[tvirus]

Hello everyone.

In chemistry we are doing an Extended Experimental Investigation to see how mixing a 0.25M lead nitrate solution with other aqueous solutions affects the amount of precipitate that is produced. (Our constant varibale is lead nitrate, our independent variables are sodium iodide, hydrochloric acid and sodium sulphate. Volume, molarity etc. are being controlled).

I was wondering, can anyone point me in a direction for what I need to research in order to explain why for example, sodium iodide mixed with lead nitrate produces much more precipitate than hydrochloric acid mixed with lead nitrate. How come certain aqueous solutions produce more precipitate when mixed with lead nitrate compared to others? We haven't been taught any of the theory behind this, and my research through google on "factors affecting precipitate" has been fruitless. Can someone please just tell me some of the chemical forces that are involved here so I know what to research?

[renge ishyo]

You have observed correctly that lead iodide forms more precipitate than lead chloride under equal conditions. The quantitative measure of this comes from their K_sp values which are as follows for the halogens and sulfate (source: http://www.ktf-split.hr/periodni/en/abc/kpt.html):

PbI₂ = 9.8x10^-9
PbBr₂ = 6.60x10^-6
PbCl₂ = 1.70x10^-5
PbF₂ = 3.3x10^-8

PbSO₄ = 2.53x10^-8

The way to interpret these values is that the smaller the K_sp the more readily the precipitate forms. For instance, PbI₂ has a smaller K_sp (9.8x10^-9 which really means that the ratio of ions to precipitate is 0.0000000098(ions)/1(precipitate) than Chlorine (1.70x10^-5 which really means that the ratio of ions to precipitate is 0.0000170(ions)/1(precipitate)). You can see that there are way fewer ions left over for lead iodide than there are with lead chloride (the more ions that stay ions, the less precipitate that forms).

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Okay, so now the problem is why? Well, if you notice (with the exception of fluoride which is a group 2 element and has weird properties) that as you decrease the size of the halogen by moving up on the periodic table that the K_sp changes to indicate that less precipitate forms with lead as you move up. This means that iodide>bromine>chlorine when it comes to forming a precipitate. One of the reasons for this is there is a better size match between lead and iodine than there is between lead and chloride. Look at the periodic table. Lead is very close to iodine on the table indicating the the size of the two atoms are similar, and chlorine is much higher up indicating that the chloride ion is much smaller than lead. The size difference between lead and the far aware chloride can lead to weaker bonds (again, you can't take this analogy too far because that weirdo fluorine breaks the trend). Notice that Sulfate is somewhere inbetween the two? This can be explained because Sulfur is in the same row as Chlorine, but it increases its size by bonding to four oxygen atoms. In doing so the sulfate anion is  bigger than chloride so that its size matches up better with lead. The result is stronger bonds and so more precipitate forms with lead sulfate than with lead chloride.

[tvirus]

Thank you so much This is very helpful information.

Can I just ask, when you say "9.8x10-9 which really means that the ratio of ions to precipitate is 0.0000000098(ions)/1(precipitate)", what is the precipitate being measured in? Is that 9.8 x 10^-9 ions for every gram of precipitate? (that seems like too large a quanitity of precipitate). I'm just a bit confused about how that ratio works.

[renge ishyo]

Yeah, the ratio is a bit tricky to understand. It comes from the definition of K_sp, but since this was a high school forum I was hesitant before to show it and tried to go with the ratio idea because it might be easier to see it that way. Since you asked for a better explanation I'll go with the more mainstream chemical approach if you'd like. For PbI₂ the equation for equilibrium will be:

PbI₂  Pb²⁺ + 2l^-   K_sp = 9.8x10^-9

Now the solubility product is the product of the concentrations of the ions or:

9.8x10^-9 = [Pb²⁺][I^-]²

Don't worry about the details, just know that you are only allowed to have so many ions in solution such that when you multiply them together you get the answer to be 9.8x10^-9 which is a constant. So what happens if you add a larger amount of Pb²⁺ and I^- ions such that when you multiple these ion amounts together you get a BIGGER number than 9.8x10^-9? Then the extra ions you added get precipitated out as solid PbI₂ until there are a small enough amount of ions left in solution so that when you multiply their amounts together you get this 9.8x10^-9 number. In other words, if you have too many ions the extra amount will get used up to make precipitate until you are left with a small amount that is indicated by the K_sp. The smaller the K_sp number, the fewer the ions you can have before precipitate starts being made. Thus, the smaller the K_sp the more things have to precipitate out of solution to get you down to that number and the more precipitate you observe. (I'm not really sure if i am just saying the same thing over and over at this point or not  )

Hope that was better, it is late as I type this