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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?

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    $\begingroup$ You are ignoring the fact that the conc. of chloride is way too high than chromate. See this reference books.google.com/… $\endgroup$ – M. Farooq Dec 9 '19 at 4:31
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    $\begingroup$ I edited you question, chromate is $not$ a catalyst, it is an indicator in Mohr's titration. $\endgroup$ – M. Farooq Dec 9 '19 at 4:36
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    $\begingroup$ 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 '19 at 6:34
  • $\begingroup$ @M farooq oh yes i meant indicator what a horrible mistake! Thank you $\endgroup$ – user716591 Dec 9 '19 at 7:46
  • $\begingroup$ @Poutnik aha yes indeed how did i forget that! thank you!! $\endgroup$ – user716591 Dec 9 '19 at 7:48
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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.

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    $\begingroup$ Very good answer. For the OP, another alternative to Mohr's method is Vollhardt's method. $\endgroup$ – M. Farooq Dec 9 '19 at 7:42

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