Equivalent conductance of 0.1 M HA (weak acid) solution is 10 Scm2 equivalent - 1 and that at infinite dilution is 200 Scm2 equivalent - 1 . Hence pH of HA solution is

Question

Equivalent conductance of 0.1 M HA (weak acid) solution is 10 Scm2 equivalent - 1 and that at infinite dilution is 200 Scm2 equivalent - 1 . Hence pH of HA solution is (A) 1.3 (B) 1.7 (C) 2.3 (D) 3.7

Tardigrade Admin · Accepted Answer

α =( Lambda/ Lambda∈ fty)=(10/200)=(1/20) text[H text+ text] textweak textacid text= textCÎ± =0.1× ((1/20))=5× 10- 3 pH=3-log5=2.3

Q. Equivalent conductance of 0.1 M HA (weak acid) solution is 10 $S c m^{2}$ equivalent $^{- 1}$ and that at infinite dilution is 200 $S c m^{2}$ equivalent $^{- 1}$ . Hence pH of HA solution is

1.3

1.7

2.3

3.7

$α = \frac{Λ}{Λ _{\in f t y}} = \frac{10}{200} = \frac{1}{20}$
$[H^{+}]_{weak acid} = Cα$
$= 0.1 \times (\frac{1}{20}) = 5 \times 1 0^{- 3}$
$p H = 3 - l o g 5 = 2.3$

Q. Equivalent conductance of 0.1 M HA (weak acid) solution is 10 Scm2 equivalent −1 and that at infinite dilution is 200 Scm2 equivalent −1 . Hence pH of HA solution is

1.3

1.7

2.3

3.7

α=Λ∈fty​Λ​=20010​=201​ [H+]weakacid​=Cα =0.1×(201​)=5×10−3 pH=3−log5=2.3

Q. Equivalent conductance of 0.1 M HA (weak acid) solution is 10 $S c m^{2}$ equivalent $^{- 1}$ and that at infinite dilution is 200 $S c m^{2}$ equivalent $^{- 1}$ . Hence pH of HA solution is

$α = \frac{Λ}{Λ _{\in f t y}} = \frac{10}{200} = \frac{1}{20}$
$[H^{+}]_{weak acid} = Cα$
$= 0.1 \times (\frac{1}{20}) = 5 \times 1 0^{- 3}$
$p H = 3 - l o g 5 = 2.3$