# Why does silver react preferentially with chlorine instead of chromate?

Mohr's method of precipitation titration relies on the fact that silver reacts preferentially with chloride ions instead of chromate ions in solution. When I searched up why this is true, the only response was that the $$\text{Ksp of }\ce{AgCl}$$ ($$1.77\times10^{-10}$$) is greater than the $$\text{Ksp of }\ce{Ag_2CrO_4}$$ ($$1.2\times10^{-12}$$) seen here : https://www.answers.com/earth-science/Why_does_silver_nitrate_combine_first_with_chloride_ions_in_water_and_not_with_potassium_dichromate.

However this doesn't make sense to me since shouldn't the more soluble a compound is indicate that it is less likely to react and form a precipitate meaning that by this logic shouldn't $$\ce{Ag_2CrO_4}$$ be preferential not $$\ce{AgCl}$$? Could someone explain what is wrong with this reasoning and how to actually explain why silver reacts preferentially with chloride ions?

The idea of $$K_{sp}$$ is just the equilibrium constant for a dissolving reaction of a solid in liquid. $$aA_{s}\leftrightharpoons cC_{aq} + dD_{aq}$$ $$K_{sp}=[C]^c[D]^d$$ But, another concept is the solubility $$s$$ of a substance. As you calculate the solubility of a salt you should take into account the stoichiometry of the reaction. $$AgCl_{s}\rightleftharpoons Ag^+_{aq}+Cl^-_{aq}$$ $$K_{sp}=[Ag^+][Cl^-]$$ $$s^2 = K_{sp} \rightarrow s = \sqrt{K_{sp}} = \ce{1.33·10^{-5}}$$ On the other hand for the silver chromate reaction we get: $$Ag_2CrO_4\leftrightharpoons 2Ag^+_{aq}+\ce{CrO_4^-_{aq}}$$ $$K_{sp}=[2s]^2[s]=4s^3\rightarrow s = (K_{sp}/4)^{1/3}=6.69· \ce{10^{-5}}$$ And we get that the silver chloride is less soluble than the the silver chromate due to the stoichometry of the two reactions.