The conductance of 0 .0015 M aqueous solution of a weak monobasic...

The conductance of

0 .0015 M

aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized

Pt

electrodes. The distance between the electrodes is

120cm

with an area of cross-section of

1{cm}^{2}

. The conductance of this solution was found to be

5 \times {10}^{- 7}S

. The

pH

of the solution is

4

. The value of limiting molar conductivity of this weak monobasic acid in aqueous solution is

Z \times {10}^{2}S{cm}^{- 1}{mol}^{- 1} .

The value of

Z

is.

Correct answer

Step-by-step solution

For weak acid, $\left[{H}^{+}\right] = c \alpha = c \times {∧}_{m}^{c}{∧}_{m}^{\infty}$

$= c \times k \times \dfrac{1000}{c}{∧}_{m}^{\infty} = k \times 1000{∧}_{m}^{\infty} = G \times \left(\dfrac{l}{a}\right) \times 1000{∧}_{m}^{\infty}$

${10}^{- 4} = \dfrac{5 \times {10}^{- 7}\left(\dfrac{120}{1}\right) \times 1000}{Z \times {10}^{2}} ⇒ Z = 6$

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