1869 + Interview Questions in IONIC-EQUILIBRIUM IN GENERAL AWARENESS Page 19 InterviewSolution

901.	0.02 M monobasic acid dissociates 2% hence, pH of the solution isA. 0.3979B. 1.3979C. 1.699D. 3.3979
Answer» Correct Answer - D `[H^(+)] = alpha.C = (2)/(100) xx 0.2, [H^(+)] = 4 xx 10^(-4)M` `pH = -log [H^(+)] = 4 -log 4 , pH = 33979`.

Discussion

902.	pH of a solution is 9.5. The solution isA. NeutralB. AcidicC. BasicD. Amphoteric
Answer» Correct Answer - C When pH = 7 means neutral, `pH lt 7` means acidic, `pH gt 7` means basic.

Discussion

903.	Which on reaction with water will have pH less than 7A. BaOB. CaOC. `Na_(2)O`D. `P_(2)O_(5)`
Answer» Correct Answer - D `BaO, CaO` and `Na_(2)O` are shows more than 7 pH because of their basic nature.

Discussion

904.	Which of the following will decrease the pH of a 50 ml solution of `0.01 M HCI` ?A. Addition of 50 mL of 0.01 M HClB. Addition of 50 mL of 0.02 M HClC. Addition of 150 mL of 0.02 M HClD. Addition of 5 mL of 1 M HCl
Answer» Correct Answer - D For 0.01 M HCl, `\|H^(+)\| =10^(-2), pH=2` `(A) N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))` `50 xx 0.01+50 xx 0.01 =N_(3)(50+50)` or `N_(3)=(.5 +.5)/(100)=10^(-2),pH=2` (No change) `(B) 50 xx 0.01 +c0 xx 0.002=N_(3)(50+50)` or `N_(3) =(0.5 +0.1 )/(100)=6 xx 10^(-3)` `pH= - log (6 xx 10^(-3))=2.22 (`pH increases) `(C ) 50 xx 0.01 +150 xx 0.002 = N_(3)(50+150)` or `N_(3)=(.5 +.3)/(200)=4 xx 10^(-3)` `pH =-log (4 xx 10^(-3))=2.39 pH (pH ` increases ) `(D) 50 xx 0.01 +5 xx 1 =N_(3)(50+5)` or `N_(3)=(0.5 +5)/(55)=(5.5)/(5)=10^(-1)` `pH =1 (`pH decreases )

Discussion

905.	Which would be decrease the pH of `25 cm^(3)` of a 0.01 M solution of hydrochloric acidA. The addition of `25cm^(3)` 0.005 M hydrochloric acidB. The addition of `25cm^(3)` of 0.02 M hydrochloric acidC. The addition of magnesium metalD. None of these
Answer» Correct Answer - B Given concentration of HCl = 0.01 M `pH = -log 10^(-2) = 2` In option (b) Concentration of HCl = 0.02 M M.eq `= 0.02 xx 25 = .5 M`. Eq of HCl M.eq in given condition `= 0.01 xx 25 = 0.25` Total M.eq in solution `= 0.25 + 0.50 = 0.75 = (.75)/(25+25) = 1.5 xx 10^(-2)N` Then new pH will be `= -log 1.5 xx 10^(-2) = 2 - log 1.5` Which is less than 2 ltbr. So we canvery that that by adding `25cm^(3)` of 0.2 M HCl the pH of `25 cm^(3)` of 0.01 M decrease.

Discussion

906.	Which one of the following is not a buffer solutionA. `0.8 M H_(2)S + 0.8 M KHS`B. `2M C_(6)H_(5)NH_(2) + 2M C_(6)H_(5)overset(+)(N)H_(3)Br`C. `3M H_(2)CO_(3) + 3M KHCO_(3)`D. `0.05 M KClO_(4) + 0.05 M HClO_(4)`
Answer» Correct Answer - D

Discussion

907.	The mixture of `10 ml 0.5 N CH_(3)COOH` and `10 ml 0.25 M NaOH`, having `pH = 5`, then find the value of pKa?
Answer» Correct Answer - 5

Discussion

908.	Which can act as bufferA. `NH_(4)Cl + HCl`B. `CH_(3)COOH + H_(2)CO_(3)`C. `40mL` of `0.1M NaCN + 20mL` of `0.1M HCN`D. `NaCl + NaOH`
Answer» Correct Answer - C Acidic buffer is a weak acid and its salt of strong base

Discussion

909.	Assertion (A): The `pH` of an aqueous solution of `CH_(3)COOH` remains unchanged on the addition of `CH_(3)COONa`. Reason (R) : The `pH` of an aqueous solution of `CH_(3)COOH` remains unchanged on the addition of `CH_(3)COONa`. Reason (R) : The ionisation of `CH_(3)COOH` is supressed by the addition of `CH_(3)COONa`.A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct.
Answer» Correct Answer - D `(A)` is wrong. It forms acidic buffer and this `pH` changes. `(R)` is correct. Due to common ion `(CH_(3)COO^(Theta))`, ionisation of `CH_(3)COOH` is supressed.

Discussion

910.	When `CH_(3)COONa` is added to an aqueous solution of `CH_(3)COOH`A. The pH of solution decreasesB. The pH of solution increasesC. The pH of the solution remains unalteredD. An acid salt is produced
Answer» Correct Answer - B `P^(H) = P^(Ka) + "log" (["salt"])/(["acid"])`

Discussion

911.	The `[H^(+)]` in `0.2M` solution of formic acid is `6.4xx10^(-3)` mol `litre^(-1)`. To this solution formate is added so as to adjusrt the conc.of sodium formate to one mol per litre. What will be pH of this solution ? `(K_(a) for HCOOH=2.4xx10^(-4))` and degree of dissociation of `HCOONa = 0.75`)
Answer» In `0.2M HCOOH, [H^(+)]= 6.4xx10^(-3)` `Calpha= 6.4xx10^(-3)` `:. alpha= 3.2xx10^(-2)` Now sodium formate is added and the dissociation will fiurther be suppressed and therefore, new degree of dissociation `(alpha_(1))` for HCOOH in presence of HCOONa is so small that it may be neglected. `:. [HCOOH]` after dissociation = [HCOOH] before dissociation `[HCOOH]= 0.2` `{:(,HCOONahArr,HCOO^(-)+,Na^(+)),("Conc. before dissociation", 1,0,0),("Conc. after dissociartion", (1-0.75),0.75,0.75):}` So `[HCOOH]= 0.2` `:. [HCOO^(-)]= 0.75` `pH= -log 2.4xx10^(-4)+log((0.75)/(0.2))=4.19`

Discussion

912.	The degree of hydrolysis of a salt of `W_(A)` and `W_(B)` in its `0.1M` solution is `50%`. If the molarity of the solution is `0.2M`, the percentage hydrolysis of the salt woukd beA. `25%`B. `50%`C. `75%`D. `100%`
Answer» Correct Answer - B For the salt of `W_(A)//W_(B)`, the degree of hydrolysis is independent of concentration.

Discussion

913.	The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of 3% in its 0.1 M solution at `25^@C`.Given that the ionic product of water is `10^(-14)` at this temperature , what is the dissociation constant of the acid ?A. `~~1xx10^(-10)`B. `~~1xx10^(-9)`C. `3.33xx10^(-9)`D. `3.33xx10^(-10)`
Answer» `h=0.3 C=0.1M` `K_(h)=CH^(2)=9xx10^(-5)` `K_(h)=(K_(w))/(K_(a))=9xx10^(-5)rArr K_(a)=(10^(-14))/(9xx10^(-5))=1.11xx10^(-10)~~1xx10^(-10)`

Discussion

914.	The degree of hydrrolysis of a salt of weak acid and weak base in its `0.1M` solultion is found to be `50%`. If the molarity of the solution is `0.2M`, the percentage hydrolysis of the salt should be:A. `100%`B. `50%`C. `25%`D. None of these
Answer» the degree of hydrolysis of a salt of weak acid and weak base is independent of concentration of salt solution.

Discussion

915.	The pH of a solution is 2. If its pH is to be raised to 4, then the `[H^(+)]` of the original solution has to beA. DoubledB. HalvedC. Increased hundred timesD. Decreased hundred times
Answer» Correct Answer - D If pH of any solution is 2. Then `[H^(+)] = 10^(-2) M`. If pH of any solution is just double then pH = 4 and `[H^(+)]` will be `10^(-4)`.

Discussion

916.	Assuming `100%` ionization in case of strong electrolytes which of the following will have minimum pHA. 0.1 M `NH_(4)Cl`B. 0.1 M `(NH_94))_(2)SO_(4)`C. 0.1M `(NH_(4))_(3)PO_(4)`D. All will have same pH.
Answer» Correct Answer - B Minimum pH means maximum `[H^(+)]` `NH_(4)Cl rarr NH_(4)^(+) +Cl^(-)` `NH_(4)^(+) +H_(2)O hArr NH_(4)OH +H^(+)` `:. [H^(+)] =0.01 xx K_(h)` In case of `(NH_(4))_(2)SO_(4)` `[H^(+)]=0.02 xx K_(h)` However `(NH_(4))_(3)PO_(4)` will be almost neutral `(pH ~~7)` as it is a salt of a weak acid and a weak base. Thus pH will be minimum (i.e., highest `H^(+)` conc. ) in case of 0.01 M `(NH_(4))_(2)SO_(4)`

Discussion

917.	pH of one litre solution containing 40gm of `NaOH`A. 2B. 10C. 8D. 14
Answer» Correct Answer - D `[OH^(-)] = (40)/(40) = 1` `p^(OH) = 0` `p^(H) = 14`

Discussion

918.	During the titration of a weak base with a strong acid, one should use an acid-base indicator that changes colour in the :A. acid rangeB. basic rangeC. buffer rangeD. neutral range
Answer» Correct Answer - A

Discussion

919.	A solution of `pH = 9` is one thousand times as basic as a solution of pH.A. 4B. 7C. 10D. 6
Answer» Correct Answer - D For every `10` times dilution `p^(H)` value decreased by one unit

Discussion

920.	Arrhenius theory is failed to explain the acidic nature ofA. HClB. `HCOOH`C. `H_(2)SO_(4)`D. `CO_(2)`
Answer» Correct Answer - D Substance that gives `H^(+)` ions when dissolved in water

Discussion

921.	The pH of solution is `9`. It is `"………"` times more basic than a solution with `pH = 6`A. `3`B. `2`C. `4`D. `6`
Answer» Correct Answer - C For every ten times dilution `P^(H)` decreased by one unit

Discussion

922.	The `pH` of a solution increased from `3` to `6`. Its `[H^(o+)]` will beA. Reduced to halfB. DoubledC. Reduced by `1000` timesD. Increased by `1000` times
Answer» Correct Answer - C If pH is increased by `3` units then strength of the acids decreased by `1000` times

Discussion

923.	`HCl` gas cannot act as an acid inA. Polar solventsB. Highly polar solventsC. `NH_(3)` liquidD. Non-polar solvents.
Answer» Correct Answer - D In Non polar solvents `HCl` does not dissociate to give `H^(+)` ions

Discussion

924.	Which of the following is an Arrhenius acid?A. `CO_(2)`B. `SO_(2)`C. `FeCl_(2)`D. `HNO_(2)`
Answer» Correct Answer - D `HNO_(3)` produce `H^(+)` ions in water

Discussion

925.	`0.1 M HCl` solution is diluted by `100` times The pH of the solution formed isA. `3`B. `2`C. `4`D. `6`
Answer» Correct Answer - A `[H^(+)] = 0.1 xx (1)/(100)` `= 10^(-3)` `:. P^(H) = 3`

Discussion

926.	`100 ml` of a solution of `HCl` with `pH` value `3` is diluted with `400 ml` of water. The new `P^(H)` of the solution isA. `3.7`B. `5.3`C. `4.2`D. `5.6`
Answer» Correct Answer - A `100 xx 10^(-3) = 500 xx [H^(+)]`

Discussion

927.	`10^(-6)M HCl` is diluted to `100` times. Its `pH` is:A. 6B. 8C. 6.95D. 9.5
Answer» Correct Answer - c

Discussion

928.	The weakest base among the followingA. `I^(-)`B. `Cl^(-)`C. `F^(-)`D. `Br^(-)`
Answer» Correct Answer - A Strong acid conjugate base is the weakest base

Discussion

929.	Which of the following mixture is not a buffer solutionA. `100 mL` of `0.5 N CH_(3)COOH + 100 mL ` of `0.05 N NaOH`B. `100 mL` of `0.6 N HCN + 100 mL` of `0.4 NaOH`C. `100 mL` of `0.5N NH_(4)OH + 10 mL` of `0.2 NCH_(3)COOH`D. `100 mL` of `0.4 N HCL + 100 mL` of `0.4N NaOH`
Answer» Correct Answer - D A mixutre of storng acid and strong base does not acts as buffer

Discussion

930.	Which solutionwill have pH closer to `1.0` ?A. 100 mL of `(M//10) HCI + 100 mL` of `(M//10) NaOH`B. 55 mL of `(M//10) HCI +45 mL` of `(M//10) NaOH`C. 75 mL of `(M//10) HCI + 90` mL of `(M//10) NaOH`D. 75 mL of `(M//5) HCI +25 mL of `(M//5) NaOH`
Answer» Correct Answer - D

Discussion

931.	The following has no conjugate baseA. `H_(2)PO_(4)^(-)`B. `H_(2)PO_(2)^(-)`C. `H_(2)PO_(3)^(-)`D. `CH_(3)COOH`
Answer» Correct Answer - B `H_(2)PO_(2)^(-)` has no conjugate base `H-underset(H)underset(\|)overset(O)overset(\|\|)P-O^(-)`

Discussion

932.	Which solutionwill have pH closer to `1.0` ?A. `75 mL` of `M//5HC1 + 25 mL` of `M//5 NaOh`B. `10mL` of `M//10 HC1 + 90 mL` of `M//10 NaOH`C. `55mL` of `M//10 HC1` + `54mL` of `M//10 NaOH`D. `100 mL` of `M//10 HC1 + 100 mL` of `M//10 NaOH`
Answer» Correct Answer - A a.mmol of `HC1 = 75 xx (1)/(5) = 15` mmol of `NaOH = 25 xx (1)/(5) = 5` mmol of HC1 left `= 15 - 5 = 10` Volume `= 75 + 25 = 100` `[H^(o+)] = (10)/(100) = 0.1 = 10^(-1)M` `pH = 1` b. mmol of `HC1 = 10 xx (1)/(10) = 1` mmol of `NaOH = 90 xx (1)/(10) = 9` mmol of `NaOH` left `= 9 - 1 = 8` Since base is left, `:. pH gt 7` c. mmol of `HC1 = 55 xx (1)/(10) = 5.5` mmol of `NaOH = 45 xx (1)/(10) = 4.5` mmol of `HC1` left `= 5.5 - 4.5 = 1.0` Volume `= 55 + 45 = 100` `[H^(o+)] = (1)/(100) = 10^(-2)` `pH = 2` d. mmol of `HC1 = 100 xx (1)/(10) = 10` mmol of `NaOH = 100 xx (1)/(10) = 10` Both `S_(A)` and `S_(B)` are neutralised. `:. pH = 7`.

Discussion

933.	Which of the following solution will have `P^(H)` closer to `1.0`?A. `100 ml` of `M//10 HCl +` `45 ml` of `M//10 NaOH`B. `55 ml` of `M//10HCl +``45 ml` of `M//10 NaOH`C. `10 ml` of `M//10 HCl+` `90 ml` of `M//10 NaOH`D. `75ml` of `M//5 HCl+` `25 ml` of `M//5 NaOH`
Answer» Correct Answer - D `[H^(+)] = (NaVa - NbVb)/(Va + Vb)` (if `NaVa gt NbVb`) `P^(H) = -"log"^([H^(+)])` `[H^(+)] = 10^(-1), :. P^(H) = 1`

Discussion

934.	Weskest base isA. `C_(2)H^(-)`B. `NO_(3^(-))`C. `CN^(-)`D. `HS^(-)`
Answer» Correct Answer - B Strong acid conjugate base is the weakest base

Discussion

935.	Strongest Bronsted base isA. `ClO^(-)`B. `ClO_(2)^(-)`C. `ClO_(3)^(-)`D. `ClO_(4)^(-)`
Answer» Correct Answer - A Weak acid conjugate base is the strongest base

Discussion

936.	`K_(h)` of salt obtained from strong acid and weak base is `2 xx 10^(-5)`. The `K_(b)` of weak base isA. `2 xx 10^(-19)`B. `5 xx 10^(-10)`C. `2 xx 10^(-10)`D. `5 xx 10^(-9)`
Answer» Correct Answer - B `K_(h) = (K_(w))/(K_(b))`

Discussion

937.	Which of the following is not a salt?A. `NaCl`B. `Ca(OH)_(2)`C. `PbS`D. `Zn(NO_(3))_(2)`
Answer» Correct Answer - B `Ca(OH)_(2)` is a base

Discussion

938.	When `100 ML` of `1.0 M HCl` was mixed `100 MI` of `1.0 M NaOH` in an insulated beakertat constant pressure, a temperature incease of `5.7^(0)C` was measured for the beaker and its contents (Expt.1) Because th enthalpy of neutralization of a c `5.7^(0)C` strong acid with a strong base is constant `(-57.0 kJ moL- I)`, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt.2) , `100 ML` of `2.0 M` acetic acid `(K_(a) =2.0 xx 10^(-5))` was mixed with `100 ML` of `1.0 m NaOH` (under identical conditions to Expt. 1) Where a temperature rise of `5.0^(0)C` was measured. (Consider heat capcity of all solution as `4.2 Jg^(-1)` and density of all solution as `1.0 gmL^(-1)`) Enthalpy of dissociation (in `KJ Mol^(-1)`) of acetic acid obtained from Expt, `2` isA. `1.0`B. `10.0`C. `24.5`D. `51.4`
Answer» Correct Answer - A

Discussion

939.	When `100 ML` of `1.0 M HCl` was mixed `100 MI` of `1.0 M NaOH` in an insulated beakertat constant pressure, a temperature incease of `5.7^(0)C` was measured for the beaker and its contents (Expt.1) Because th enthalpy of neutralization of a c `5.7^(0)C` strong acid with a strong base is constant `(-57.0 kJ moL- I)`, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt.2) , `100 ML` of `2.0 M` acetic acid `(K_(a) =2.0 xx 10^(-5))` was mixed with `100 ML` of `1.0 m NaOH` (under identical conditions to Expt. 1) Where a temperature rise of `5.0^(0)C` was measured. (Consider heat capcity of all solution as `4.2 Jg^(-1)` and density of all solution as `1.0 gmL^(-1)`) The `p^(H)` of the solution after Expt.2 isA. `2.8`B. `4.7`C. `5.0`D. `7.0`
Answer» Correct Answer - B

Discussion

940.	Aqueous solution of which of the following shows lower `p^(H)`?A. `KNO_(3)`B. `ZnCl_(2)`C. `CH_(3)COONa`D. `CH_(3)COONH_(4)`
Answer» Correct Answer - B Cationic hydrolysis

Discussion

941.	What will the pH of the solution of the salt of weak acid and weak base? `(K_(b) = 1 xx 10^(-6)` and `K_(a) = 1 xx 10^(-4)`)
Answer» Correct Answer - 6 `pH = 1/2 (14 + pK_(a) - pK_(b))` `(1)/(2)(14+4-6) = 6`

Discussion

942.	The solubility of a salt of weak acid (AB) at pH 3 is `Y xx 10^(-3) mol L^(-1)`. The value of Y is ___________ (Given that the value of solubility product of `AB(K_(sp)) = 2 xx 10^(-10)` and the value of ionization constant of `HB(K_(a)) = 1 xx 10^(-8))`
Answer» Correct Answer - 4.47 `S = sqrt(K_(sp)(([H^(+)])/(K_(a))+1)) = sqrt(2 xx10^(-10)((10^(-3))/(10^(-8))+1)) = sqrt(2 xx 10^(-5))` `= 4.47 xx 10^(-3) M`

Discussion

943.	Assertion `(A):` Equivalent conductance increase with dilution for an electrolyte solution. Reason `(R):` The number of ions per litre of electrolyte increases with dilution.A. S is correct but E is wrong.B. S is wrong but E is correc.C. Both S and E are correct and E is correct explanation of S.D. Both S and E are correct but E is correct explanation of S.
Answer» Correct Answer - A It is a fact.

Discussion

944.	In which ratio of volumes 0.4 M HCl and 0.9 M HCl are to be mixed such that the concentration of the resultant solution becomes 0.7 M ?A. `4:9`B. `2:3`C. `3:2`D. `1:1`
Answer» Correct Answer - B `0.4 xx 6+0.9 (1000-y)=0.7 xx 1000` `0.4 y +900-0.9 y =700` `0.5 y =700 -900= -200` `0.5 y=200` or `y=400` The ratio of volumes of 0.4 M HCl and 0.9 M HCl to be mixed to get 0.7 M concentration of the solution is `400 : 600` i.e., `2:3`

Discussion

945.	The pH of `M//100` NaOH solution is `:`A. 2B. 10C. 6D. 12
Answer» Correct Answer - D `[OH^(-)]=[NaOH]=10^(-2)M` `[H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(10^(-2))=10^(-12)` `pH=2`

Discussion

946.	A certain buffer solution sontains equal concentration of `X^(-)` and `HX`. The `K_(a)` for `HX` is `10^(-8)`. The of the buffer isA. 3B. 4C. 11D. 14
Answer» Correct Answer - B `pH=pK_(a)+log.(["Salt"])/(["Acid"])` `=-log (1.0 xx 10^(-8)) +log. (0.1)/(0.1)` `-log 10^(-8) +0=8`

Discussion

947.	A certain weak acid has a dissociation constant of `1.0 xx 10^(-4)`. The equilibrium constant for its reaction with a strong base isA. `1.0 xx 10^(-4)`B. `1.0 xx 10^(-10)`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(14)`
Answer» Correct Answer - C The equilibrium constant for the neutralization of a weak acid with a strong base, `K = (K_(a))/(K_(w)) = (1.0 xx 10^(-4))/(1.0 xx 10^(-14)) = 1.0 xx 10^(10)`

Discussion

948.	A certain buffer solutions contains equal concentration of `X^(-)` and HX. The `K_(b)` for HX is `10^(-10)`. The pH of the buffer isA. 4B. 7C. 10D. 14
Answer» Correct Answer - A `X^(-) + H_(2)O hArr OH^(-) + HX` `K_(b) = ([OH^(-)][HX])/([X^(-)])` `HX hArr H^(+) + X^(-)` `K_(a) = ([H^(+)][X^(-)])/([HX]), :. K_(a) xx K_(b) = [H^(+)][OH^(-)] = K_(w) = 10^(-14)` Hence `K_(a) = 10^(-4)` Now as `[X^(-)] = [HX], pH = pK_(a) = 4`.

Discussion

949.	Calculate the volume of water required to dissolve`0.1g` lead (II) chloride to get a saturaed solution `(K_(sp)` of `PbCI_(2) = 3.2 xx 10^(-8)`, atomic mass of `Pb = 207 u)`. Multiply your answer with 10 to get answer.
Answer» Correct Answer - `0002`

Discussion

950.	0.023 g of sodium metal is reacted with 100`cm^(3)` of water. The pH of the resulting solution isA. 10B. 11C. 9D. 12
Answer» Correct Answer - D `Na+H_(2)O rarr NaOH +(1)/(2) H_(2)` 1 mol of Na produces 1 mole of NaOH 0.023 g Na`=(0.023)/(23)=10^(-3) ` mole `:. ` NaOH produced `10^(-3)` mole `:. \|NaOH \|=(10^(-3))/(100)xx 1000=10^(-2)M` `pOH =2` or `pH =14-2 =12`

Discussion

Explore topic-wise InterviewSolutions in .

0.02 M monobasic acid dissociates 2% hence, pH of the solution isA. 0.3979B. 1.3979C. 1.699D. 3.3979

pH of a solution is 9.5. The solution isA. NeutralB. AcidicC. BasicD. Amphoteric

Which on reaction with water will have pH less than 7A. BaOB. CaOC. `Na_(2)O`D. `P_(2)O_(5)`

Which of the following will decrease the pH of a 50 ml solution of `0.01 M HCI` ?A. Addition of 50 mL of 0.01 M HClB. Addition of 50 mL of 0.02 M HClC. Addition of 150 mL of 0.02 M HClD. Addition of 5 mL of 1 M HCl

Which would be decrease the pH of `25 cm^(3)` of a 0.01 M solution of hydrochloric acidA. The addition of `25cm^(3)` 0.005 M hydrochloric acidB. The addition of `25cm^(3)` of 0.02 M hydrochloric acidC. The addition of magnesium metalD. None of these

Which one of the following is not a buffer solutionA. `0.8 M H_(2)S + 0.8 M KHS`B. `2M C_(6)H_(5)NH_(2) + 2M C_(6)H_(5)overset(+)(N)H_(3)Br`C. `3M H_(2)CO_(3) + 3M KHCO_(3)`D. `0.05 M KClO_(4) + 0.05 M HClO_(4)`

The mixture of `10 ml 0.5 N CH_(3)COOH` and `10 ml 0.25 M NaOH`, having `pH = 5`, then find the value of pKa?

Which can act as bufferA. `NH_(4)Cl + HCl`B. `CH_(3)COOH + H_(2)CO_(3)`C. `40mL` of `0.1M NaCN + 20mL` of `0.1M HCN`D. `NaCl + NaOH`

When `CH_(3)COONa` is added to an aqueous solution of `CH_(3)COOH`A. The pH of solution decreasesB. The pH of solution increasesC. The pH of the solution remains unalteredD. An acid salt is produced

The `[H^(+)]` in `0.2M` solution of formic acid is `6.4xx10^(-3)` mol `litre^(-1)`. To this solution formate is added so as to adjusrt the conc.of sodium formate to one mol per litre. What will be pH of this solution ? `(K_(a) for HCOOH=2.4xx10^(-4))` and degree of dissociation of `HCOONa = 0.75`)

The degree of hydrolysis of a salt of `W_(A)` and `W_(B)` in its `0.1M` solution is `50%`. If the molarity of the solution is `0.2M`, the percentage hydrolysis of the salt woukd beA. `25%`B. `50%`C. `75%`D. `100%`

The degree of hydrrolysis of a salt of weak acid and weak base in its `0.1M` solultion is found to be `50%`. If the molarity of the solution is `0.2M`, the percentage hydrolysis of the salt should be:A. `100%`B. `50%`C. `25%`D. None of these

The pH of a solution is 2. If its pH is to be raised to 4, then the `[H^(+)]` of the original solution has to beA. DoubledB. HalvedC. Increased hundred timesD. Decreased hundred times

Assuming `100%` ionization in case of strong electrolytes which of the following will have minimum pHA. 0.1 M `NH_(4)Cl`B. 0.1 M `(NH_94))_(2)SO_(4)`C. 0.1M `(NH_(4))_(3)PO_(4)`D. All will have same pH.

pH of one litre solution containing 40gm of `NaOH`A. 2B. 10C. 8D. 14

During the titration of a weak base with a strong acid, one should use an acid-base indicator that changes colour in the :A. acid rangeB. basic rangeC. buffer rangeD. neutral range

A solution of `pH = 9` is one thousand times as basic as a solution of pH.A. 4B. 7C. 10D. 6

Arrhenius theory is failed to explain the acidic nature ofA. HClB. `HCOOH`C. `H_(2)SO_(4)`D. `CO_(2)`

The pH of solution is `9`. It is `"………"` times more basic than a solution with `pH = 6`A. `3`B. `2`C. `4`D. `6`

The `pH` of a solution increased from `3` to `6`. Its `[H^(o+)]` will beA. Reduced to halfB. DoubledC. Reduced by `1000` timesD. Increased by `1000` times

`HCl` gas cannot act as an acid inA. Polar solventsB. Highly polar solventsC. `NH_(3)` liquidD. Non-polar solvents.

Which of the following is an Arrhenius acid?A. `CO_(2)`B. `SO_(2)`C. `FeCl_(2)`D. `HNO_(2)`

`0.1 M HCl` solution is diluted by `100` times The pH of the solution formed isA. `3`B. `2`C. `4`D. `6`

`100 ml` of a solution of `HCl` with `pH` value `3` is diluted with `400 ml` of water. The new `P^(H)` of the solution isA. `3.7`B. `5.3`C. `4.2`D. `5.6`

`10^(-6)M HCl` is diluted to `100` times. Its `pH` is:A. 6B. 8C. 6.95D. 9.5

The weakest base among the followingA. `I^(-)`B. `Cl^(-)`C. `F^(-)`D. `Br^(-)`

Which of the following mixture is not a buffer solutionA. `100 mL` of `0.5 N CH_(3)COOH + 100 mL ` of `0.05 N NaOH`B. `100 mL` of `0.6 N HCN + 100 mL` of `0.4 NaOH`C. `100 mL` of `0.5N NH_(4)OH + 10 mL` of `0.2 NCH_(3)COOH`D. `100 mL` of `0.4 N HCL + 100 mL` of `0.4N NaOH`

Which solutionwill have pH closer to `1.0` ?A. 100 mL of `(M//10) HCI + 100 mL` of `(M//10) NaOH`B. 55 mL of `(M//10) HCI +45 mL` of `(M//10) NaOH`C. 75 mL of `(M//10) HCI + 90` mL of `(M//10) NaOH`D. 75 mL of `(M//5) HCI +25 mL of `(M//5) NaOH`

The following has no conjugate baseA. `H_(2)PO_(4)^(-)`B. `H_(2)PO_(2)^(-)`C. `H_(2)PO_(3)^(-)`D. `CH_(3)COOH`

Which solutionwill have pH closer to `1.0` ?A. `75 mL` of `M//5HC1 + 25 mL` of `M//5 NaOh`B. `10mL` of `M//10 HC1 + 90 mL` of `M//10 NaOH`C. `55mL` of `M//10 HC1` + `54mL` of `M//10 NaOH`D. `100 mL` of `M//10 HC1 + 100 mL` of `M//10 NaOH`

Which of the following solution will have `P^(H)` closer to `1.0`?A. `100 ml` of `M//10 HCl +` `45 ml` of `M//10 NaOH`B. `55 ml` of `M//10HCl +``45 ml` of `M//10 NaOH`C. `10 ml` of `M//10 HCl+` `90 ml` of `M//10 NaOH`D. `75ml` of `M//5 HCl+` `25 ml` of `M//5 NaOH`

Weskest base isA. `C_(2)H^(-)`B. `NO_(3^(-))`C. `CN^(-)`D. `HS^(-)`

Strongest Bronsted base isA. `ClO^(-)`B. `ClO_(2)^(-)`C. `ClO_(3)^(-)`D. `ClO_(4)^(-)`

`K_(h)` of salt obtained from strong acid and weak base is `2 xx 10^(-5)`. The `K_(b)` of weak base isA. `2 xx 10^(-19)`B. `5 xx 10^(-10)`C. `2 xx 10^(-10)`D. `5 xx 10^(-9)`

Which of the following is not a salt?A. `NaCl`B. `Ca(OH)_(2)`C. `PbS`D. `Zn(NO_(3))_(2)`

Aqueous solution of which of the following shows lower `p^(H)`?A. `KNO_(3)`B. `ZnCl_(2)`C. `CH_(3)COONa`D. `CH_(3)COONH_(4)`

What will the pH of the solution of the salt of weak acid and weak base? `(K_(b) = 1 xx 10^(-6)` and `K_(a) = 1 xx 10^(-4)`)

The solubility of a salt of weak acid (AB) at pH 3 is `Y xx 10^(-3) mol L^(-1)`. The value of Y is ___________ (Given that the value of solubility product of `AB(K_(sp)) = 2 xx 10^(-10)` and the value of ionization constant of `HB(K_(a)) = 1 xx 10^(-8))`

In which ratio of volumes 0.4 M HCl and 0.9 M HCl are to be mixed such that the concentration of the resultant solution becomes 0.7 M ?A. `4:9`B. `2:3`C. `3:2`D. `1:1`

The pH of `M//100` NaOH solution is `:`A. 2B. 10C. 6D. 12

A certain buffer solution sontains equal concentration of `X^(-)` and `HX`. The `K_(a)` for `HX` is `10^(-8)`. The of the buffer isA. 3B. 4C. 11D. 14

A certain weak acid has a dissociation constant of `1.0 xx 10^(-4)`. The equilibrium constant for its reaction with a strong base isA. `1.0 xx 10^(-4)`B. `1.0 xx 10^(-10)`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(14)`

A certain buffer solutions contains equal concentration of `X^(-)` and HX. The `K_(b)` for HX is `10^(-10)`. The pH of the buffer isA. 4B. 7C. 10D. 14

Calculate the volume of water required to dissolve`0.1g` lead (II) chloride to get a saturaed solution `(K_(sp)` of `PbCI_(2) = 3.2 xx 10^(-8)`, atomic mass of `Pb = 207 u)`. Multiply your answer with 10 to get answer.

0.023 g of sodium metal is reacted with 100`cm^(3)` of water. The pH of the resulting solution isA. 10B. 11C. 9D. 12