Calculating amount of precipitate that will form

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Discussion Overview

The discussion revolves around calculating the amount of precipitate, specifically zinc sulfide (ZnS), that forms when zinc nitrate and sodium sulfide solutions are mixed. Participants explore the stoichiometry involved in the reaction, the concept of limiting reagents, and the implications of the solubility product constant (Ksp) on the precipitation process.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the mass of ZnS precipitate, suggesting 4.85 mg based on new concentrations after mixing.
  • Another participant questions the initial calculations and emphasizes the importance of determining the limiting reagent and the significance of the solubility product constant in the context of the reaction equilibrium.
  • A participant provides a formula for calculating the concentration of zinc and sulfide ions after dilution, leading to a discussion about the limiting reagent.
  • There is a correction regarding the method of calculating the mass of precipitate, with a focus on the need to consider the final volume of the solution.
  • Some participants express uncertainty about the correct approach to find the mass of the precipitate, with one suggesting that the amount of precipitate is equal to the moles of the limiting reagent.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method for calculating the amount of precipitate. There are competing views on the limiting reagent and the appropriate calculations to use.

Contextual Notes

There are unresolved aspects regarding the assumptions made about the limiting reagents and the calculations involving concentrations and final volumes. The discussion reflects varying levels of understanding of the stoichiometric principles involved.

Who May Find This Useful

This discussion may be useful for students or individuals studying chemistry, particularly those interested in precipitation reactions, stoichiometry, and the application of solubility product constants in calculations.

ultimateguy
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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?
 
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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:
  • #10
Ah I see it now. Thanks. (Now how about that other thread? :P)
 

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