The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is text0 text.005 textS textm text2 textmo textl text-1 and the limiting molar conductivity of HA is text0 text.005 textS textm text2 textmo textl text-1 at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka) of HA at this temperature is

Question

The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is text0 text.005 textS textm text2 textmo textl text-1 and the limiting molar conductivity of HA is text0 text.005 textS textm text2 textmo textl text-1 at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka) of HA at this temperature is (A)  1× 10-4  (B)  0.1  (C)  9× 10-4  (D)  1.1× 10-5

Tardigrade Admin · Accepted Answer

λ mo=0.05Sm2mol-1 C= 0.009 m λ m=0.05S m2mol-1 α =(λ m/λ mo)=0.1 For weak acid, Ka=(α 2c/1-α )=((0.1)2× 0.009/(1-0.1))=10-4

Q. The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is $0 .005 S m^{2} mo l^{-1}$ and the limiting molar conductivity of HA is $0 .005 S m^{2} mo l^{-1}$ at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant $(K_{a})$ of HA at this temperature is

$1 \times 10^{- 4}$

$0.1$

$9 \times 10^{- 4}$

$1.1 \times 10^{- 5}$

$λ_{m^{o}} = 0.05 S m^{2} m o l^{- 1}$ C= 0.009 m $λ_{m} = 0.05 S m^{2} m o l^{- 1}$ $α = \frac{λ _{m}}{λ _{m^{o}}} = 0.1$ For weak acid, $K_{a} = \frac{α ^{2} c}{1 - α} = \frac{( 0.1 ) ^{2} \times 0.009}{( 1 - 0.1 )} = 10^{- 4}$

Q. The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is 0.005Sm2mol-1 and the limiting molar conductivity of HA is 0.005Sm2mol-1 at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka​) of HA at this temperature is

1×10−4

0.1

9×10−4

1.1×10−5