To titrate $\ce{Cl-}$ with $\ce{Ag+}$ we use chromate $\ce{CrO4^2-}$ as an indicator. The titration reaction is:

$$\ce{Ag+ + Cl- <=> AgCl}\tag{R1}$$

$$K_1 = \frac{1}{K_\mathrm{sp}(\ce{AgCl})} = \frac{1}{1.8×10^{-10}} = 5.56×10^9\tag{1}$$

The theory says that after all $\ce{Ag+}$ are reacted with $\ce{Cl-}$ the end point of titration is detected when excess $\ce{Ag+}$ reacts with the indicator chromate to form silver chromate:

$$\ce{2 Ag+ + CrO4^2- <=> Ag2CrO4}\tag{R2}$$

$$K_2 = \frac{1}{K_\mathrm{sp}(\ce{Ag2CrO4})} = \frac{1}{1.1×10^{-12}} = 9.1×10^{11}\tag{2}$$

However, as you see, $K_1 < 100K_2,$ so when both $\ce{Cl-}$ and $\ce{CrO4^2-}$ are present, $\ce{Ag+}$ will react with $\ce{CrO4^2-}$ and not with $\ce{Cl-}$.

But our teacher and everywhere on Google they say $\ce{AgCl}$ precipitates before $\ce{AgCrO4}$. And that should be true since this method of titration (Mohr's method) has been used long ago.

But, how can that be true? I don't understand why. Where have I mistaken?

  • 1
    $\begingroup$ You are ignoring the fact that the conc. of chloride is way too high than chromate. See this reference books.google.com/… $\endgroup$
    – AChem
    Dec 9, 2019 at 4:31
  • 1
    $\begingroup$ I edited you question, chromate is $not$ a catalyst, it is an indicator in Mohr's titration. $\endgroup$
    – AChem
    Dec 9, 2019 at 4:36
  • 1
    $\begingroup$ Be aware you compare Ksp of a binary and a ternary product. In such case, lower Ksp of the latter does not automatically mean it is less soluble. Do calculations for real analysis and you will see. $\endgroup$
    – Poutnik
    Dec 9, 2019 at 6:34
  • $\begingroup$ @M farooq oh yes i meant indicator what a horrible mistake! Thank you $\endgroup$
    – user208973
    Dec 9, 2019 at 7:46
  • $\begingroup$ @Poutnik aha yes indeed how did i forget that! thank you!! $\endgroup$
    – user208973
    Dec 9, 2019 at 7:48

1 Answer 1


You got the solubility part reversed. The solubility of $\ce{AgCl}$ is lower than the solubility of $\ce{Ag2CrO4}:$

$$s(\ce{AgCl}) = \sqrt{K_\mathrm{sp}(\ce{AgCl})} = \sqrt{\pu{1.8E-10 mol2 L-2}} = \pu{1.34E-5 mol L-1}$$

$$s(\ce{Ag2CrO4}) = \sqrt[3]{\frac{K_\mathrm{sp}(\ce{Ag2CrO4})}{4}} = \sqrt[3]{\frac{\pu{1.1E-12 mol3 L-3}}{4}} = \pu{6.50E-5 mol L-1}$$

Therefore, if the $\ce{AgNO3}$ solution is gradually added to the solution containing the both $\ce{Cl-}$ and $\ce{CrO4^2-}$ ions, then initially the formation of a sparingly soluble $\ce{AgCl}$ salt occurs. After the $\ce{Cl-}$ ions are almost completely isolated in the form of $\ce{AgCl},$ the $\ce{Ag2CrO4}$ precipitation starts to occur, signifying the equivalence point is reached.

The same reasoning can also be applied to the titration of even less soluble silver bromide $\ce{AgBr}$ with $K_\mathrm{sp}(\ce{AgBr}) = \pu{5.3E-13}$ (try it yourself).

Note, however, that while using Mohr's method it's imperative to titrate halide salts solutions with $\ce{AgNO3}$ and not vice versa. Otherwise the precipitation condition

$$c(\ce{Ag+}) · c(\ce{Cl-}) > K_\mathrm{sp}(\ce{AgCl})$$

will be overridden by

$$c(\ce{Ag+})^2 · c(\ce{CrO4^2-}) > K_\mathrm{sp}(\ce{Ag2CrO4})$$

due to high concentration of silver ions in solution, favoring silver chromate precipitation (note squared term $c(\ce{Ag+})^2$) and thus shifting the equivalence point.

  • 1
    $\begingroup$ Very good answer. For the OP, another alternative to Mohr's method is Vollhardt's method. $\endgroup$
    – AChem
    Dec 9, 2019 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.