During titration of a weak acid with a strong base, the conductance increases due to the formation of a salt and further ionization of the strong base.




Conductometric titration is a method used to measure the electrical conductivity of a solution as a reagent is added. In this context, we're looking at the conductometric titration of benzoic acid with sodium hydroxide.
Conductometric titration involves observing changes in conductivity as the titration progresses. For the titration of a weak acid like benzoic acid with a strong base like sodium hydroxide, the curve typically shows a decrease in conductivity followed by an increase.
From the options given, the correct graph representation would depict an initial decrease followed by an increase after the equivalence point. Thus, the option with correctly represents this behavior.
The correct answer is (C) : 

(A) → (B) Free \(H^+\) ions are replaced by \(Na^⊕\) which decreases conductance.
(B) → (C) Un-dissociated benzoic acid reacts with \(NaOH\) and forms salt which increases ions & conductance increases.
(C) → (D) After equivalence point at (3), \(NaOH\) added further increases \(Na^⊕\) & \(OH^⊙\) ions which further increases the conductance.
This graph logically represents the conductometric titration process for benzoic acid against sodium hydroxide, making Option 3 the correct answer.
Rods $x$ and $y$ of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^\circ$C and $40^\circ$C respectively. Given the thermal conductivity of rod $x$ is three times of that of rod $y$, the temperature at junction points $B$ and $E$ are (close to): 
Given below are two statements:
Statement I: Mohr's salt is composed of only three types of ions—ferrous, ammonium, and sulphate.
Statement II: If the molar conductance at infinite dilution of ferrous, ammonium, and sulphate ions are $ x_1 $, $ x_2 $, and $ x_3 $ $ \text{S cm}^2 \, \text{mol}^{-1} $, respectively, then the molar conductance for Mohr's salt solution at infinite dilution would be given by $ x_1 + x_2 + 2x_3 $.