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.

Define bond order.

Answer»

Solution :BOND ORDER is half of the difference between the NUMBER of electrons in BMO and number of electrons n ABMO.
`:.BO=(1)/(2)[e_(BMO)-e_(ABMO)]`
Discussion
2.

Define bond length.

Answer»

Solution :It is defined as AVERAGE DISTANCE between the CENTERS of nuclei of the TWO bonded atoms in a molecule.
Discussion
3.

Define bond energy.

Answer»

Solution :Covalent bonds are directional in NATURE and are oriented in specific direction in space. This directional nature CREATES a FIXED angle between TWO covalent bonds in a MOLECULE and this angle is termed as bond angle.
Discussion
4.

Definethe avogadro's number :

Answer»

Solution :AVOGADRO number is the number of ATOMS present in one MOLE of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = `6.023xx10^(23) `.
Discussion
5.

Define Avogadro constant (N_(A)).

Answer»

SOLUTION :The number of molecules in ONE MOLE of a gas has been determined to be `6.022xx10^(23)` and is know as Avogadro CONSTANT.
Discussion
6.

Define auto -oxidation reaction and its examples.

Answer»

Solution :
Displacement reaction : Redox reactions in which an ion ( or an atom) in a compound is replaced by an ion (or atom) of another element are called displacement reactions. They are further classified into (i) METAL displacement reactions (ii) non-metal displacement reactions.
(I) Metal displacement reactions : `ast` PLACE a zinc metal STRIP in an aqueous copper sulphate solution taken in a beaker. Observe the solution, the INTENSITY of blue colour of the solution slowly reduced and finally disappeared.
`ast` The zinc metal strip because coated with brownish metallic copper. This is due to the following metal displacement reaction .
Discussion
7.

Define Arrhenius acid-base theory with one example.

Answer»

Solution :An ELECTROLYTE which when DISSOLVED in water, PRODUCES `H^(+)` ion is called an ACID.
An electrolyte which when dissolved in water, produces `OH^(-)` ion is called a base.
`HCNoverset("water")toH^(+)+CN^(-)""thereforeHCN" is an acid"`
`NaOHoverset(H_(2)O)toNa^(+)+OH^(-)""thereforeNaOH" is an base"`.
Discussion
8.

Define antibonding molecular orbitals .

Answer»

Solution :The MOLECULAR ORBITALS formed by the subtractive effect of the ELECTRON WAVES of the combining ATOMIC
orbitals is called antibonding molecular orbital .
Discussion
9.

Define an ideal gas.

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SOLUTION :IDEAL gas is gas which follows all the gas LAWS at TEMPERATURE and pressure.
Discussion
10.

Define an atomic orbital. What does angular momentum quantum number tell about an orbital ?

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Solution :For definition, Angular momentum quantum number tells about the type of orbital OCCUPIED by the ELECTRON, i.e., s, p, d or f depending UPON VALUE of l, viz, 0, 1, 2, 3 respectively
Discussion
11.

Define - Acid rain .

Answer»

SOLUTION :Rain water normally has a pH of 5 . 6 due to dissolution of ATMOSPHERIC `CO_(2)` into It . Oxides of sulphur and nitrogen in the atmosphere MAY be absorbed by droplets of water that make up clouds and get CHEMICALLY converted into sulphuric acid and nitric acid respectively as a results of pH of rain water drops to the level 5 . 6 . HENCE it is called acid rain .
Discussion
12.

Define - Acid rain.

Answer»

Solution :RAIN water has a pH of 5.6 due to the DISSOLUTION of `CO_(2)` into it. Oxides of sulphur and NITROGEN in the atmosphere· MAY be absorbed by droplets of water that make up the clouds and get chemically converted into sulphuric acid and nitric acid.. Due to this the pH of rain water drops below the LEVEL of 5.6. Hence it is called acid rain.
Discussion
13.

Define absolute zero. Is it possible to attain a further lower temperatuer ? Comment on your answer.

Answer»

Solution :(i) The LOWEST hypothetical or imaginary temperature at which GASES are supposed to OCCUPY zero volume is called absolute zero.
(ii) Temperature below `-273.15^(@)C` (absolute zero) is not POSSIBLE, because it corresponds to NEGATIVE volume, which is meaningless.
Discussion
14.

Define about nature of matter.

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Solution :Anything which has mass and occupies space is called matter. For EXAMPLE, book, pen, pencil, water, air all living beings ETC. All are composed of matter and they have mass and they OCCUPY space.
The matter can be classified into two categories :
(i) ACCORDING to PHYSICAL state
(ii) According to chemical state
Discussion
15.

Definesolubility

Answer»

SOLUTION :The solubility of a substances is DEFINED as the AMOUNT of the solute that can be dissolved in 100 G of the solvent at a given TEMPERATURE to form a saturated solution.
Discussion
16.

Define a non-spontaneous process.

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Solution :A process which does not TAKE place by iteself but takes place only with the help of some external AID under the given condition is called a non-spontaneous process.
EXAMPLE : Flow of HEAT energy lower to higher temperatures.
Discussion
17.

Define : (a) Enthalpy of atomization (b) Lactic enthalpy.

Answer»

Solution :Enthalpy of atomization : It is defined as the enthalpy change accompanying the breaking of one mole of a SUBSTANCE completely into its atoms in the GAS phase.
`H_(2)(g) rarr 2H(g) , Delta_(s)H^(@) = 435.0 kJ mol^(-1)`
LATTICE enthalpy : It is the enthalpy change which occurs when one mole of an IONIC compound dissociates into its gaseous ionic state.
`NaCl(s) rarr Na^(+) (g) + Cl^(-) (g) , Delta H = +788 ks//mol`.
Discussion
18.

Definciencyof calciumcauses _______.

Answer»

RICKETS
CONVULSION
HYPERTENSION
circulatoryfailure

ANSWER :A
Discussion
19.

Deep-sea divers use air diluted with helium gas in their tanks. Why ? (or) Justify this statement.

Answer»

Solution :(i) Deep-sea divers carry a compressed air tank for breating at hgh pressure under water. This air tank contains nitrogen and OXYGEN which are not very soluble in blood and other body fluids at normal pressure.
(ii) As the pressure at the depth is far GREATER than the surface atmospheric pressure, more nitrogen dissolves in the blood when the diver breathers from tank.
(iii) When the divers ascends to the surface, the pressure DECREASES, the dissolved nitrogen comes out of the blood quickly FORMING bubbles in the blood stream. These bubbles restrict blood flow, affect the transmission of never IMPULSES and can even burst the capillaries or block them. This condition is called "the bends" which are painful and dangerous to life.
(iv) To avoid such dangerous condition they use air diluted with helium gas (`11.7%` helium, `56.2` nitrogen and `32.1%` oxygen) of lower solubility of helium in the blood than nitrogen.
Discussion
20.

Deep sea divers ascend slowly and breath contentiously by the time they reach the surface. Give reason.

Answer»

Solution :(i) For every 10m of depth a DIVER EXPERIENCES an additional 1 atm of pressure due to the weight of water surrounding him.
(ii) At 20m, the diver experiences a total, pressure of 3atm. So the most important rule is DIVING is never hold BRATH.
(iii) Divers must ascend slowly and breath continously allowing the reuglator to bring the air pressure in their LUNGS to 1 atm by time they reach the surface.
Discussion
21.

Deduce the Vant Hoff's equation.

Answer»

Solution :This equation gives the equntitaive temperature dependance of equilibrium constant K.
The relation between standard free ENERGY change `DeltaG^@` and equilibrium constant is
`DeltaG^@ = -RT InK`…(1)
We know that, `DeltaG^@ = DELTAH^@ -TDeltaS^@`...(2)
Substituting (1) in equation (2)
`-RT InK= DeltaH^@ - TDeltaS^@`
Rearranging, `In K= (-DeltaH^@)/(RT)+ (DeltaS^@)/R`...(3)
DIFFERENTIATING equation (3) with respect to temperature
`(d(InK))/(dT) = (DeltaH^@)/(RT^2)` ...(4)
Equation (4) is known as differential from of van.t Hoff equation.
On integrating the equation 4, between `T_1 and T_2` with their respective equilibrium constant `K_1 and K_2`
`underset(K_1)OVERSET(K_2)intd(InK) = (DeltaH^@)/(R)underset(T_1)overset(T_2)int (dT)/T^2`
`[InK]_(K_1)^(K_2) = (DeltaH^@)/R [-1/T]_(T_1)^(T_2)`
`InK_2 - In K_1 = (DeltaH^@)/R [-1/T_2+1/T_1]`
`In K_2/K_1 = (DeltaH^@)/R[(T_2-T_1)/(T_2T_1)]`
`log""(K_2)/K_1 = (DeltaH^@)/(2.303R) [(T_2-T_1)/(T_2T_1)]`...(5)
Equation (5) is known as integrated from of van.t Hoff equation.
Discussion
22.

Deduce the Vant Hoff equation.

Answer»

Solution :This equation GIVES the quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy CHANGE `(DeltaG^(@))` and equilibrium constant is
`DeltaG^(@)=-RTlnK`. . . (1)
We know that
`DeltaG^(@)=DELTAH^(@)-TDeltaS^(@)`. . . (2)
Substituting (1) in equation (2)
`-RTlnK=DeltaH^(@)-TDeltaS^(@)`
Rearranging
In `K=(-DeltaH^(@))/(RT)+(DeltaS^(@))/(R)`. . . (3)
Dierentiating equation (3) with respect to temperature,
`(d("In K"))/(dT)=(DeltaH^(@))/(RT^(2))`. . . .(4)
Equation 4 is known as differential from of Van't Hoff equation.
On integrating the equation 4, between `T_(1)` and `T_(2)` with integrating the equation 4, between `T_(1)` and `T_(2)` with their respective equilibrium CONSTANTS `K_(1)` and `K_(2)`.
`int_(K_(1))^(K_(2))d("In K")=(DeltaH^(@))/(R)int_(T_(1))^(T_(2))(dT)/(T^(2))`
`["In K"]_(K_(1))^(K_(2))=(DeltaH^(@))/(R)[-(1)/(T)]_(T_(1))^(T_(2))`
`"In "K_(2)-"In "K_(1)=(DeltaH^(@))/(R)-[(1)/(T_(2))+(1)/(T_(1))]`
`"In"(K_(2))/(K_(1))=(DeltaH^(@))/(R)[(T_(2)-T_(1))/(T_(2)T_(1))]`
`"log"(K_(2))/(K_(1))=(DeltaH^(@))/(2.303R)[(T_(2)-T_(1))/(T_(2)T_(1))]`. . . (5)
Equation 5 is known as integrated from of Van't Hoff equation.
Discussion
23.

Deduce the Van der Waals equation for real gases at (a) low presssure and (b) high pressure.

Answer»

Solution :Van der Waals EQUATION for one mole of real gas is `(P + (a)/(V_(2))) (V-b) = RT`
(a) At low PRESSURE where MOLAR volume of the gas is very high, b is neglible when compared with molar volume V.
`V - b = V`
`(P + (a)/(V_(2))) V = RT (or) PV + (a)/(V) = RT `
`(PV)/(RT) + (a)/(RTV) =1`
`(PV)/(RT) = 1 - (a)/(RTV) (or) z = 1 - (a)/(RTV)`
(b) At high pressure, `(a)/(V^(2))` can be NEGLECTED when compared with P
`P + (a)/(V^(2)) = P (or) P (V-b) = RT`
`PV - Pb = RT (or) (PV)/(RT) = 1 + (pb)/(RT)`
Discussion
24.

Deduce structure of (A) C_(8)H_(10)(A) overset(KMnO_4)to C_8H_(6) O_4(B) underset(Fe)overset(Br_2)to C_8H_5BrO_4(C) if (C) has onely one possible structure (One - product only)

Answer»





SOLUTION :
Discussion
25.

Deduce Hendersons equation for an acidic buffer.

Answer»

Solution :Consider the acidic buffer, a mixture of acetic acid and sodium acetate.
`CH_(3)COOHhArrCH_(3)COO^(-)+H^(+)to(1)`
`CH_(3)COONatoCH_(3)COO^(-)+Na^(+)to(2)`
Applying law of mass action to (1), `K_(a)=([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])to(3)`
where, `K_(a)` is DISSOCIATION CONSTANT of weak acid.
In this solution, `[CH_(3)COO^(-)]=["Salt"]to(4)`
`[CH_(3)COOH^(-)]=["Acid"]to(5)`
Substituting for (4) and (5) in (3), `K_(a)=(["Salt"][H^(+)])/(["Acid"])RARR[H^(+)]=K_(a)(["Acid"])/(["Salt"])`
Taking `-log_(10)` on both the sides, `-log_(10)[H^(+)]=-log_(10)K_(a)+(-log_(10))(["Acid"])/(["Salt"])`
`pH=pK_(a)+"log"(["Salt"])/(["Acid"])rArr""thereforepH=pK_(a)+"log"([S])/([A])`.
Discussion
26.

Deduce Handerson's equation for a basic buffer.

Answer»

Solution :CONSIDER the basic buffer, a mixture of ammonium hydroxide and ammonium chloride,
`NH_(4)OHhArrNH_(4)^(+)+OH^(-)to(1)`
`NH_(4)ClhArrNH_(4)^(+)+Cl^(-)to(2)`
Applying law of mass ACTION for (1) `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(4)OH])to(3)`
where `K_(b)` = DISSOCIATION constant of weak base.
In this solution, `[NH_(4)^(+)]=["Salt"]to(4)`
`[NH_(4)OH]=["Base"]to(5)`
Substituting for (4) and (5) in (3), we get, `K_(b)=(["Salt"][OH^(-)])/(["base"]),[OH^(-)]=K_(b)(["base"])/(["salt"])`
Taking `-log_(10)` on both the sides, `-log_(10)[OH^(-)]=-log_(10)K_(b)+(-log_(10))(["base"])/(["salt"])`
`pOH=pK_(b)+"log"(["Salt"])/(["Base"])rArrthereforepOH=pK_(b)+"log"([S])/([B])`.
Discussion
27.

Decribe how would yor prepare the following solution from solute and solvent 566 mL "of " 6.0% (V//V) methanol solution

Answer»

Solution :500 mL of 6.0% (V/V) methanol solution
` (V_a)/( V_a+V_b) XX 100 = 60% `
`V_a=( 6)/( 100) = 30 mL `
Discussion
28.

Decrecying order of stability

Answer»

SOLUTION :`(I) LT (II) lt (III)`
Discussion
29.

Decreasing order of stability of O_(2) , O_(2)^(-), O_(2)^(+) andO_(2)^(2-) is

Answer»

`O_(2)^(2-) GT O_(2)^(-) gt O_(2) gt O_(2)^(+)`
`O_(2) gt O_(2)^(+) gt O_(2)^(2-) gt O_(2)^(-)`
`O_(2)^(-) gt O_(2)^(2-) gt O_(2)^(+) gt O_(2)`
`O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)`

Solution :`{:("",O_(2) , O_(2)^(-), O^(2-),O_(2)^(+)),("BOND order :",2,1.5,1,2.5):}`
Greater the bond order, greateris the STABILITY
Hence , decreasing order of stability is
` O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)`
Discussion
30.

Decreasing order of stability of O_(2), O_(2)^(-) , O_(2)^(+) and O_(2)^(2-) is :

Answer»

`O_(2) gt O_(2)^(+) gt O_(2)^(2-) gt O_(2)^(-)`
`O_(2)^(-) gt O_(2)^(2-) gt O_(2)^(+) gt O_(2)`
`O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)`
`O_(2)^(2-) gt O_(2)^(-) gt O_(2) gt O_(2)^(+)`

SOLUTION :`O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)`
GIVEN species : `O_(2), O_(2)^(-1), O_(2)^(+1) , O_(2)^(2-)`
`"" O_(2)^(+1) ""O_(2) "" O_(2)^(-1) "" O_(2)^(-2)`
Number of `e^(-) " :15161718 " `
Bond ORDER `":2.521.51 " `
Stability `XX` B.O.
* Stability order `[ O_(2)^(+1) gt O_(2) gt O_(2)^(-1) gt O_(2)^(2-) ]`
Discussion
31.

Decreasing order of stability of following alkenes is (i) CH_(3)-CH=CH_(2) (ii) CH_(3)-CH=CH-CH_(3) (iv) CH_(3)-underset(CH_(3))underset(|)(C)=CH-CH_(3) (iv) CH_(3)-overset(CH_(3))overset(|)(C)=overset(CH_(3))overset(|)(C)-CH_(3)

Answer»

(i)gt(II)gt(iii)gt(iv)
(iv)gt(iii)gt(ii)gt(i)
(iii)gt(ii)gt(i)gt(iv)
(ii)gt(iii)gt(iv)gt(i)

Solution :Higher the NUMBER of `alpha-H`,more the hyperconjugating STRUCTURES, more the stability of the COMPOUND.
Discussion
32.

Decreasing order of potential energy of the following cations is :

Answer»

`IIgtIgtIII`
`IgtIIgtIII`
`IIIgtIIgtI`
`IIIgtIIgtI`

ANSWER :A
Discussion
33.

Decreasing order of "p" orbital character in the following (i) SiO_(2)(ii) CO_(2)(iii) Graphite

Answer»

`I GTII GT III`
`II gt I gt III`
`II gt III gt I`
`I gt III gt II`

Solution :i=`sp^(3)` = 75%,,ii `= sp^(3) = 50%`
`iii=sp^(2) = 66.66%`
Discussion
34.

Decreasing order of nucleophilicity is ........

Answer»

`OH ^(- )gtNH_(2)^(-)gt ""^(-)OCH_3gt RNH_2`
`NH_2^(-) gt OH^(-)gt ""^(-)OCI_(3)gt RNH_2`
`NH_2^(-)gtCH_3 O^(-)gtOH^(-)RNH_2`
`CH_3 O^(-) gtNH_(2 )^(-)gtRNH_2`

Solution :`NH_2^(-) gt OH^(-)gt ""^(-)OCI_(3)gt RNH_2`
Discussion
35.

Decreasingorder of nucleophilicity is

Answer»

`OH^(-)gt NH_(2)^(-) gt ^(-)OCH_(3) gt RNH_(2)`
`NH_(2)^(-) gt OH^(-) gt ^(-)OCH_(3) gt RNH_(2)`
`NH_(2)^(-) gt CH_(3)O^(-) gt OH^(-) gt RNH_(2)`
`CH_(3)O^(-) gtNH_(2)^(-) gt OH^(-) gtRNH_(2)`

Answer :B
Discussion
36.

Decreasing order of hydration energy of the following cations is

Answer»

`Be^(+2)gtMg^(+2)gtCa^(+2)BA^(+2)`
`Be^(+2)ltMg^(+2)LTCA^(+2)ltBa^(+2)`
`CA^(+2)ltMg^(+2)ltBe^(+2)ltBa^(+2)`
`Be^(+2)ltMg^(+2)gtCa^(+2)ltBa^(+2_`

Solution :DUE to increases in ionic radii hydration energies DECREASES from `Be^(+2)` to `Ba^(+2)`
Discussion
37.

Decreasing order of enol content of the following.(along with proper explanation).

Answer»


Answer :`(##BSL_CHM_ISO_E01_151_A01##)`
Tightly on stable KETO due to repulsion between `alpha-CO` groups has 100 % ENOL. >
(B) Active 'H' atoms/Acids 'H' ATOM so has more enolic content (enol stablise by resonance & intra molecular H-bonding) >
(c) Enolic contents decreases with introduction of `e^(-)` donator group which causes repulsion in enolic form
(d) Due to ester group acidc structure of active H decreases & C=C of enol UNDERGOES cross resonance >
(e) Lowest enolic content because `(##BSL_CHM_ISO_E01_151_A01##)` is more stable than `(##BSL_CHM_ISO_E01_151_A02##) `Bond ]
Discussion
38.

Decreasing order of enol content of the following compound in liquid phase I) CH_(3)-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-O-EtII) CH_(3)-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-CH_(3) III) Ph-overset(O)overset(||)C-CH_(2)-overset(O)overset(||)C-Ph

Answer»

I `GT` II `gt` III `gt` IV
III `gt` II `gt` I `gt` IV
III `gt` II `gt` IV `gt` I
II `gt` III `gt` I `gt` IV

Solution :`UNDERSET("less stable")(CH_(3)-overset(OH)overset(|)C=CH-overset(O)overset(||)C-O)et`
`underset("stable")(CH_(3)-overset(OH)overset(|)C=CH-overset(O)overset(||)C-)CH_(3),""underset("More stable")(Ph-overset(OH)overset(|)C=CH-overset(O)overset(||)C-)Ph`

Anti -Aromatic
LEAST Stable
KETO group is more with drawing group than ester group.
Discussion
39.

Decreasing order of acidic strength of the following compounds is

Answer»

`I gt III gt I gt IV`
`II gt I gt IV gt III`
`I gt III gt IV gt II`
`II gt IV gt I gt III`

Solution :Generally acidic strength `uarr` when electron WITHDRAWING GROUP on phenol (or) Benzene ring &`DARR` electron DONATING group
Based on this the order of acidic strength is
Discussion
40.

Decreasing -I effect of given groups is : (i) -CN"(ii)"-NO_(2)"(iii)"-NH_(2)"(iv)"-F

Answer»

`iiigtiigtigtiv`
`iigtiiigtivgti`
`iiigtiigtivgti`
`iigtigtivgtiii`

ANSWER :D
Discussion
41.

Deconposition of H_(2)O_(2) can be prevented by addition of

Answer»

FERROUS SULPHATE
oxalic acid
OZONE
Glycerol

Solution :DECOMPOSITION of `H_(2)O_(2)` can be prevented by addition of because it can form H-bond with glycerol.
Discussion
42.

Decomposition temperatures of saline hydrides are very high. Why?

Answer»

Solution :Cations of group 1 and 2 FORM saline hydrides SIZE of these cations is LESS. Lattice energy of the hydride is high. Hence the decomposition temperatue is also high.
Discussion
43.

Decomposition of oxalic acid in the presence of conc. H_(2)SO_(4) gives

Answer»

CO
`CO_(2)`
Formic acid
`H_(2)O`

SOLUTION :`UNDERSET("COOH")underset(|)("COOH")+underset(Delta)overset(H_(2)SO_(4))(rarr)CO + CO_(2)+H_(2)O`
Discussion
44.

Decomposition of hydrogen peroxide is prevented by

Answer»

NaOH
`MnO_2`
Glycerol
oxalic aid

Solution :DECOMPOSITION of hydrogen PEROXIDE is PREVENTED by Glycerol
Discussion
45.

Decomposition of H_(2)O_(2) is an example of

Answer»

NEUTRALISATION
precipitation
disproportionation
hydrolysis

Answer :C
Discussion
46.

Decomposition of H_(2)O_(2) in accompained by

Answer»

DECREASE in free energy
increase in free energy
no CHANGE in freemergy
evolutium of HEAT

Solution :Decomposition of `H_(2)O_(2)` is accompained by decrease in free energy
Discussion
47.

Decolourization of coloured material is called as "_____________".

Answer»

BLEACHING
corrosion
combustion
respiration

Answer :A
Discussion
48.

Decimolar weak monoprotic acid (pK_a = 4.8) is titrated with decimolar potash solution. What is the pH at half-equilvalence point ?

Answer»

Solution :At half the equivalence point 50% of the acid is consumed for salt FORMATION and REMAINING 50% of acid is unconsumed.
This is a good buffer, (salt] = [acid]
PH at half-equilvalence point = `PK_a` = 4.8
Discussion
49.

Decimolar NH_4 OH (PK_a = 4.8) is titrated with decimolar HC1. What are the pH values at equivalence point and at one-half of the equivalence point ?

Answer»

SOLUTION :`5.25 ,9.2`
Discussion
50.

Decimolar weak monoprotic acid (pK_a = 4.8) is titrated with decimolar potash solution. What is the pH at equivalence point?

Answer»

Solution :At EQUIVALENCE point, HA+KOH `to`KA+`H_2 O`
SALT hydrolysis of salt of weak ACID and STRONG base takes place
`pH = 1/2 (PK_w+ PK_a+ log C)`, where .C. is one-half of that of the acid, because equal volume of base is added for its neutralisation.
` pH = 1/2 ( 14+4.8 + log5 xx 10^2 ) = 1/2(18.8 +0.7 -2 )= 8.75 `
Discussion