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In Mohr Titration, Why does AgCl and not Ag2CrO4 Precipitate First?

March 30th, 2008

Chemistry Tutor Site - Question

In Mohr titration using Ag ion, why does AgCl and not Ag₂CrO₄ that precipitates first, although the Ksp value for AgCl (10^-10) is greater than Ksp Ag₂CrO₄ ( 10^-12)? If Ag₂CrO₄ is less soluble than AgCl why this happen when titrated with silver and chromate as indicator?

Chemistry Tutor Site – Answer

You can not use the Ksp value for determine which ones of the species are less soluble than the other, and conclude that the smaller Ksp value is will precipitate fisrt. The best way to compare the solubility of the species is using the “solubility” rather than Ksp value.

For the practical approach when you do the Mohr titration, you must see that AgCl will precipitate first rather than Ag₂CrO₄, this circumstance giving us a hypothesis that silver chromate is more soluble than silver chloride, but how we can make a theoretical calculation to show that our hypothesis is right? You can do this by calculating the solubility (s) for each species:

Ksp Ag₂CrO₄ = [Ag⁺]²[CrO4^-2] = 1.1×10^-12

let x be the molarity of CrO4^-2, we know then that the molarity of the silver would be twice that value. Therefore

[2x]²[x] = 1.1×10^-12 , solve for x
4x².x = 1.1×10-12
4x³=1.1×10^-12
x= 6.51×10^-5 M

and therefore is the same as the number of moles of Ag2CrO4 that is in solution at the saturation point. This equal to:

=6.51×10^-5 mole Ag2CrO4/liter * 331.73 g/mole
= 2.15×10^-2 g, that is about 20 mg in a liter.

What about AgCL solubilty?

Ksp AgCl = 10^-10, that works out to be 1.26×10^-5 Molar, and the mass per litter is

= (1.26×10^-5 moles/liter * 143.32g/mole )
= 1.81×10^-3 g, that is almost 2 mg in a liter.

from the calculation above we approve that AgCrO4 is less soluble than AgCl and because of these AgCl will be precipitate first when we perform Mohr titration.

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