JEE MAIN - Chemistry (2019 - 12th April Evening Slot - No. 17)
The decreasing order of electrical conductivity of the following aqueous solutions is :
0.1 M Formic acid (A),
0.1 M Acetic acid (B),
0.1 M Benzoic acid (C)
0.1 M Formic acid (A),
0.1 M Acetic acid (B),
0.1 M Benzoic acid (C)
A > B > C
C > B > A
A > C > B
C > A > B
Explanation
0.1 M Formic acid(HCOOH) (A),
0.1 M Acetic acid(CH3COOH) (B),
0.1 M Benzoic acid(C6H5COOH) (C)
Conductivity (K) depends on no of ions present in unit volume of solution. When no of ions increases conductivity also increases.
Also no of ions depends on degree of dissociation($$\alpha $$).
We know, $$\alpha $$ = $$\sqrt {{{{K_a}} \over C}} $$
For all solutions C = 0.1 M
So, $$\alpha $$ depends on only K$$a$$ here.
K$$a$$ of HCOOH = 10-4
K$$a$$ of CH3COOH = 1.8 $$ \times $$ 10-5
and K$$a$$ of C6H5COOH = 6 $$ \times $$ 10-5
$$ \therefore $$ $$\alpha $$(HCOOH) > $$\alpha $$(C6H5COOH) > $$\alpha $$(CH3COOH)
Therefore conductivity order = A > C > B
0.1 M Acetic acid(CH3COOH) (B),
0.1 M Benzoic acid(C6H5COOH) (C)
Conductivity (K) depends on no of ions present in unit volume of solution. When no of ions increases conductivity also increases.
Also no of ions depends on degree of dissociation($$\alpha $$).
We know, $$\alpha $$ = $$\sqrt {{{{K_a}} \over C}} $$
For all solutions C = 0.1 M
So, $$\alpha $$ depends on only K$$a$$ here.
K$$a$$ of HCOOH = 10-4
K$$a$$ of CH3COOH = 1.8 $$ \times $$ 10-5
and K$$a$$ of C6H5COOH = 6 $$ \times $$ 10-5
$$ \therefore $$ $$\alpha $$(HCOOH) > $$\alpha $$(C6H5COOH) > $$\alpha $$(CH3COOH)
Therefore conductivity order = A > C > B
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