March 1st, 2018, 09:15 AM  #1 
Newbie Joined: Feb 2018 From: Iran Posts: 16 Thanks: 2  Solubility equilibrium I am new to solubility equilbrium Will this is a worked example in my book and I 've got confused by something 1:if Q is the product of concentrations of Ag and Cl in the solution and ksp is the product of concentration of dissolved Ag and Cl why the difference of these two doesn't give us the concentration of Ag and Cl which didnt dissolved (precipitated ) 2: I don't really get the part b I don't understand what is going on there if we weren't given the value of ksp can we find it from part b? (Sorry if my questions are a little stupid because I am new to this subject and I have no teacher so I have to study by myself) 
March 1st, 2018, 09:41 AM  #2 
Senior Member Joined: Jun 2015 From: England Posts: 764 Thanks: 220 
If you would like to post a large enough scan or post it in larger pieces then someone might be able to help.

March 1st, 2018, 08:18 PM  #3  
Newbie Joined: Feb 2018 From: Iran Posts: 16 Thanks: 2  Quote:
Last edited by Elize; March 1st, 2018 at 08:30 PM.  
March 3rd, 2018, 02:26 PM  #4  
Senior Member Joined: Jun 2015 From: England Posts: 764 Thanks: 220  Quote:
I am not suprised you got confused by it. I am going to show you a standard way that will work for many other calculations besides solubility products. I have summarised the working in the attachment and provide extended explanation in this text. So you need both together. I hope you can read my writing, but you would have to wait longer if I typed out all those equations. So the first thing is to decide what chemical equation you need to work on. This is at the top of the sheet and I repeat it in words Add 10ml of silver nitrate solution to 10 ml of sodium chloride solution to precipitate g grams of silver chloride, leaving silver, sodium, chloride and nitrate ions in solution. Now I have written it as though you add the solid salts to 20 ml of water for a reason. You don't want to complicate things by considering before and after concentrations. That would add 2 extra equations to be solved for no benefit. Looking first at the right hand side of this equation we have unknown concentrations of 4 ions plus an unknown weight of precipitate. That is 5 unknowns. To this we add another unknown the solubility product Ksp. This brings the total up to six unknowns. I hope you know enough maths to understand that in general you require the same number of equations as unknowns to solve for the unknowns. So we are looking for 6 equations. However things are not as bad as that sounds as we shall see. All sodium compounds and silver nitrate are very soluble. So both the silver nitrate and the sodium chloride ionise completely in solution. Since all sodium compounds are soluble no sodium compound is precipitated so the sodium remains in solution. However the volume of that solution is doubled so the final concentration is half the original. This is expressed by equation number 1, solving immediately for one of the unknowns on the right hand side of the chemical equation. Similarly for the nitrate ion  we have just ionised it so it will not precipitate with the silver or the sodium. This is expressed by equation 2 So now we have 2 of our 6 unknowns. Equation 3 is just the definition of the solubility product and is expressed in terms of two more of the unknowns  the concentrations of the silver and the chloride ions. this tells us we need to find them. Now we need to look at other properties to generate more equations. The solution is electrically neutral and equation 4 says just that. The sum of the positive charges (ions) = the sum of the negative charges (ions) or the concentration of the positive ions = the concentration of the negative ones. This is called the charge balance condition Another property is that the mass of each species at the start must equal the its mass at the end. So applying this to the silver we have equation 5 The original amount (in grams) of silver (on the LHS of the original chemical equation) is the concentration x atomic mass of silver which must equate to the mass on the right hand side which is made up of two parts The atomic mass times the concentration of silver left in solution and the fraction of the precipitate that is silver. Equation 6 is formed in the same way for the chloride. You can easily see that equations 5 and 6 give us values for the concentrations of the remaining silver and chloride, in terms of g, the number of grams of precipitate. substituting values from equations 1,2, 5 and 6 into equation 4 will give an equation in a single variable, g. Solving this will tell is if there is a precipitate (g is positive) and how much (in grams) Back substituting this value of g into 5 an 6 will give the values of the silver and chloride concentrations left in solution. Finally substituting these values into equation 3, which as been sitting there quietly will the solubility product.  

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equilibrium, solubility 
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