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How to Determine Solubility

February 21, 2024 by admin Category: How To

You are viewing the article How to Determine Solubility  at Tnhelearning.edu.vn you can quickly access the necessary information in the table of contents of the article below.

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This article was co-written by Bess Ruff, MA. Bess Ruff is a graduate student in geography at Florida. She received her Master’s degree in Environmental Science and Management from the Bren School of Environmental Science & Management, UC Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the coastal area. Caribbean and support research as a contributor to the Sustainable Fisheries Group.

There are 9 references cited in this article that you can view at the bottom of the page.

This article has been viewed 11,557 times.

In chemistry, solubility is used to describe the property of a solid compound when completely dissolved in a liquid without leaving any insoluble residue. Only ionic compounds (with an electrical charge) are soluble. In fact, just memorizing a few principles or looking up the literature is enough to know if an ionic compound will stay solid when put in water or a large amount will dissolve. Actually a certain number of molecules will dissolve even if you don’t see a change, so for an accurate experiment you must know how to calculate this amount of solute.

Table of Contents

  • Steps
    • Use quick rules
    • Calculate the solubility from the constant K sp
  • Things you need
  • Advice
  • Warning

Steps

Use quick rules

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Learn about ionic compounds. Each atom usually has a certain number of electrons, but sometimes it gains or gives away an electron. This process causes it to become a charged ion . When an ion with a negative charge (one electron excess) meets a positively charged ion (one electron missing), they bond like the cathode and anode of two magnets. The result is an ionic compound.

  • An ion with a negative charge is called an anion , and an ion with a positive charge is called a cation .
  • Normally the number of electrons in an atom is equal to the number of protons, so it has no charge.
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Understanding solubility. The water molecule (H 2 O) has an unusual structure so it is similar to a magnet: one end has a positive charge and the other end has a negative charge. When you put an ionic compound in water, these water “magnets” gather around it, trying to pull the positive and negative ions apart. [1] X Research Source

  • Some ionic compounds are not very strongly attracted to each other, they are considered soluble because they will separate and dissolve when added to water. Other compounds with stronger bonds do not dissolve because the ions are strongly attracted to each other regardless of the attraction of the water molecule.
  • Some compounds have a binding force comparable to that of a water molecule. They are considered sparingly soluble because most of the compound will be separated but the rest will still be attracted to each other.
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Learn the principle of dissolution. Because the interactions between atoms are quite complex, you can’t rely solely on intuition to distinguish which compounds can and cannot dissolve. Look up the first ion in the compound on the list below for its usual properties, and then check for outliers to make sure the second ion doesn’t have unusual interactions with it.

  • For example, to test for strontium chloride (SrCl 2 ), look up Sr or Cl in the steps in bold below. Cl is “generally soluble” so you check the exceptions below it. Sr is not on the exception list so SrCl 2 must be soluble.
  • The most common exceptions to each rule are written below that rule. There are other exceptions, but these are unlikely to happen during normal chemistry or lab hours.
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Compounds are soluble when they contain alkali metals such as Li + , Na + , K + , Rb + and Cs + . These metals are also known as Group IA elements: lithium, sodium, potassium, rubidium, and cesium. Almost any compound containing one of these ions is soluble.

  • Exception: Li 3 PO 4 is insoluble.
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Compounds of NO 3– , C 2 H 3 O 2– , NO 2– , ClO 3– and ClO 4– are soluble. The names corresponding to the above ions are nitrate, acetate, nitrite, chlorate and perchlorate. Note that acetate is often abbreviated as OAc. [2] X Research Source

  • Exception: Ag(OAc) (silver acetate) and Hg(OAc) 2 (mercury acetate) are insoluble.
  • AgNO 2– and KClO 4– are only “slightly soluble”.
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Compounds of Cl – , Br – and I – are generally soluble. Chloride, bromide, and iodide ions almost always form soluble compounds, called halogen salts.

  • Exception: If any of these ions combine with silver ions Ag + , mercury Hg 22+ , or lead Pb 2+ , an insoluble compound will form. The same is true for the less common compounds formed when combined with copper Cu + and thallium Tl + .
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Compounds containing SO42- are generally soluble. Sulfate ions usually form soluble compounds, but there are many exceptions.

  • Exception: The sulfate ion forms insoluble compounds with the following ions: strontium Sr 2+ , barium Ba 2+ , lead Pb 2+ , silver Ag + , calcium Ca 2+ , radium Ra 2+ , and diatomic silver Ag 22+ . Note that silver sulfate and calcium sulfate are only moderately soluble so some consider them sparingly soluble.
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Compounds containing OH – or S 2- are insoluble. The names of these ions are hydroxide and sulfide, respectively.

  • Exception: Do you remember the alkali metals (Group IA) and their preference to form soluble compounds? Li + , Na + , K + , Rb + and Cs + all form compounds that are soluble with hydroxide or sulfite ions. In addition, hydroxide forms soluble salts with alkaline earth metal ions (Group II-A): calcium Ca 2+ , strontium Sr 2+ , and barium Ba 2+ . Note: compounds composed of hydroxides and alkaline earth metals actually have a significant number of molecules that remain bound together so are sometimes referred to as “slightly soluble”.
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Compounds containing CO 32- or PO 4-3- are insoluble. One last check for carbonate and phosphate ions, and you’ll know your compound is soluble.

  • Exception: These ions form compounds that are soluble with alkali metals such as Li + , Na + , K + , Rb + and Cs + , as well as with the ammonium ion NH 4+ .
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Calculate the solubility from the constant K sp

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Look up tan product constant K sp . This constant is different for each compound, so you have to look it up on a graph in your textbook or look it up online. Since these values are determined experimentally and can vary considerably between graphs, it is best to use textbook graphs if available. Unless otherwise specified, most graphs assume an experimental temperature of 25ºC.

  • For example, assuming you are dissolving lead iodide with the formula PbI 2 , write down its solubility product constant. If you refer to the graph at bilbo.chm.uri.edu, you use the constant 7.1×10 –9 .
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Write a chemical equation. First, determine how the compound dissociates into ions when dissolved. Then write the equation with K sp on one side and the component ions on the other side.

  • For example, a PbI 2 molecule dissociates into Pb 2+ , I – , and I – ions. (You only need to know or look up the charge of an ion, since the whole compound is always electrically neutral.)
  • Write the equation 7.1×10 –9 = [Pb 2+ ][I – ] 2
  • This equation is the solubility constant, which you can find out for the 2 ions in the solubility graph. Since there are 2 l- ions, l- must be to the power of two.
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Transform the equation to use variables. Rewrite the equation algebraically using the information you know about the number of molecules and ions. Set x equal to the mass of the compound that will dissolve, and rewrite the equation with x representing the amount of each ion.

  • In this example, we need to rewrite the equation 7.1×10 –9 = [Pb 2+ ][I – ] 2
  • Since there is only one lead ion (Pb 2+ ) in the compound, the number of dissolved molecules will be equal to the number of free lead ions. We can therefore set [Pb 2+ ] to be x.
  • Since there are two iodine ions (I – ) for each lead ion, we set the number of iodine atoms to 2x.
  • Now the equation becomes 7.1×10 –9 = (x)(2x) 2
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Take into account common ions if present. Skip this step if you are dissolving the compound in distilled water. If the compound is dissolved in a solution that already has one or more component ions (“common ions”), the solubility of the compound will decrease significantly. [3] X Source of Study The effect of the common ion will be most pronounced for compounds that are almost insoluble, and in this case you can assume that most of the ions at equilibrium are pre-existing ions. it in solution. Rewrite the equation to account for the concentration mp (mp per liter or M) of the ions already present in the solution, substituting this value for the x variable that you use for that ion. [4] X Research Sources

  • For example, if lead iodide was dissolved in 0.2M lead chloride solution (PbCl 2 ), we would rewrite the equation as 7.1×10 –9 = (0.2M+x)(2x) 2 . Since 0.2M is a higher concentration than x, we can rewrite it as 7.1×10 –9 = (0.2M)(2x) 2 .
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    Solve the equation. Solve for x, and you will know the solubility of the compound. By definition of the solubility constant, you must write your answer in terms of the number of mp of compound dissolved per liter of water. You may have to use a calculator to find the final answer.

    • The following example is solubility in distilled water without any common ions.
    • 7.1×10 –9 = (x)(2x) 2
    • 7.1×10 –9 = (x)(4x 2 )
    • 7.1×10 –9 = 4x 3
    • (7.1×10 –9 ) 4 = x 3
    • x = ((7.1×10 –9 ) 4)
    • x = 1.2 x 10 -3 mp per liter will dissolve . This is a very small mass, so the compound is almost insoluble.
  • Things you need

    • Table of solubility product constants of compounds (K sp )

    Advice

    • If you have empirical data on the amount of compound dissolved then the same equation can be used to solve for the solubility constant K sp . [5] X Research Sources

    Warning

    • There is no consensus on the definition of these terms, but chemists agree on the majority of compounds. Some special compounds in which soluble and insoluble molecules both make up a significant part, each solubility table describes these compounds differently.
    • Some old textbooks consider NH 4 OH a soluble compound. This is not true; Small amounts of NH 4+ and OH – ions were detected, but these two ions could not combine to form a compound. [6] X Research Source
    X

    This article was co-written by Bess Ruff, MA. Bess Ruff is a graduate student in geography at Florida. She received her Master’s degree in Environmental Science and Management from the Bren School of Environmental Science & Management, UC Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the coastal area. Caribbean and support research as a contributor to the Sustainable Fisheries Group.

    There are 9 references cited in this article that you can view at the bottom of the page.

    This article has been viewed 11,557 times.

    In chemistry, solubility is used to describe the property of a solid compound when completely dissolved in a liquid without leaving any insoluble residue. Only ionic compounds (with an electrical charge) are soluble. In fact, just memorizing a few principles or looking up the literature is enough to know if an ionic compound will stay solid when put in water or a large amount will dissolve. Actually a certain number of molecules will dissolve even if you don’t see a change, so for an accurate experiment you must know how to calculate this amount of solute.

    Thank you for reading this post How to Determine Solubility at Tnhelearning.edu.vn You can comment, see more related articles below and hope to help you with interesting information.

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