# How to determine the dynamic of progression of two simultaneous precipitation (double replacement) reactions? [duplicate]

We have a solution of mineral water with potassium chromate added to it, and want to titrate the chloride ions in the mineral water. We were told to use silver nitrate for this, which would produce the two following reactions:

\begin{align} \ce{AgNO3 _{(aq)} + Cl- _{(aq)} & <--> AgCl_{(s)}} + \text{some nitrates} \\ \ce{2AgNO3 _{(aq)} + K2CrO4 _{(aq)} & <--> Ag2CrO4_{(s)} + 2KNO3_{(s)}} \end{align}

And namely we interest in:

\begin{align} \ce{Ag+ _{(aq)} + Cl- _{(aq)} & <--> AgCl_{(s)} v} \label{AgClX} \\ \ce{2Ag+ _{(aq)} + CrO4^{2-} _{(aq)} & <--> Ag2CrO4_{(s)} v} \label{AgCrX} \end{align}

This titration hinges on two facts:

1. The first reaction will progress to near completion in the forwards direction.
2. The first reaction will nearly finish before the second one progresses appreciably, so we can observe equivalence by noticing the color of the silver chromate precipitate (brick-red).

I would like to understand why are these two facts true? How do I show that the solubility of silver chromate and silver nitrate causes this behavior?

Consider that solubility of $$\ce{AgCl}$$ is 1.9 mg/L and of $$\ce{Ag2CrO4}$$ is 22 mg/L, both at 25 °C.
When just a little $$\ce{AgNO3}$$ has been added, the $$\ce{Ag2CrO4}$$ is not concentrated enough to precipitate -- it's still in solution. The $$\ce{AgCl}$$ falls out of solution at a far lower concentration.
BTW, it's possible to dissolve some of the precipitated $$\ce{AgCl}$$ by adding more $$\ce{Cl-}$$, e.g., more $$\ce{NaCl}$$, forming the soluble $$\ce{AgCl2-}$$ ion. That principle was used to "fix" photographic images by washing with excess salt water, e.g., when at sea.