. In the example (barium sulfate) in the introduction, what is the justification
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Question
. In the example (barium sulfate) in the introduction, what is the justification for assuming that S is much smaller than 0.1?
Introduction-
III. Introduction:
A precipitate of a slightly soluble salt is in equilibrium with its ions that are in solution. A double arrow is used to indicate this, where a two-component salt (A & B) is assumed:
In the example (barium sulfate) in the introduction, what is the justification for assuming that S is much smaller than 0.1? Introduction- III. Introduction: A precipitate of a slightly soluble salt is in equilibrium with its ions that are in solution. A double arrow is used to indicate this, where a two-component salt (A & B) is assumed: where I have assumed that S is so much smaller than 0.1 that the right side is essentially (S)(0.1). Notice that the common ion effect has greatly reduced the amount of barium that can go into solution - the new concentration of the barium ion is four factors of ten less - ten thousand times less! Thus, 1.05 x 10-5 moles of barium sulfate will dissolve in a liter of pure water at 25 C to make a saturated solution. Now suppose we dissolve barium sulfate in a 0.1 M solution of sodium sulfate. Sulfate is the common ion. Since there a??s no distinction between sulfate ions, irregardless of where they come from, the sulfate already present will inhibit additional sulfate from the BaSO4. For this case (dissolving BaSO4 in a 0.1 M solution of Na2SO4) we haveThus, if the concentration of either ion can be measured, its square is the value of the solubility product constant. For the case above, the KHT was dissolved in pure water. Instead, suppose it is dissolved in a dilute solution of KCl? In that case, there is a common ion (potassium). To discuss this situation with a little simpler molecule, let's consider another slightly soluble salt, barium sulfate. At 25 C, Ksp for barium sulfate is 1.10 x 10-10 and, since the concentrations of the barium and the sulfate are the same (S): For convenience, this compound will be referred to as a??KHT. a?? While you might think that the equilibrium expression should contain the solid, remember that its concentration remains the same (please see the discussion in section 14.5 of Tro, Second Custom Edition, page 661). Since the equation above shows that equal numbers of moles of the ions are produced, letting S = those moles, For potassium hydrogen tartrate, this isExplanation / Answer
This is because the common ion effect makes lower the solubility of a compound. So, the amount of SO4-2 added to the solution is insignificant.
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