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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.
| 1801. |
If `pK_(b)` for floride at `25^(0)C` is `11`, the ionisation constant of hydrofluoric acid in water at this temperature isA. `10^(-3)`B. `10^(-4)`C. `10^(-2)`D. `10^(-5)` |
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Answer» Correct Answer - A `K_(w) = K_(a)xxK_(b) , K_(a) = (K_(w))/(K_(b))` |
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| 1802. |
The solubility of `C_(6)H_(5)NH_(3^(+))Cl^(-)` would be highest among the following solvents in:A. Acidic buffer of pH = 3B. Basic buffer of pH = 10C. Neutral buffer of pH = 7D. Pure water |
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Answer» Correct Answer - B The solubility of `C_(6)H_(5)NH_(3)Cl^(-)` highest in basic buffer with more `P^(H)`. |
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| 1803. |
The gaseous phase neutralisatin reaction can be explanied only on the basis ofA. Arrhenius theoryB. Bronsted theoryC. Lewis theoryD. Bohrs theory |
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Answer» Correct Answer - C The gaseous phase neutralisation reaction can be explained by lewis theory |
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| 1804. |
`H^(+)` is an acid, accoding toA. Arrhenius theoryB. Bronsted theoryC. Lewis theoryD. All theories |
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Answer» Correct Answer - C According to lewis theory `H^(+)` is an acid |
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| 1805. |
The solubility product of `SrF_(2)` in water is `8xx10^(-10)`. Calculate its solubility in `0.1M` NaF aqueous solution. |
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Answer» `K_(SP)=[Sr^(2+)][F^(-)]^(2)` `8xx10^(-10)=S[2S+0.1]^(2)` `:. S= (8xx10^(-10))/((0.1)^(2))=8xx10^(-8)mol//litre` |
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| 1806. |
Two buffer, (X) and (Y) of pH `4.0` and `6.0` respectively are prepared from acid HA and the salt NaA. Both the buffers are `0.50` M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers ? `(K_(HA)=1.0xx10^(-5))` |
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Answer» pH of buffer is given by: `pH= -log K_(a)+log ((["Salt"])/(["Acid"]))` For I: `4= - log 1.0xx10^(-5)+log ((["Salt"])/([0.5]))` `:. Log ((["Salt"])/(0.5))= -1` or `["Salt"]= 0.1xx0.5=0.05M` For II: `6= - log 1.0xx10^(-5)+log ((["Salt"])/(0.5))` `:. log ((["Salt"])/(0.5))=1` `:. ["Salt"]= 10xx0.5=5M` Now the two buffer `[(I NaA = 0.05M and HA = 0.5M)` and II `(NaA= 5M and HA= 0.5M)]` are mixed in equal proportion. Thus, new conc. of NaA is mixed buffer `=(0.05xxV+5xxV)/(2V)=(5.05)/(2)` New conc. of HA in mixed buffer `=(0.5xxN+0.5xxV)/(2V)=0.5M` Thus, `pH= -log 1.0xx10^(-5)+log (([5.05//2])/([0.5]))` `pH=5+0.7033= 5.7033` |
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| 1807. |
`100mL` of a buffer solution contains `0.1M` each of weak acid `HA` and salt `NaA`. How many gram of `NaOH` should be added to the buffer so that it `pH` will be `6`? `(K_(a)` of `HA = 10^(-5))`.A. `0.328`B. `0458`C. `4.19`D. None |
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Answer» Correct Answer - A For acidic buffer, `pH = pK_(a) + "log" (0.1)/(0.1)` `pH = pK_(a) = - log (10^(-5)) = 5`. Rule: `ABC` (In acidic buffer `(A)`, on addition of `S_(B)(B)`, the concentration of `W_(A)(A)` decreases and that of salt increases. Let `x M` of `NaOH` is added. `pH_(new) = 5 + log((0.1 +x)/(0.1-x))` `6-5 = log ((0.1 +x)/(0.1 - x))` `((0.1 +x)/(0.1 - x)) = Antilog (1) = 10` Solve for `x:` `x = 0.082 M = (0.082)/(1000) xx 100` `= 0.0082 mol (100 mL)^(-1)` `= 0.0082 xx 40 g(100 mL)^(-1)` `= 0.328 g` |
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| 1808. |
Two buffer, (X) and (Y) of pH `4.0` and `6.0` respectively are prepared from acid HA and the salt NaA. Both the buffers are `0.50` M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers ? `(K_(HA)=1.0xx10^(-5))`A. `4.7033`B. `5.7033`C. `6.7033`D. `8.7033` |
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Answer» Correct Answer - B `pH` of buffer is given by : `pH = - "log" K_(a) + "log" (["Salt"])/(["Acid"])` Case I : `4 = - "log" 1.0 xx 10^(-5) + "log" (["Salt"])/((0.5))` or [Salt] `= 0.1 xx 0.5 = 0.05 M` Case II : `6 = - "log" 1.0 xx 10^(-5) + "log" (["Salt"])/((0.5))` `"log" (["Salt"])/((0.05)) = 1` `["Salt"] = 10 xx 0.5 = 5 M` Now the two buffer `[(I.NaA = 0.05 M` and `HA = 0.5 M`) and (II. `NaA = 5M` and `HA = 0.5 M`)] are mixed in equal proportion. Thus, new conc. of NaA in mixed buffer `= (0.5 xx V + 5 xx V)/(2V) = 0.5 M` Thus, `pH = - log(1.0 xx 10^(-3)) + "log" ([0.05//2])/([0.5])` `pH = 5 + 0.7033 = 5.7033` |
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| 1809. |
A buffer solution is made by mixing a weak acid HA `(K_(a) =10^(-6))` with its salt NaA in equal amounts. What should be amount of acid or salt that should be added to make 90 mL of buffer solution in which if 0.1 mole of strong acid are added into 1 L of this buffer solution then change in pH is unity ?A. 10 m molesB. 22 m molesC. 9 m molesD. 11 m moles |
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Answer» Correct Answer - d |
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| 1810. |
The `P^(H)` of a buffer solution is `5`. It consists of a weak acied `HA` and its conjugate base `A^(-)`. `P^(K_(a))` of `HA` is `4.699`. Which of the following is correct combination of solutions present in the buffer solution?A. `100 mL` of `1.1 M HA` and `200 mL` of `0.1 MA^(-)`B. `200 mL` of `0.1M HA` and `100 mL` of `0.1 MA^(-)`C. `100 mL` of `0.1M HA` and `200 mL` of `0.2 MA^(-)`D. `100 mL` of `0.2M HA` and `200 mL` of `0.1 MA^(-)` |
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Answer» Correct Answer - A for an acidic buffer `:. P^(H) = P^(ka) + "log" (["salt"])/(["acid"])` `5 = 4.699 + "log" (["salt"])/(["acid"])` `0.310 = "log"(["salt"])/(["acid"])` `(["salt"])/(["acid"]) = 2` `(200 xx 0.1)/(100 xx 0.1) = 2` |
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| 1811. |
A basic buffer contains `0.8` mole of `NH_(4)Cl` and `0.2` mole of `NH_(4)OH` for litre of a solution the `K_(b)` of base is `1.8 xx 10^(-5)`. Then the `P^(H)` of the buffer solution is `(log 1.8 = 0.2553)`A. `8.6532`B. `6.345`C. `2.301`D. `7.635` |
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Answer» Correct Answer - A `P^(H) = 14 - P^(k_(b)) - "log" (["salt"])/(["Base"])` |
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| 1812. |
The hydrolysis constant of `NaX` (`K_(a)` of HX is `2 xx 10^(-6)`) isA. `5 xx 10^(-9)`B. `2 xx 10^(-8)`C. `5 xx 10^(-6)`D. `10^(-7)` |
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Answer» Correct Answer - A `K_(h) = (kw)/(K_(a)) = (10^(-14))/(2 xx 10^(-6))` |
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| 1813. |
Among the following a) On dilution, the `P^(H)` of an acid increases b) A solutions with `P^(H) = 61000` times more basic than a solution with `P^(H) = 3` c) A Solution with `p^(H) = 9` is `1000` times more acidic than a solution with `P^(H) = 12` d) The `P^(H)` of `10^(9) M NaOH` is slightly greater tha n 7A. a,b are correct onlyB. a.d are ony correctC. a,b,c are only correctD. All are correct |
| Answer» Correct Answer - D | |
| 1814. |
The aqueous solution of potash alum is acidic due to hydrolysis ofA. `K^(+)`B. `Al^(3+)`C. `SO_(4)^(2-)`D. `Na^(+)` |
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Answer» Correct Answer - B `Al^(3+)` is a Cation of weak base |
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| 1815. |
`pK_(a)` of a weak acid is defined asA. `log_(10)K_(a)`B. `(1)/(log_(10) K_(a))`C. `log_(10). (1)/(K_(a))`D. `-log_(10).(1)/(K_(a))` |
| Answer» Correct Answer - C | |
| 1816. |
In the reaction `AlCl_(3) + overset(-)(C)l rarr AlCl_(4)^(-), AlCl_(3)` isA. Lewis acidB. Lewis baseC. Lewis saltD. Arrhenius acid |
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Answer» Correct Answer - A Lone pair donor is `L.A.` |
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| 1817. |
Water is aA. Amphoteric acidB. Aprotic acidC. Protophobic solventD. None of these |
| Answer» Correct Answer - D | |
| 1818. |
The aqueous solution of `AlCl_(3)` is acidic due to the hydrolysis ofA. Aluminium ionB. Chloride ionC. Both aluminium and chloride ionD. None of these |
| Answer» Correct Answer - A | |
| 1819. |
Which of the following ions or compounds in a solutions tends to produe an acidic, a basic, or a neutral solution. a. i. `C_(2)H_(3)O_(2)^(Θ)` ii. `Na^(o+)` iii. `So_(3)^(2-)` iv. `F^(Θ)` v. `NH_(4)^(o+)` b. i. `CH_(3)COONa` ii. `ZnCI_(2)` iii. `KNO_(3)` iv. `NH_(4)CI` c. i. `NaCN` ii. `K_(2)CO_(3)` iii. `H_(3)PO_(4)` iv. `NaF` d. i. `NH_(4)NO_(3)` ii. `Ba_(2)CO_(3)` iii. `NaHSO_(4)` iv. `NaOCI` v. `HOCI` vi. `AI(NO_(3))_(3)` |
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Answer» a. (i), (iii), and (iv) are basic (since each is conjegate base of a weak acid) (ii) Neutral (it is the conjugate acid of a strong base) (v) Acidic (it is the conjugate base of a weak base) b. (i) Basic (salt of `W_(A)//S_(B))` (ii) and (iv) are Acidic (salt of `W_(B)//W_(A))` (iii) neutral (salt of `S_(A)//W_(B))` c. (i),(ii), and (iv) are basic (salt of `S_(B)//W_(A))` (iii) is acidic (it is a weak acid) d. (i),(iii), `(v)` and (vi) are acidic (expect (v), all are salt of `W_(B)//S_(A))` But (v) is a weak acid. (ii) and (iv) are basic (salt of `S_(B)//W_(A))` |
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| 1820. |
The different colours of litmus in acidic, neutral and basic solutions are, respectively.A. Red, orange and blueB. Blue, violet and redC. Red, colourless and blueD. Red, violet and blue |
| Answer» Correct Answer - D | |
| 1821. |
Calculate the `pH` of a solution which contains `100mL` of `0.1 M HC1` and `9.9 mL` of `1.0 M NaOH`. |
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Answer» Correct Answer - C::D `{:(,HC1+,NaOHrarr,NaC1+,H_(2)O),("mEq before reaction",100xx0.1,9.9xx1,,),(,=0,=9.9,,),("mEq after reaction",0.1,0,9.9,9.9):}` `:. [H^(o+)]` left from `HC1 =(0.01)/(109.9) = 9.099 xx 10^(-4)M` `:. pH =- log [H^(o+)] =- log (9.099 xx 10^(-4))` `pH = 3.0409` |
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| 1822. |
`pH` of a solution made by mixing `200mL` of `0.0657M NaOH, 140 mL` of `0.107M HC1` and `160mL` of `H_(2)O` isA. `3.04`B. `2.43`C. `2.74`D. `2.27` |
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Answer» Correct Answer - B mmol of `NaOH = 200 xx 0.0657 = 13.14` mmol of `HC1 = 140 xx 0.107 = 14.98` mmol of `NaOH` left `= 14.98 - 13.14 = 1.84` Total volume `= (200 + 140 + 160) mL = 500 mL` `[H^(o+)] = (1.84 m mol)/(500 mL) = 0.00368 = 368 xx 10^(-5)` `pH =- log (368 xx 10^(-5)) =- 2.5658 +5` `= 2.4343 ~~ 2.43` `pH = 2.43`. |
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| 1823. |
The value of ionic product of water at 393K isA. less than `1 xx 10^(-14)`B. greater than `1 xx 10^(-14)`C. equal to `1 xx 10^(-14)`D. equal to `1 xx 10^(-7)` |
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Answer» Correct Answer - B Ionic product of water increases with increase of temperature. `:. `At 393, `K_(w) gt 1 xx 10^(-14)`. |
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| 1824. |
`pH` of an aqueous solution of `0.6M NH_(3)` and `0.4M NH_(4)CI` is `9.4 (pK_(b) = 4.74)`. The new `pH` when `0.1M Ca(OH)_(2)` solution is added to it.A. `9.86`B. `10.14`C. `10.2`D. `10.86` |
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Answer» Correct Answer - A Since `Ca(OH)_(2)` is completely ionised. `{:(,Ca(OH)_(2)rarr,Ca^(2+)+,2overset(Theta)OH,),("Initla",0.1,0,0,),("Final",c,0,2xx0.1=0.1M,):}` BBB Rule: On adding base, to the basic buffer, concentration of base increases and salt decreases. `:.` New concentration of base and salt are: `["Base"] = [NH_(3)] = 0.6 xx 0.2 = 0.8M` `["Salt"] = [NH_(4)CI] = 0.4 - 0.2 = 0.2M` `:. pOH = pK_(b) + log [("Salt")/("Base")]` `pOH = 4.74 + log ((0.2)/(0.8))` `= 4.74 - 2 log2 = 4.74 - 2 xx 0.30 = 4.14` `pH = 14- 4.14 = 9.86` |
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| 1825. |
Both HCOOH and `CH_(3)COOH` solutions have equal pH. If `K_(1) // K_(2)` ( ration of acid ionisation constants ) of these acids is 4, their molar concentration ratio will beA. 2B. 0.5C. 0.05D. 0.25 |
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Answer» Correct Answer - D As pH of HCOOH sol. `= pH` of `CH_(3)COOH` sol. `:. [H^(+)] ` in HCOOH sol. `=[H^(+)]` in `CH_(3)COOH` sol. `K_(1)=([HCOO^(-)][H^(+)])/([HOOH])=([H^(+)]^(2))/([HCOOH])` `K_(2)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])=([H^(+)]^(2))/([CH_(3)COOH])` `K_(1)[HCOOH]=K_(2)[CH_(3)COOH]` `([HCOOH])/([CH_(3)COOH])=(K_(2))/(K_(1))=(1)/(4)=0.25` |
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| 1826. |
What is the concentration of `CH_(3)COOH` which can be added to `0.5M HCOOH` solution so that dissociation of both is same. `K_(CH_(3)COOH) = 1.8 xx 10^(-5), K_(HCOOH) = 2.4 xx 10^(-4)` |
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Answer» `HCOOH hArr H^(o+) + HCOO^(Theta)` `CH_(3)COOH hArr H^(o+) + CH_(3)COO^(Theta)` Since both are present in a mixture and therefore, `[HCOO^(Theta)]` by `HCOOH = [CH_(3)COO^(Theta)]` by `CH_(3)COOH` `C_(1)alpha_(1) = C_(2)alpha_(2)` `sqrt((K_(a_(1))C_(1))) = sqrt((K_(a_(2))C_(2)))` `:. sqrt((2.4 xx 10^(-4) xx 0.5)) = sqrt((1.8 xx 10^(-5)xxC))` `:. C_(CH_(3)COOH) = 6.66M` |
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| 1827. |
The `pK_(a)` of a weak acid (HA) is 4.5 the pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized isA. 4.5B. 2.5C. 9.5D. `7.0` |
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Answer» Correct Answer - C For buffer solution `pH = pK_(a) + log.(["Salt"])/(["Acid"]) = 4.5 + log.(["Salt"])/(["Acid"])` as HA is 50% ionized so [Salt] = [Acid] pH = 4.5 pH + pOH = 14 `rArr pOH = 14 - 4.5 = 9.5`. |
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| 1828. |
`HCOOH` and `CH_(3)COOH` solutions have equal `pH.` If `K_(1)//K_(2)` is `4`, the ratio of their molar concentration will beA. `0.25`B. `0.5`C. `2`D. `4` |
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Answer» Correct Answer - A Since `pH` is same for two acids. `:. [H^(o+)]_(1) = [H^(o+)]_(2)` Thus, `C_(1)alpha_(1) = C_(2)alpha_(2)` `(K_(a_(1)))/(K_(a_(2))) = (C_(1)alpha_(1)^(2))/(C_(1)alpha_(2)^(2)) = (alpha_(1))/(alpha_(2)) …(i)` `alpha_(1) = sqrt((K_(a_(1)))/(C_(1))), alpha_(2) = sqrt((K_(a_(2)))/(C_(2)))` `:. (alpha_(1))/(alpha_(2)) = sqrt((K_(a_(1))C_(2))/(K_(a_(2))C_(1)))` Substituting from equation (i) the value of `alpha_(1)//alpha_(2)` `(K_(a_(1)))/(K_(a_(2))) = sqrt((K_(a_(1))C_(2))/(K_(a_(2))C_(1)))` `:. (K_(a_(1)))/(K_(a_(2))) = (C_(2))/(C_(1))` `4 = (C_(2))/(C_(1))` `:. C_(1) : C_(2) = (1)/(4) = 0.25` |
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| 1829. |
The `pK_(a)` of a weak acid `(HA)` is `4.5`. The `pOH` of an aqueous buffered solution of `HA` in which `50%` of the acid is ionized is:A. `4.5`B. `2.5`C. `9.5`D. `7.0` |
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Answer» Correct Answer - C For buffer solution : `pH= pK_(a)+log (["Salt"])/(["Acid"])= 4.5+log (["Salt"])/(["Acid"])` As HA is `50%` ionised, so [Salt] = [Acid] `pH = 4.5` `pH+pOH=14` `:. pOH= 14-4.5 = 9.5` |
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| 1830. |
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^(-14)`B. `10^(-8)`C. `10^(-6)`D. `10^(-12)` |
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Answer» Correct Answer - D `P^(H)` depends on concentration but not volume |
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| 1831. |
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 `K_(w) = [H_(3)O^(o+)][overset(Theta)OH]` `=10^(-6.7) xx 10^(-6.7) = 10^(-13.4)` |
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| 1832. |
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 `K_(w) = [H^(o+)] [overset(Theta)OH] = 10^(-6) xx 10^(-6) = 10^(-12)` |
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| 1833. |
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^(-12)`C. `10^(-14)`D. `10^(-8)` |
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Answer» Correct Answer - B `[H_(3)O^(o+)] = 10^(-6) = [overset(Theta)OH]` Because in pure water, `[H_(3)O^(o+)] = [overset(Theta)OH]` `K_(w) = [H_(3)O^(o+)] = [overset(Theta)OH]` `= 10^(-6) xx 10^(-6)` `= 10^(-12)` |
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| 1834. |
When equal volumes of the following solutions are mixed, precipitation of `AgCI (K_(sp) = 1.8 xx 10^(-10))` will occur only wityA. `10^(-4)M (Ag^(o+))` and `10^(-4)M (CI^(Theta))`B. `10^(-5)M (Ag^(o+))` and `10^(-5)M (CI^(Theta))`C. `10^(-5)M(Ag^(o+))` and `10^(-6)M(CI^(Theta))`D. `10^(-4)M (Ag^(o+))` and `10^(-10)M (CI^(Theta))` |
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Answer» Correct Answer - A Since equal volumes are added a. Ionic product `=(10^(-4))/(2) xx (10^(-4))/(2) = (10^(-8))/(4)` b. Ionic product `= (10^(-5))/(2) xx (10^(-5))/(2) = (10^(-10))/(4)` c. Ionic produt `= (10^(-5))/(2) xx (10^(-6))/(2) = (10^(-11))/(4)` d. Ionic product `= (10^(-10))/(2) xx (10^(-10))/(2) = (10^(-20))/(4)` Only in (a) Ionic product `((10^(-8))/(4)) gt K_(sp) (1.8 xx 10^(-10))` So precipitate will take place only with (a) |
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| 1835. |
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 strength?A. `ivltiiltiiilti`B. `iiltiiiltiltiv`C. `iltiiiltiiltiv`D. `iiiltiltivltii` |
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Answer» Correct Answer - C The `pK_(a)` values of `HCO_(3)^(-), H_(3)O^(+)` and `HSO_(4)^(-)` are `10.25, -1.74` and `1.92` respectively and `HSO_(3)F` is super acid. `(iv)gt(ii)gt(iii)gt(i)` |
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| 1836. |
Equal volumes of the following `Ca^(2+)` and `F^(-)` solutions are mixed. In which of the solutions will precipitations occur ? `[K_(sp)` of `CaF_(2)=1.7 xx 10^(-10)]` 1. `10^(-2) M Ca^(2+)+10^(-5)MF^(-)` 2. `10^(-3) M Ca^(2)+10^(-3)MF^(-)` 3. `10^(-4) MCa^(2)+10^(-2)MF^(-)` 4. `10^(-2) MCa^(2)+10^(-3)MF^(-)` Select the correct answer using the codes given belowA. In 4 onlyB. In 1 and 2C. In 3 and 4D. In 2,3 and 4 |
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Answer» Correct Answer - C In case I `[Ca^(2+)]` in reaction mixture `=0.5 xx 10^(-2)M` `[F^(-)]` in reaction mixture `=0.5 xx 10^(-5)M` `:. ` Ionic product `=[Ca^(2+)][F^(-)]^(2)=(0.5 xx 10^(-2))xx (0.5 xx10^(-5))^(2)` `=125 xx 10^(-15)=1.25 xx 10^(-13)xx1.7 xx 10^(-10)` `:. `No. ppt is formed in this case In case 2 `[Ca^(2+)]` in reaction mixture `=0.5 xx 10^(-3)M` `[F^(-)]` in reaction mixture `=0.5 xx 10^(-3)M` `:. `Ionic product `=(0.5 xx 10^(-3))(0.5 xx 10^(-3))^(2)` `=125 xx 10^(-12)=1.25 xx 10^(-10)lt 1.7 xx 10^(-10)` No. ppt. is formed in this case In case 3 `[Ca^(2+)]` in reaction mixture `=0.5 xx 10^(-4)M` `[F^(-)]` in reaction mixture `=0.5 xx 10^(-2)M` `:. ` Ionic product `=[Ca^(2+)]xx[F^(-)]^(2)` `=[0.5 xx 10^(-4)] xx [0.5 xx 10^(-2)]^(2)` `=125 xx 10^(-11)` `=1.25 xx 10^(-9) gt 1.7 xx 10^(-10)` Thus a ppt. of `CaF_(2)` will be formed In case 4 `[Ca^(2+)]` in reaction mixture `=0.5 xx 10^(-2)M` `[F^(-)]` in reaction mixture `=0.5 xx 10^(-3)M` Ionic product `[Ca^(2+)][F^(-)]^(2)` `=[0.5 xx 10^(-2)][0.5 xx 10^(-3)]^(2)=125 xx 10^(-11)` `=1.25 xx 10^(-9) gt 1.7 xx 10^(-10)` Thus a ppt of `CaF_(2)` will be formed. |
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| 1837. |
For which of the following equilibria does decrease in pressure not favour the forward reaction ?A. `CaCO_(3)(s) hArr CaO(s) +CO_(2)(g)`B. `CO(g)+2H_(2)(g) hArr CH_(3)OH (g)`C. `NH_(4)Cl(s) hArr NH_(3)(g)+HCl(g)`D. `2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g)` |
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Answer» Correct Answer - B Decrease in pressure shifts equilibrium in a direction in which number of moles decrease. |
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| 1838. |
When solid silver chloride (MM=143.4) is added to 100mL of `H_(2)O,1.9xx10^(-4)` grams dissolves. What is the `K_(sp)` for silver chloride?A. `1.3xx10^(-5)`B. `3.7xx10^(-6)`C. `3.7xx10^(-8)`D. `1.8xx10^(-10)` |
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Answer» Correct Answer - d |
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| 1839. |
Calculate pH of a saturated solution of `Mg(OH)_(2)`. `(K_(SP)for Mg(OH)_(2)=8.9xx10^(-12))` |
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Answer» Let solubility of `Mg(OH)_(2)` be S mol `litre ^(-1)` `Mg(OH)_(2)hArr underset(S)(Mg^(2+))+underset(2S)(2OH^(-))` `:. [Mg^(2+)][OH]^(2)=K_(SP)` `4S^(3)= 8.9xx10^(-12)` `:. S = 1.305xx10^(-4)"mol liltre"^(-1)` `:. [OH^(-)]=2xx1.305xx10^(-4)"mol litre"^(-1)` `:. pOH=3.5832` `pH = 10.4168` |
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| 1840. |
A saturated solution of a salt MX exhibits an osmotic pressure of `74.4mm` Hg at `25^(@)C`. Assuming `100%` ionisation of MX, calculate `K_(SP)` of MX. |
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Answer» `piV=nST(1+alpha)` `pi=ST(1+alpha)` `(74.4)/(760)=Cxx0.0821xx298xx2` , `(.: alpha=1)` `:. C= 2.0xx10^(-3)"mol litre"^(-1)` `:. K_(SP)of MX= C^(2)=4xx10^(-6)` |
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| 1841. |
Which of the following has maximum pH ?A. 1 N `CH_(3)COOH`B. 1N HClC. 1 N `H_(2)SO_(4)`D. 1 N `HNO_(3)` |
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Answer» Correct Answer - A `CH_(3)COOH` is the weakest acid out of these and hence it has maximum pH value. |
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| 1842. |
20 ml of 0.5 N HCl and 35 ml of 0.1 NaOH are mixed. The resulting solution willA. Be neutralB. Be basicC. Turn phenolphthalein solution pinkD. Turn methyl orange red |
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Answer» Correct Answer - D (i) 20 ml of 0.5 N HCl `0.5 N rArr 1000 m` 0.5 mole HCl is present in 20 ml `= (20 xx 0.5)/(1000) = 1.0 xx 10^(-2)` (ii) 35 ml of 0.1 NaOH `0.1 N rArr 1000 ml` of 0.1 mole NaOH is 35 ml `= (35 xx 0.1)/(1000) = 0.35 xx 10^(-2)` `rArr (1.0 - 0.35) 10^(-2) = 0.65 xx 10^(-2)` mole HCl `HCl = H^(+) + Cl^(-) rArr [HCl] = [H^(+)] + [Cl^(-)]` 55 ml contains `0.65 xx 10^(-2)` mole of `H^(+)` ions `1000 ml - (0.65 xx 10^(-2) xx 10^(3))/(55) = (6.5)/(55)` `pH = -log [H^(+)] = -log (6.5//55)` `= log 55 - log 6.5 = 9.2` Due to acidic nature of solutions the colour of methylorange becomes red. |
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| 1843. |
`50ml N//10 NaOH` solution is mixed with `50 ml N//20 HCl` solution. The resulting solution will `(a)` Turns phenolphthalein solution pink`" " (b)` Turns blue litmus red `(c )` Turns methyl orange red `" "(d) [H^(+)] & [OH^(-)]`A. only `(a)` is correctB. `(a) & (b)` are correctC. `(a) & (d)` are correctD. all are correct |
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Answer» Correct Answer - C Since solution is basic in nature hence, `[H^(+)]lt[OH^(-)]` and phenolphathalein will turn the solution pink, Ans `(a) & (d)`. |
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| 1844. |
50 mL of 0.1 M HCl and 50 mL of 2.0 M NaOH are mixed. The pH of the resulting solution isA. `1.30`B. `4.2`C. `12.70`D. `11.70` |
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Answer» Correct Answer - C Mollimoles of HCl `=0.1 xx 50 =5 m mol` Millimoles of NaOH `=0.2 xx 50 = 10 m mol` `:. `Resulting solution contains 5 m mole of NaOH IN `(50+50)100` solution `:. [NaOH ]=(5 xx 10^(-3))/(100xx10^(-2))=5 xx 10^(-2)M` `[OH^(-)]=5 xx 10^(-2)M` `[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(5 xx 10^(-3))=(1)/(5) xx 10^(-12)` `=(10)/(5)xx 10^(-13) = 2 xx 10^(-13)` `pH =-log [H^(+)]=- log (2 xx 10^(-13))` `=13-log 2=13-0.3010=12.6990` |
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| 1845. |
The number of `H^(o+)` ions present in `1mL` of solution having `pH = 13` isA. `6.023 xx 10^(10)`B. `6.023 xx 10^(7)`C. `6.023 xx 10^(13)`D. `10^(13)` |
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Answer» Correct Answer - B `pH = 13` `[H^(o+)] = 10^(-13)M = 10^(-13) mol (1000 mL)^(-1)` `= (10^(-13))/(1000) = 10^(-16) mol mL^(-1)` Number of `[H^(o+)] = 10^(-16) xx 6.023 xx 10^(23)` `= 6.023 xx 10^(7) ions`. |
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| 1846. |
Three suggestion are made for ways to removes silver ions from solution: a. Make the solution `0.01M`in `Nal`. b. Buffer the solution at `pH = 13`. c. Make the solution `0.01M Na_(2)S`. What will be the equilibrium silver ion concentartion in each case? which course of action is most effective in removing `Ag^(o+)` ions? `K_(sp) (AgI) = 8.5 xx 10^(-17), K_(sp) (AgOH) = 2 xx 10^(-8)`, `K_(sp) (Ag_(2)S) = 5.5 xx 10^(-51)` |
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Answer» `Ag1 hArr Ag^(o+) + I^(Θ)`, `AgOH hArr Ag^(o+) + overset(Θ)OH` `Ag_(2)S hArr 2Ag^(o+) + S^(2-)` a. `[Ag^(o+)] = (8.5 xx 10^(-17))/(10^(-2)) = 8.5 xx 10^(-15) M` b. `pH = 13 :. [H^(o+)] = 10^(-13), [overset(Θ)OH] = 10^(-1)` `[Ag^(o+)] = (2xx10^(-8))/(10^(-1)) = 2xx 10^(-7)M` c. `[Ag^(o+)] = sqrt((5.5 xx 10^(-51))/(10^(-3))) = 2.345 xx 10^(-24) M` |
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| 1847. |
Calculate pH at which an acid indicator Hin with concentration `0.1M` changes its colour (`K_(a)` for `"Hin" = 1 xx 10^(-5)`) |
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Answer» Correct Answer - 5 Indicator equation is : `pH = pK_(In) + "log"([In^(-)])/([Hln])` Colour change takes place when `[In^(-)] = [HIn]`i.e.,when `pH = pK_(In) = 5` |
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| 1848. |
When 20 mL `M//20` NaOH is added to 10 ml of `M//10` HCl. The resulting has pHA. `lt7`B. `=7`C. `gt7`D. `~~2` |
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Answer» Correct Answer - B Millimoles of HCl `=(1)/(20)xx 10 =1` Millimoles of NaOH `=(1)/(10)xx10=1` Since millimoles are equal, the resulting solution is neutral and has pH `=7` |
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| 1849. |
The number of `H^(+)` ions in `1 cc` of a solution of `pH = 13`A. `6.023 xx 10^(7)`B. `1 xx 10^(-13)`C. `6.023 xx 10^(13)`D. `1 xx 10^(16)` |
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Answer» Correct Answer - A `[H^(+)] = 10^(-13)` `10^(-13) = nxx (1000)/(1)` Number of moles `= (10^(-13))/(10^(3)) = 10^(-16)` Number of `H^(+) = 10^(-16) xx 6.023 xx 10^(23)` `6.023 xx 10^(7)` |
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| 1850. |
50 mL of `H_(2)O` is added to `50 mL` of `1 xx 10^(-3)M` barium hydroxide solution. What is the pH of the resulting solution?A. 3B. 3.3C. 11D. 11.7 |
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Answer» Correct Answer - C `V_(1)N_(1) = V_(2)N_(2)` `100 xx N = 50 xx 10^(-3) 2` `N = 10^(-3)` `pOH = 3, pH = 11` |
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