# Is chromate a suitable indicator for the titration of Ag⁺ with Cl⁻?

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?

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

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

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

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

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.

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