Explore topic-wise InterviewSolutions in Current Affairs.

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.

Ph -underset(darr Cao//Delta)(CO_(2))H + CH_(3) CO_(2)H

Answer»

`PH-overset(O)overset(||)(C)-PH`
`PH-overset(O)overset(||)(C)-CH_(3)`

All

Answer :D
Discussion
2.

Ph-underset(14)overset(O)overset(||)(C)-CHN_2-underset(H_2O)overset(Ag_2O)(rarr)(P). Product (P) is

Answer»

`Ph-overset(14)(C)H_2-CO_2H`
`Ph-CH_2-overset(14)(C)O_2H`
`Ph-overset(14)(C)O_2H`
`Ph-CO_2H`

ANSWER :B
Discussion
3.

pH scale was introduced by

Answer»

Arrhenius
Sorensen
Lewis
Lowry

Answer :B
Discussion
4.

pH of water is 7. When a substance Y is dissolved in water, the pH becomes 13. The substance Y is a salt by

Answer»

STRONG acid and strong BASE
Weak acid and weak base
Strong acid and weak base
Weak acid and strong base

Solution :If we MIXED any substance into the solution. Then the value of pH is INCREASED these substance is a SALT of weak acid and strong base.
Discussion
5.

Ph-overset(Ph)overset(|)underset(H)underset(|)C-overset(H)overset(|)underset(CH_3)underset(|)C-CH_2-OH overset(H^+)underset(Delta)to Product Select the correct P.E. graph for major product.

Answer»




None

Answer :B
Discussion
6.

Ph-overset(O)overset("||")C-Phoverset(PhCO_(3)H)rarrAoverset(LiAlH_(4))rarrB+C B and C are respectively :

Answer»

`Ph-CH_(2)-OH`
`Ph-OVERSET(O)overset("||")C-O-Ph`
`Ph- OH`
`Ph-overset(O)overset("||")C-CH_(2)-Ph`

ANSWER :A::C
Discussion
7.

Ph-overset(O)overset(||)C-O-CPh_(3)-overset(H_(3)overset(18)O^(o+))rarrProduct of this reaction is :

Answer»

`Ph-OVERSET(O)overset(||)C-OH+Ph_(3)C-overset(18)OH`
`Ph-overset(O^(18))overset(||)C-OH+Ph_(3)C-OH`
`Ph-overset(O)overset(||)C-overset(18)OH+Ph_(3)C-overset(18)OH`
`Ph-overset(O)overset(||)C-overset(18)OH+Ph_(3)C-OH`

Solution :Esters which give stable CARBOCATION reacts by `A_(AL^(1))`mechanical.
Discussion
8.

Ph-overset(O)overset(||)(C)-CH_(3)+underset("(excess)")(CH_(2)=O) overset(OH^(Theta))toPh-overset(O)overset(||)(C)-C(CH_(2)OH)_(3) In the above reaction how times of aldol reaction takes place ?

Answer»


SOLUTION :
Discussion
9.

Ph-overset(O)overset(||) underset(14)(C)-OH overset(NaHCO_(3)//Delta)to(A)

Answer»

`CO_(2)`
`OVERSET(14)(CO_(2))`
`H_(2)`
`NH_(3)`

ANSWER :A
Discussion
10.

pH of which solution is not affected by dilution

Answer»

` 0.01 M CH_3 COONH_4`
`0.01 M NaH_2PO_4`
`0.01 M NACL`
`0.01 M NaHCO_3`

SOLUTION :Inaqueoussolution, a saltof weakacidandaweakbaseacts asabuffersolution. THEPH ofbuffersolutiondo not changeon DILUTION`NaHCO_3`and ` H_2 PO_4` from`NaH_2 PO_4` areamphiporticanions. ThepHof solutioncontainingamphiproticanionsdo notchangeondilution .
Discussion
11.

pH of which solution is not affected by dilution ?

Answer»

0.01 M `CH_(3) COONH_(4)`
0.01 M `CH_(3)COONa`
A mixture of 00.1 M `CH_(3)COOH+ 0.01M CH_(3) COONa`
0.01 M `NH_(4)CL`

ANSWER :A,C
Discussion
12.

pH of water is 7.0 at 25^(@)C. If water is heated to 80^(@)C

Answer»

pH will increase
pH will decrease
pH REMAINS 7.0
`H^(+)` ion concentration will increase but `OH^(-)` ion concentration will decrease

Solution :pH of WATER is 7 at `25^(@)C` as temp. increases ionic product of water increase so pH decreases.
Discussion
13.

pH of two solutions : I. 50 mL of 0.2 MHCl + 50 mL of 0.2 MHA (K_(a)=1.0xx10^(-5)) and II. 50 mL of 0.2 M HCl +50 mL of 0.2 M NaA will be respectively

Answer»

0.70 and 2.85
1 and 2.85
1 and 3
3 and 1

Solution :[HCl] in the mixture `=(50xx0.2)/(100)=0.1M`
In PRESENCE of STRONG acid HCl, the weak acid HA remains PRACTICALLY unionized. Hence
`[H^+]=[H^(+)]_(HCl)=0.1M,pH=1`
(II) `NaA+HCl""toNaCl+HA`
`[HA]=(50xx0.2)/(100)=0.1M`
No HCl is left. Hence `[H^(+)]`
`=sqrt(K_(a)C)=sqrt(1.0xx10^(-5)xx0.1)=1.0xx10^(-3),pH=3`
Discussion
14.

pH of the solution in the anode compartment of the following cell at 25^@C is x when Eced E_("cell")-E_("cell")^(@)=0.0591V. underset("1 atm")(PT (H_(2))|pH=Xx|| Ni^(2+)(M)|Ni Find the value of x.

Answer»


Solution :`NI^(2+) +H_(2) to Ni+2H^(-1)`
`K=([H^(+)]^(2))/([Ni^(2+)])=[H^(+)]^(2)`
`E_("CELL")-E_("cell")^(@)=0.0591 XX pH`
pH=1
Discussion
15.

pH of soluiotn obtained by mixing 50ml of 0.1M Na_(2)CO_(3) solution and 50 ml of 0.05 M HCl solution is where K_(a_(1)) and K_(a_(2)) are first and second dissociation constants of H_(2)CO_(3) respectively.

Answer»

`( pKa_(a_(1))+pK_(a_(2)))/(2)`
`pK_(a_(2)) - log_(10) 2`
`pK_(a_(2))`
`pK_(a_(1))= log_(10) 2`

Answer :C
Discussion
16.

pH of saturated solution of Ca(OH)_2 is 9. The solubility product (K_(sp)) of Ca(OH)_2

Answer»

`0.5xx10^(-15)`
`0.25xx10^(-10)`
`0.125xx10^(-15)`
`0.5xx10^(-10)`

Solution :`0.5xx10^(-15)`
`Ca(OH)_(2)hArr Ca^(2+)+2OH^(-)`
Given that `pH=9`
`pOH=14-9=5`
`[pOH=-log_(10)[OH^(-)]]`
`:.[OH^(-)]=10^(-pOH)`
`[OH^(-)]=10^(-5)M`
`K_(SP)=[Ca^(2+)][OH^(-)]^(2)`
`=(10^(-5))/(2)XX(10^(-5))^(2)=0.5xx10^(-15)`
Discussion
17.

PH of pure water is _____ at 22^@C.

Answer»


ANSWER :7
Discussion
18.

pH of NaCl solution is

Answer»

7
Zero
`gt 7`
`LT 7`

Solution :Because of NaCl is a salt of strong ACID and strong base. So that it is NEUTRAL.
Discussion
19.

pH of human blood is 7.4. Then H^(+) concentration will be

Answer»

`4 XX 10^(-8)`
`2 xx 10^(-8)`
`4 xx 10^(-4)`
`2 xx 10^(-4)`

Solution :`pH = - LOG[H^(+)], 7.4 = -log[H^(+)], [H^(+)] = 4 xx 10^(-8) M`.
Discussion
20.

pH of HCl(10^(-12)M) is

Answer»

12
`-12`
`~~7`
14

SOLUTION :AS the solution is ACIDIC `pH lt 7`. This is because `[H^(+)]` from `H_(2)O[10^(-7)]` cannot be NEGLECTED in comparison to `10^(-12) M`.
Discussion
21.

pH of completely dissociated 0.005 M H_(2)SO_(4) is

Answer»

3
4
2
5

Solution :`H_(2)SO_(4)` IONIZED in TWO STEP.
Discussion
22.

pH of buffer depends upon concentration of

Answer»

acid `(H^+)`
CONJUGATE BASE `(OH^-)`
Salt
Both (a) and (B )

Answer :D
Discussion
23.

pH of blood is maintained constant by mechanisms of

Answer»

COMMON ION effect
Buffer
Solubility
All of these

Answer :B
Discussion
24.

pH of aqucous solution of 0.1M, NH_(4),CI is found to be 5. The equilibrium constant for the neutralization of NH_(4),OH by HCl is 10^(y). The value of 'y' is

Answer»


ANSWER :9
Discussion
25.

How does PH of a solution vary with temperature ?

Answer»


ANSWER :DECREASES
Discussion
26.

pH of a solution produced when an aqueous solution of pH 6 is mixed with an equal volume of an aqueous solution of pH 3 is about

Answer»

3.3
4.3
`4.0`
4.5

Answer :A
Discussion
27.

pH of a solution of 10 ml. 1 N sodium acetate and 50 ml 2N acetic acid (K_(a) = 1.8 xx 10^(-5)), is approximately

Answer»

4
5
6
7

Solution :`pH = pK_(a) + LOG.(["SALT"])/(["Acid"])`
`pH = - log (1.8 xx 10^(-5)) + log.([10])/([100])`
`= -log 1.8 + 5 + log 10^(-1)`
`= 0.2553 + 5 - 1 = 3.7447` or = 4.
Discussion
28.

pH of a solution of 0.1 M [CH_3COONH_4(aq)] is [given: K_a(CH_3COOH) = K_b(NH_4OH) = 1.8 x 10^-5)]

Answer»

1
9
7
13

Answer :1
Discussion
29.

pH of a solution is 9.5. The solution is

Answer»

Neutral
Acidic
Basic
AMPHOTERIC

SOLUTION :When pH = 7 means neutral, `pH lt 7` means acidic, `pH GT 7` means basic.
Discussion
30.

pH of a solution is 5.5 at 25^(@)C. Calculate its [OH^(-)]

Answer»

SOLUTION :`pH+POH=14.0`
`THEREFORE pOH=14.0-pH`
`=14.0 -5.5 =8.50`
`pOH=8.5""therefore "ANTILOG" [-pOH]`
`therefore [OH^-]= "antilog" [-8.5]=3.2xx10^-9 M`.
Discussion
31.

pH of a solution is 5. Thus. the concentration of hydroxyl ion in the solution is

Answer»

`9 MOL L^-1`
`5 mol L^-1`
`10^-5 mol L^-1`
`10^-9 mol L^-1`

ANSWER :D
Discussion
32.

pH of a solution is 5. Its hydroxyl ion concentration is…….

Answer»

5
10
`10^-5`
`10^-9`

Solution :`pH=5` MEANS `[H^+]=10^-5`
`pOH=14-pH=14-5=9`
`[OH^-]=10^(-pOH)=10^-9`
Discussion
33.

pH of a solution having 0.00001 gm ions of ions of hydrogen per litre is

Answer»

5
4
`10^(-5)`
`10^(-4)`

SOLUTION :`[H^(+)] = 1 XX 10^(-5) M`
`pH = - log 10^(-5)`
pH = 5 log 10 = 5.
Discussion
34.

pH of a saturated solution of Ba(OH)_(2) is 12. The value of solubility product (K_(sp)) of Ba(OH)_(2) is

Answer»

`3.3 xx 10^(-7)`
`5.0 xx 10^(-7)`
`4.0 xx 10^(-6)`
`5.0 xx 10^(-6)`

Solution :`{:(Ba(OH)_(2),HARR,Ba^(2+)+,2OH^(-)),(,,s,2S):}`
`[OH^(-)] = 10^(-2)`
`2s = 10^(-2)`
`s = (10^(-2))/(2) K_(sp) = 4s^(3) = 4 xx ((10^(-2))/(2))^(3) = 5 xx 10^(-7)`
Discussion
35.

pH of a solution can be expressed as

Answer»

`-log_(e)(H^(+))`
`-log_(10)(H^(+))`
`log_(e)(H^(+))`
`log_(10)(H^(+))`

SOLUTION :`PH = -LOG [H^(+)]`.
Discussion
36.

pH of a saturated solution of Ca(OH)_2 is 9. The solubility product (K_(sp)) of Ca(OH)_2……..

Answer»

`0.5 times 10^-15`
`0.25 times 10^-10`
`0.125 times 10^-15`
`0.5 times 10^-10`

Solution :`Ca(OH)_2 leftrightarrowCa^(2+)+2OH^-`
Given that pH=9
`POH=14-9=5`
`[pOH=-log_10[OH^-]]`
`therefore[OH^-]=10^(-pOH)`
`[OH^-]=10^-5 M`
`K_(SP)=[Ca^(2+)][OH^-]^2`
`=10^-5/2 times (10^-5)^2=0.5 times 10^-15`
Discussion
37.

PH value of a saturated solution Ba(OH)_2 is 12. Calculate solubility product K_sp for Ba (OH)_2(3+2

Answer»

`3.3 xx 10^(-7)`
`5.0 xx 10^(-7)`
`4.0 xx 10^(-6)`
`5.0 xx 10^(-6)`

ANSWER :B
Discussion
38.

pH of a saturated solution of Al(OH)_3, is 9. The value of solubility product (K_sp) of Al(OH)_3 is

Answer»

`1/2*1^-20`
`1/3*1^-20`
`1/3*1^-15`
`1/4*1^-15`

ANSWER :2
Discussion
39.

pH of a neutral solution is equal to 7. Prove it.

Answer»

Solution :(i) In the neutral solution the CONCENTRATIONS of `[H_3O^+]` as well as `[OH-]` are EQUAL to `1 TIMES 10^-7M` at `25^@C`
(ii) The pH of a neutral solution can be calculated by substituting this `[H_3O^+]` concentration in the expression.
`pH=-log_10[H_3O^+]`
`=-log_10[1 times 10^-7]`
`=-(-7)log""10/10=+7(1)=7`
(III) pH=7 for a neutral solution.
Discussion
40.

pH of a lemon water bottle is 3 and that of orange juice is 4. Calculate the pH of final solution obtained by mixing above solutions in volume ratio of 1 : 2 respectively. (Assuming that no change in volume on mixing and no chemical reaction is taking place) [Given : log_(10)2 =0.30]

Answer»


Solution :`V xx 10^(-3) + 2V xx 10^(-4) = 3V xx [H^(+)]`
`[H^(+)] = (10^(-3) + 2 xx 10^(-4))/(3)= 4 xx 10^(-4)`
`PH= 4- LOG 4= 4- 6= 3.40`
Discussion
41.

pH of a human blood is 7.4. Then H^+ concentration will be……

Answer»

`4 times 10^-8`
`2 times 10^-8`
`4 times 10^-4`
`2 times 10^-4`

SOLUTION :`pH=-LOG[H^+]`
`7.4=-log[H^+]`
`7.4= log1-log[H^+]`
`log[H^+]=log 1-7.4`
`log[H^+]=8.6`
`THEREFORE [H^+]=ANTILOG of 8.6`
`=4 times 10^-8`
Discussion
42.

pH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic and potassium acetate at 27^(@)C. The buffer capacity of the solution is

Answer»

0.1
10
1
0.4

Answer :D
Discussion
43.

pH of a buffer solution decreases by 0.02 units when 0.12g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27^(@)C. The buffer capacity of the solution is

Answer»

`0.1`
10
1
`0.4`

ANSWER :D
Discussion
44.

pH of 10^(-8) M HNO_(3) is (nearly)

Answer»

7
6.96
7.2
6

Solution :`{:(,HNO_(3),RARR,H^(+),+,NO_(3)^(-)),("Conc. before ionisation",10^(-8)M,,0,,0),("Conc. after ionisation",0,,10^(-8),,10^(-8)):}`
`:. [H^(+)] = 10^(-8) m` but pH = 8 is not POSSIBLE because it is acid. Now `[H^(+)] = 10^(-7) M` are already present in solution and SINCE `10^(-8) lt 10^(-7)` and thus, it should not be NEGLECTED.
`:. [H^(+)] = 10^(-8) + 10^(-7) = 10^(-7)(1.1) M`
`= 1.1 xx 10^(-7)M`
`:. pH = 6.9586`.
Discussion
45.

pH of 1M HA (weak acid) is 2. Hence van't Hoff factor is -

Answer»

1.2
1.02
1.1
1.01

Answer :A
Discussion
46.

pH of a 0.1 M monobasic acid is found to be 2 .Hence , its osmotic pressure at a given temperature T K is :

Answer»

0.1 RT
0.11 RT
1.1 RT
0.01 RT

Answer :B
Discussion
47.

pH of a 0.001 M NaOH solution will be

Answer»

9
3
11
12

Answer :C
Discussion
48.

pH of10^-8 M solution of HCl in water is:

Answer»

8
-8
Between 7 and 8
Between 6 and 7

Answer :D
Discussion
49.

pH of 0.1M solution of NaA (sodium salt of weak acid HA) is 8.92.Calculate pK_(a) of HA.If a drop HPh(pK_("in")=9.52) be added to the above solution.predict whether the pink colour will visible or not the under the medical fact that our eyes can see the pink colour if the mole % of ionised form of indicated is 25% or more.

Answer»

Solution :`pH=(1)/(2)(pK_(a)+pK_(w)+logc)`
`8.92=7+(1)/(2)pK_(a)+(1)/(2)(-1)`
`(1)/(2)pK_(a)=8.92-6.5=2.42`
`pK_(a)` of `HA=4.84`
`:.pK_(a)=pK_("in")+LOG""([LN^(-)])/([HLN])`
`=8.92=9.52+log""([ln^(-)])/([Hln])`
`=8.92=9.52+log"([ln^(-)])/([Hln])`
`([ln^(-)])/([Hln])=(ionised)/(unionised)=(1)/(4)`
Since ionised from `%=(1)/(1+4)xx100=20%` PINK colour will not be visible.
Discussion
50.

pH of 0.1 M solution of a weak acid (HA) is 4.50. It is neutralised with NaOH solution to decrease the acid contant to half pH of the resulting solutin

Answer»

`4.50`
`8.00`
`7.00`
`10.00`

ANSWER :B
Discussion