# Calculating amount of precipitate that will form

• ultimateguy
In summary, there is a zero chance of guessing the answer because the problem gives no information about how to calculate the amount of precipitate that will form.

#### ultimateguy

20,0ml of 0,150M Zn(NO3)2 are mixed with 20,0ml of 1,00 x10^-4
M Na2S. Calculate the amount of milligrams of
ZnS (s) that will precipitate. (Ksp = 1,6 x10^-24
M^2 )

a) 2,68mg
b) 0,347mg
c) 4,85mg
d) 0,194mg

There are zero examples in my textbook or notes that deal with calculating the amount of precipitate that will form. Can someone help?

Does the problem description give a large excess of one of the reactants? What can this do the the equilibrium of the reaction? Is the product very very insoluble based on the size of the solubility product constant?

1) Yes,

2) Makes it small

3) Yes

I think I got it. I calculated 4.85mg by finding the new concentrations with the new volume and then using the smaller concentration to find the amount in grams.

Show your calculations. Remember you only have a 25% chance if you quess the answer.

[Zn] = (0.15M x 0.02L)/(0.04L) = 0.075M

= (1x10^-4M x 0.02L)/(0.04L) = 5 x 10^-5M

Then multiplying 5 x 10^-5M by molar mass of ZnS and finding the amount of grams in 0.04L I got 4.85mg.

Try again. Why are you multiplying a concentration by a formula wt?

Units would be grams/liter...

If you have already decided that one of the reagents is limiting and that the concentration of ZnS in solution is insignificant (something times 10^-23 mol/L, I believe), then the amount of precipitate will essentially be equal to the number of moles of the limiting reagent.

chemisttree said:
If you have already decided that one of the reagents is limiting and that the concentration of ZnS in solution is insignificant (something times 10^-23 mol/L, I believe), then the amount of precipitate will essentially be equal to the number of moles of the limiting reagent.

Isn't that what I did? S is limiting, so I took the mass of that, which is 4.85mg.

moles S = (1 X 10^-4 mol/L)*(0.02L) = 2 X 10^-6 mol S

mass of S (g) = (2 X 10^-6 mol S)*(FW ZnS)

mass of ZnS (mg) = 1000 * mass ZnS (g)

You found the amount in grams in 0.02 L. The final volume is 0.04 L.

Last edited:
Ah I see it now. Thanks. (Now how about that other thread? :P)

## 1. How do you calculate the amount of precipitate that will form?

The amount of precipitate that will form can be calculated using the formula: mass of precipitate = (volume of solution) x (concentration of precipitating reagent) x (molar mass of precipitate).

## 2. What is the purpose of calculating the amount of precipitate that will form?

Calculating the amount of precipitate that will form helps determine the efficiency of a chemical reaction and can also help with determining the appropriate amount of reagents to use in a reaction.

## 3. What factors can affect the amount of precipitate that will form?

The amount of precipitate that will form can be affected by factors such as the concentration of the reactants, temperature, and pH of the solution.

## 4. Can the amount of precipitate that will form be accurately predicted?

While the amount of precipitate that will form can be estimated, it is not always possible to accurately predict the exact amount due to factors such as human error and variations in experimental conditions.

## 5. Is it possible for no precipitate to form in a reaction?

Yes, it is possible for no precipitate to form in a reaction if the reactants are not in the correct ratios or if the conditions are not suitable for a precipitation reaction to occur.