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I don’t understand why the conductivity doesn’t change after the equivalence point. To my understanding, at the equivalence point there is $\ce{NH3}$, $\ce{NH4+}$ and $\ce{Cl-}$ in the solution. As more $\ce{NH3}$ is added it only partially dissociates, thus conductivity stays the same. But doesn’t the concentration of $\ce{Cl-}$ ions also contribute to the conductivity, and its conductivity decreases are more aqueous $\ce{NH3}$ is added as it becomes more dilute? Where is the issue in my understanding. Thanks for the help.

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You are missing an important point. Conductometric titration is not just plotting the conductance after adding the titrant. One cannot plot conductance as it is, because you are significantly diluting the solution. You have to apply a correction to take dilution into account. Once you apply a correction, the curve will remain flat because further addition of ammonia will not enhance conductivity.

See for example,

Conductometric titrations

Throughout a titration the volume of the solution is always increasing, unless the conductance is corrected for this effect, non linear titration curves result. The correction can be accomplished by multiplying the observed conductance either by total volume (V+V') or by the factor (V+ V')/V, where V is the initial volume of solution and V' is the total volume of the reagent added. The correction presupposes that the conductivity is a linear function of dilution, this is true only to a first approximation.

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    $\begingroup$ Thank you. We learnt it very briefly at school and this helped me make sense of it. $\endgroup$
    – Subbota
    Commented Jan 14, 2021 at 9:04
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After equivalence point you add more and more NH3 solution (remember NH4OH does not exist). So you add a negligible amount of ions. As a consequence, the conductance does not change significantly. This is what your diagram shows.

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  • $\begingroup$ But wouldn’t conductivity decrease because of lowered Cl - concentration? $\endgroup$
    – Subbota
    Commented Jan 13, 2021 at 12:58
  • $\begingroup$ @Subbota It would and will, but see the other comment. Its decrease is insignificant, compared to replacing highly mobile hydronium ions by relatively slow ammonium ions. $\endgroup$
    – Poutnik
    Commented Jan 13, 2021 at 14:02

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