Solveeit Logo

Question

Question: Correct order of \(p{K_a}\) is: (i)![](https://www.vedantu.com/question-sets/abfe1ace-9f7a-4712-8c...

Correct order of pKap{K_a} is:
(i)
(ii)
(iii)

A.(i)>(ii)>(iii)(i) > \,(ii) > \,(iii)
B.(ii)>(i)>(iii)(ii) > \,(i) > \,(iii)
C.(iii)>(ii)>(i)(iii) > \,(ii) > \,(i)
D.(ii)>(iii)>(i)(ii) > \,(iii) > \,(i)

Explanation

Solution

pKap{K_a} value is inversely proportional to the acidic strength but directly proportional to basic strength. So we find out the order of pKap{K_a} value if we know the acidic strength order. Reversing the order of the acidic strength we can easily write the pKap{K_a} order. Thus we can write as pKaα1Acidicstrengthp{K_a}\,\alpha \,\,\dfrac{1}{{Acidic\,strength}} and also pKaαBasicstrengthp{K_a}\,\alpha \,\,Basic\,strength .

Complete step-by-step answer: For finding the acidic strength order, we have to first make the conjugate base by removing a proton from acid and then compare the stability of conjugate base. If the conjugate base is stable it means it acts as a weak base thus the acid from it is made up of is strong in nature. Let’s take an example of option (i) which is ethyl alcohol.

Here, the conjugate base of ethyl alcohol is having one (CH3)( - C{H_3}) group which shows (+I)effect( + I)\,effect thus the conjugate base is less stable hence ethyl alcohol is strong in nature. The comparison will show the effect of all other options.
If we see option (ii), here we have propan2olpropan - 2 - ol after making its conjugate base we see that there are two methyl groups (CH3)( - C{H_3}) which shows (+I)effect( + I)\,effect and thus due to which the ()ve( - )ve\, charge is more unstable than ethyl alcohol.
In the last option we have isopropyl alcohol which is having three methyl groups (CH3)( - C{H_3}) having (+I)effect( + I)\,effect by which electron density increases more on oxygen atom thus here also negative charge is not so stable.

Therefore, we can say that conjugate base is least stable. Among the three alcohols we can say the conjugate base is very stable in option (i) and thus ethyl alcohol is least acidic. While in isopropyl alcohol, the conjugate base is least stable and that is strongly acidic in nature.
Orderofacidicstrength:(iii)>(ii)>(i)Order\,of\,acidic\,strength:\,(iii) > \,(ii) > \,(i)
Thus the order of pKap{K_a} will be reverse of it by this formula pKaα1Acidicstrengthp{K_a}\,\alpha \,\,\dfrac{1}{{Acidic\,strength}}
OrderofpKa:(i)>(ii)>(iii)Order\,of\,p{K_a}:\,(i) > \,(ii) > \,(iii) Final answer.

Note: In case of amines we usually find out the basic strength, which is the ability to donate lone pairs of electrons on nitrogen. The amine which donates lone pair easily will be strong in nature and the amine which does not donate will have least basic strength. We can find the acidic strength on making the conjugate base and then checking the stability of the conjugate base.