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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?

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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.

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