# How much Silver Nitrate to produce a precipitate?

• tobywashere
In summary, to produce a precipitate in a solution with 2.50L of 0.020M K2CrO4, a minimum of 0.0090 grams of AgNO3 needs to be added. This can be determined by using the solubility product constant, Ksp Ag2CrO4, and the stoichiometry of the precipitate reaction.
tobywashere

## Homework Statement

What is the minimum amount of AgNO3, in grams, that has to be added to 2.50L of 0.020M K2CrO4 to produce a precipitate?

## Homework Equations

Ksp Ag2CrO4 = 9.0 x 10-12
[Ag+]2[CrO4-2] = 9.0x10-12

## The Attempt at a Solution

2AgNO3(aq) + K2CrO4(aq) --> Ag2CrO4(s) + 2KNO3(aq)
Ag2CrO4(s) is the precipitate
The net ionic equation is:
Ag2CrO4(s) --> 2Ag+(aq) + CrO4-2(aq)
Let y represent the concentration of CrO4-2 at equilibrium. The concentration of Ag+ will be 2y due to the 2:1 stoichiometry. Plugging into the solubility equilibrium expression:
[Ag+]2[CrO4-2] = 9.0x10-12
(2y)2(y) = 9.0x10-12
Therefore, y = 1.31 x 10-4 M
So the equilibrium concentration of CrO4-2 is 1.31 x 10-4 M, and the equilibrium concentration of Ag+ is double that, or 2y, or 2.62x10-4M
The initial concentration of CrO4-2 is 0.020M (from the 0.020M K2CrO4). The decrease in concentration of CrO4-2 to form the precipitate is:
0.020M - 1.31 x 10-4M = 0.01987M
The initial concentration of Ag+ (what we're trying to find) decreases by twice this amount, 0.03974M, to reach an equilibrium concentration of 2.62x10-4M (the 2y found earlier).
Initial concentration of Ag+ - 0.03974M = 2.62x10-4M.
Initial concentration = 2.62x10-4M + 0.0397M = 0.040M
Find mass:
0.040M x 2.5L x 170g / mol = 17g
Therefore, 17 g of AgNO3 needs to be added to form a precipitate.
Did I do this correctly? I'm really not sure.<br>

No, it is wrong. You add silver to the solution that contains chromate. Nothing consumes chromate, so its concentration stays constant, and concentration of Ag+ goes up - till you reach solubility product.

Technically when adding silver nitrate as a solution you would dilute chromate a little bit, but we can assume you add it as a solid, not changing solution volume.

Oh ok. So you just need 2.62x10-4M of AgNO3.
2.62x10-4M x 2.50L x 170g/mol = 0.11135 grams?

No, concentration of silver is much lower than that.

Ksp defines saturated solution concentrations, it doesn't mean ratio of concentrations is given by precipitate stoichiometry. Ksp for AgCl is about 10-10. If you put solid AgCl into water, saturated solution will be 10-5M both in [Ag+] and [Cl-]. However, saturated solution can be also 1M in [Ag+] and 10-10M in [Cl-], or - say - 10-3M in [Ag+] and 10-7M in [Cl-] - if it was prepared by mixing solutions containing these ions in correct quantities.

Oh so I made this problem unnecessarily complicated.
[0.020M][Ag+]2 = 9.0x10-12
[Ag+]2 = 2.12x10-5M
2.12x10-5M x 2.5L x 170g/mol = 0.0090 g

Is that right?
10 char

Yes.

## 1. How is the amount of silver nitrate needed to produce a precipitate determined?

The amount of silver nitrate needed to produce a precipitate is determined by the amount of the other reactant present, as well as the stoichiometric ratio between the two reactants. This can be calculated using the molar mass and molar ratio of the reactants.

## 2. Can the amount of silver nitrate needed to produce a precipitate vary depending on the reaction conditions?

Yes, the amount of silver nitrate needed to produce a precipitate can vary depending on factors such as temperature, pressure, and the presence of catalysts. These factors can affect the rate of the reaction and the efficiency of the precipitation process.

## 3. Is it possible to use too much silver nitrate in the precipitation process?

Yes, using too much silver nitrate can result in excess unreacted silver nitrate in the final product, which can be costly and inefficient. It is important to use the correct amount of silver nitrate to ensure maximum yield and purity of the precipitate.

## 4. How can I accurately measure the amount of silver nitrate needed for precipitation?

The most accurate way to measure the amount of silver nitrate needed for precipitation is by using a balance to measure the mass of the silver nitrate. Alternatively, volumetric methods such as titration can also be used to accurately measure the amount of silver nitrate needed.

## 5. Can the amount of silver nitrate needed for precipitation be determined by visual observation?

No, the amount of silver nitrate needed for precipitation cannot be accurately determined by visual observation. It is important to use precise measurements and calculations to ensure the proper amount of silver nitrate is added for the desired result.

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