# Will a precipitate form? Molar solubility question

• BrettJimison
In summary: The second step is to figure out what the OH concentration would be if some of the OH reacted with Al and Co. For that, you need to consider the equilibration of the reaction NH3+H2O=NH4 + OH. ChetThe second step is to figure out what the OH concentration would be if some of the OH reacted with Al and Co. For that, you need to consider the equilibration of the reaction NH3+H2O=NH4 + OH. ChetActually it is pretty simple - you just have to realize these are two problems, each of which can be solved separately.Thanks guys, I got it. Found [OH-] then just
BrettJimison

## Homework Statement

Hello all,
This is a problem on a worksheet I was given and I am stumped!
Statement:
Will a ppt form when 35.00 mL of 1.0x10^-3 M CoSO4 is mixed with 15.00 mL of 7.50x10^-4 M Al2(SO4)3 and 200 ml of a buffer which is .200 M NH3 and .200 M NH4Cl?

I do not need the answer, I just need a nudge in the right direction.

## Homework Equations

Ksp Al(OH)3 = 5x10^-33
Ksp Co(OH)2 = 3x10-16

## The Attempt at a Solution

I'm not even sure where to start. I would usually write out the net ionic eqn in this situation then use an ICE table but I'm not sure even what the ionic equation will look like here. We didnt go over molar solubility that much in my lecture, so all the problems were relatively straight forward till I got to this one. There just seems to be a lot going on. Any help? Thanks in advance.

BrettJimison said:

## Homework Statement

Hello all,
This is a problem on a worksheet I was given and I am stumped!
Statement:
Will a ppt form when 35.00 mL of 1.0x10^-3 M CoSO4 is mixed with 15.00 mL of 7.50x10^-4 M Al2(SO4)3 and 200 ml of a buffer which is .200 M NH3 and .200 M NH4Cl?

I do not need the answer, I just need a nudge in the right direction.

## Homework Equations

Ksp Al(OH)3 = 5x10^-33
Ksp Co(OH)2 = 3x10-16

## The Attempt at a Solution

I'm not even sure where to start. I would usually write out the net ionic eqn in this situation then use an ICE table but I'm not sure even what the ionic equation will look like here. We didnt go over molar solubility that much in my lecture, so all the problems were relatively straight forward till I got to this one. There just seems to be a lot going on. Any help? Thanks in advance.
The first step is to figure out what the OH concentration would be if none of the OH reacted with Al and Co. For that, you need to consider the equilibration of the reaction NH3+H2O=NH4 + OH. Don't forget that the two smaller solutions dilute the 200 ml solution to 260 ml.

Chet

Actually it is pretty simple - you just have to realize these are two problems, each of which can be solved separately.

Thanks guys, I got it. Found [OH-] then just used ice tables/ Ksp values to see which would precipitate first. Just needed a nudge in the right direction. Turns out Co(OH)2 ppts
-Brett

Based on the given information, it is difficult to determine if a precipitate will form without knowing the solubility products (Ksp) of the compounds involved. The Ksp values provided are for Al(OH)3 and Co(OH)2, but the compounds in the problem are CoSO4 and Al2(SO4)3. It is important to note that the solubility of a compound is not only dependent on its Ksp value, but also on its concentration and the concentrations of other ions in solution.

To solve this problem, you will need to calculate the molar solubility of each compound in the given solution. This can be done by using the Ksp values and the concentrations of the compounds in the solution. Once you have calculated the molar solubility of each compound, you can compare it to the initial concentrations given in the problem to determine if a precipitate will form.

In general, if the molar solubility of a compound is greater than its initial concentration, a precipitate will not form. However, if the molar solubility is less than the initial concentration, a precipitate may form. It is important to also consider the common ion effect, which states that the presence of a common ion (in this case, NH4+) can decrease the solubility of a compound.

Overall, to solve this problem, you will need to use the Ksp values, concentrations, and the common ion effect to determine the molar solubility of each compound and then compare it to the initial concentrations to determine if a precipitate will form.

## 1. What factors affect the formation of a precipitate?

Several factors can affect the formation of a precipitate, including temperature, concentration of the solutes, and the solubility product constant for the specific solute.

## 2. How can I predict if a precipitate will form?

To predict if a precipitate will form, you can use the solubility product constant (Ksp) and compare it to the ion product (IP) of the solution. If the IP is greater than the Ksp, a precipitate will form.

## 3. What is molar solubility and how is it calculated?

Molar solubility is the maximum amount of a solute that can dissolve in a specific amount of solvent at a given temperature. It is calculated by dividing the molar mass of the solute by the solubility in grams per liter (g/L).

## 4. How does the common ion effect impact molar solubility?

The common ion effect can decrease the molar solubility of a solute by increasing the concentration of one of its ions in the solution. This shifts the equilibrium towards the formation of a precipitate, resulting in a lower molar solubility.

## 5. Does the pH of a solution affect the molar solubility of a solute?

Yes, the pH of a solution can affect the molar solubility of a solute. This is because the pH can impact the solubility product constant (Ksp) and the concentration of ions in the solution, both of which play a role in determining the molar solubility.

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