During the conductometric titration of HCl against NaOH, the conductance decreases before the end point because

A $H X^{+}$ ions of high mobility are replaced by $Na X^{+}$ ions of much lower mobility
Bthe number of $Cl X^{-}$ ions in the solution decreases
Cthe water formed in the reaction dilutes the solution appreciably
D $Na X^{+}$ ions have higher mobility than $H X^{+}$ ions

Answer: A

H X^{+} + OH X^{-} H X_{2} O

removes one

H X^{+}

for every

Na X^{+}

added, so the total number of ions stays constant while

H X^{+}

(

λ^{\circ} \approx 350 S c m^{2} mo l^{- 1}

) is exchanged for

Na X^{+}

(

λ^{\circ} \approx 50

). The

Cl X^{-}

count does not change, and the water formed is negligible compared with the solvent.

Na X^{+}

is much less mobile than

H X^{+}

, so (D) is the reverse of the truth.