My experiment involves a potentiometric titration, and I have used $\ce{AgNO3}$ as a titrant and KCl/KI/deionised water only as an analyte.

In the image below of my data, the potential difference as recorded by a multimeter drops from 0.6V to 0.15V.

Volume of Silver Nitrate added/cm^3 along the x axis.

What causes the drop in potential difference when the silver nitrate and whatever analyte (all exhibited the same pattern of a sudden drop in PD) reaches its endpoint in the titration?

Why is there a higher potential difference when using KI as analyte than KCl?

  • $\begingroup$ What is Your problem? Conductivity as stated in the headline or potential? $\endgroup$
    – Georg
    Feb 2, 2015 at 11:29
  • $\begingroup$ Both essentially, how do they correlate to each other? $\endgroup$
    – Talisman
    Feb 2, 2015 at 13:19

1 Answer 1


As you add $\rm AgNO_3$ to $\rm KCl$, solid $\rm AgCl$ forms. The silver ion you add will at first be completely consumed by the chloride. Once all of the chloride is locked up in the precipitate, the silver ion concentration will rapidly increase. The Nernst equation shows how the potential is expected to change when this happens. If you're using a silver/silver chloride electrode with a reference electrode, the indicator potential is

$$\rm E = E^\circ_{\rm AgCl,Ag} - \frac{R T}{\cal F}\ln\frac{K_{sp}}{a_{\rm Ag^+}}$$

To see whether you'll get a drop or a rise in $E$, ask yourself what happens when silver concentration rises. If you've got the leads hooked up to your multimeter properly (negative lead to the silver cathode, positive lead to your anode), shouldn't you be seeing an increase in potential? What exactly are your $x$ and $y$ axes on that graph?

You'll get a bigger change when you form $\rm AgI$, because $\rm AgCl$ is about a thousand times more soluble than $\rm AgI$. The $\rm AgI$ will lock up relatively more ions when it forms (the $K_{sp}$'s of $\rm AgCl$ and $\rm AgI$ are $1.8\times 10^{-10}$ and $1.5\times 10^{-16}$, respectively).

Addendum 1: You say you're just using a silver electrode; the indicator potential is

$$\rm E = E^\circ_{\rm Ag^+,Ag} - \frac{R T}{\cal F}\ln\frac{1}{a_{\rm Ag^+}}$$

(but if you dip a silver wire into a chloride solution you'll get a very thin layer of AgCl and it'll begin to be a silver/silver chloride electrode (Source: G. D. Christian, Analytical Chemistry, 4th ed., Wiley 1986, p. 282)).

You also say you don't have a true reference electrode, just a graphite electrode... the fact that its potential can change as well complicates things. Your cell potential will depend on what the oxidation process is at the anode.

Addendum 2: You asked for a source for the $K_{sp}$'s. CRC Handbook 58th edition gives $K_{sp}$'s at $25^\circ$C for AgCl and AgI as $1.56\times 10^{-10}$ and $1.5\times 10^{-16}$; the 84th edition gives $1.77\times 10^{-10}$ and $8.52\times 10^{-17}$; Keller's Basic Tables in Chemistry gives them as $1.0\times 10^{-10}$ and $1.0\times 10^{-16}$ for "temperatures near 298 K". They're very temperature dependent and they depend on conditions like ionic strength as well, so it's not surprising that the values you're finding on the web are all over the place. I took the values above from a genchem text that happened to be lying on my desk.

  • $\begingroup$ Thanks for the response Fred. So what does the chloride ion have that the nitrate ion doesn't? Otherwise, there'd be no change surely? $\endgroup$
    – Talisman
    Jan 31, 2015 at 19:22
  • $\begingroup$ Do you mean, why does AgCl precipitate when AgNO$_3$ doesn't? $\endgroup$ Jan 31, 2015 at 19:32
  • $\begingroup$ No, why can the nitrate ions not conduct the electrical current like the chloride ions can? $\endgroup$
    – Talisman
    Jan 31, 2015 at 19:45
  • $\begingroup$ OK, the issue is not conductivity so much as what ions your electrodes can sense directly to produce potential changes. If you're using a silver indicator electrode it mostly only "sees" changes in the concentration of the $\rm Ag^+$ ions. $\endgroup$ Jan 31, 2015 at 19:57
  • 1
    $\begingroup$ ...added to answer. $\endgroup$ Feb 1, 2015 at 19:55

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