_(3)` `overset(H_(2)O)underset(H^(+))rarr` `CH_(3)COOH+CH_(3)OH` `HCl+NaOH rarr NaCl+H_(2)O` `CH_(3)COOH+NaOH rarr CH_(3)COONa` |
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| 49. |
When `10 ml` of `0.1 M` acitec acid `(pk_(a)=5.0)` is titrated against `10 ml` of `0.1M` ammonia solution `(pk_(b)=5.0)`,the equivalence point occurs at `pH`A. `5.0`B. `6.0`C. `7.0`D. `9.0` |
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Answer» Correct Answer - C `pK_(a)= -log K_(a)`, `pK_(b)= -log K_(b)` `pH=(1)/(2)[logK_(a)+logK_(w)-logK_(b)]` `=-(1)/(2)[-5+log(1xx10^(-14))-(-5)]` `= -(1)/(2)[-5-14+5]= -(1)/(2)(-14)=7` |
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| 50. |
Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest `pH` value?A. `LiCl`B. `BeCl_(2)`C. `BaCl_(2)`D. `AlCl_(3)` |
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Answer» Correct Answer - C `BaCl_(2)`, a salt of strong acid `+` strong base, is not hydrolysed and thus `pH` is maximum, i.e., `7`. Rest all being salt of strong acid `+` weak base are hydrolysed to give `pH` less than `7`. |
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