She has an interest in astrobiology and manned spaceflight. The solubility of CaF2 (molar mass 78.1) at 18°C is reported to be 1.6 mg per 100 mL of water. It has long been known that the solubility of a sparingly soluble ionic substance is markedly decreased in a solution of another ionic compound when the two substances have an ion in common. What's different about the plot on the right? Solvent properties: A polar solvent will best dissolve a polar compound while a nonpolar solvent will best dissolve a nonpolar compound. This means that [Ag+] = [Cl-]. Example 1: The K sp for AgI is 8.5 x 10-17 at 25 °C. Example #1: Determine the K sp of silver bromide, given that its molar solubility is 5.71 x 10¯ 7 moles per liter. If you look carefully at the scales, you will see that this one is plotted logarithmically (that is, in powers of 10.) Estimate the solubility of La(IO3)3 and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which Ks = 6.2 × 10–12. Solubility is the amount of reagent that will be consumed to saturate the solution or reach the equilibrium of the dissociation reaction. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag+ or Cl– in the saturated solution. moles of solute in 100 mL; S = 0.0016 g / 78.1 g/mol = \(2.05 \times 10^{-5}\) mol, \[S = \dfrac{2.05 \times 10^{ –5} mol}{0.100\; L} = 2.05 \times 10^{-4} M\], \[K_{sp}= [Ca^{2+}][F^–]^2 = (S)(2S)^2 = 4 × (2.05 \times 10^{–4})^3 = 3.44 \times 10^{–11}\]. For example, let us denote the solubility of Ag2CrO4 as S mol L–1. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag + or Cl – in the saturated solution. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The resulting K value is called Ksp or the solubility product: Ksp is a function of temperature. The common-ion effect describes a decrease in solubility of an ionic compound when the solution already contains an ion that is the same as one from the compound. This is just what would be expected on the basis of the Le Châtelier Principle; whenever the process, \[CaF_{2(s)} \rightleftharpoons Ca^{2+} + 2 F^– \label{7}\], is in equilibrium, addition of more fluoride ion (in the form of highly soluble NaF) will shift the composition to the left, reducing the concentration of Ca2+, and thus effectively reducing the solubility of the solid. Legal. Riti Gupta holds a Honors Bachelors degree in Biochemistry from the University of Oregon and a PhD in biology from Johns Hopkins University. If 1000 L of a certain wastewater contains Cd2+ at a concentration of 1.6E–5 M, what concentration of Cd2+ would remain after addition of 10 L of 4 M NaOH solution? Calculate the solubility of strontium sulfate (Ks = 2.8 × 10–7) in (a) pure water and (b) in a 0.10 mol L–1 solution of Na2SO4. Volume of treated water: 1000 L + 10 L = 1010 L. Concentration of OH– on addition to 1000 L of pure water: Initial concentration of Cd2+ in 1010 L of water: \[(1.6 \times 10^{–5}\; M) \left( \dfrac{100}{101} \right) \approx 1.6 \times 10^{–5}\; M\]. … 5 × 10-º for Pbl2 Solution: 1) When AgBr dissolves, it dissociates like this: AgBr(s) ⇌ Ag + (aq) + Br¯(aq) 2) The K sp expression is: K sp = [Ag +] [Br¯] 3) There is a 1:1 molar ratio between the AgBr that dissolves and Ag + that is in solution. Calculate the solubility of both compounds. Solutions . so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released. With this information, you can find the molar solubility which is the number of moles that can be dissolved per liter solution until the solution becomes saturated. Calculate the value of Ks under these conditions. \label{9a}\]. Now "turn on the equilibrium" — find the concentration of Cd2+ that can exist in a 0.04M OH– solution: Substitute these values into the solubility product expression: Note that the effluent will now be very alkaline: pH = 14 + log .04 = 12.6, Taking the square root of both sides allows you to solve for the molar solubility: Thus, the molar solubility of both Ag + and Cl - is thus 1.3 x 10 -5 M. Finding Ksp From Molar Solubility Given molar solubility for the ions in question and the balanced equation, you can find the K sp. For this reason it is meaningless to compare the solubilities of two salts having the formulas A2B and AB2, say, on the basis of their Ks values. Ksp = 8. The point of showing this pair of plots is to illustrate the great utility of log-concentration plots in equilibrium calculations in which simple approximations (such as that made in Equation \(\ref{9b}\) can yield straight-lines within the range of values for which the approximation is valid. Missed the LibreFest? Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. It is meaningless to compare the solubilities of two salts having different formulas on the basis of their Ks values. so S ≈ (2.8 × 10–7) / 0.10M = 2.8 × 10–6 M — which is roughly 100 times smaller than the result from (a). Have questions or comments? Now, given that the molar solubility is 2.2 x 10-3 M, you can plug this into the equation for both [F-] and [Ca2+]. The solubility product is written as such: Remember: The [F-] must be raised to the second power due to the coefficient in the balanced equation. What is the molar solubility? The relation between the molar solubility and the solubility product means that one can be used to find the other. She currently teaches classes in biochemistry, biology, biophysics, astrobiology, as well as high school AP Biology and Chemistry test prep. Chemistry LibreTexts: Solubility and Factors Affecting Solubility, Chemistry LibreTexts: 17.2 Molar Solubility and Ksp. The solubility (by which we usually mean the molar solubility) of a solid is expressed as the concentration of the "dissolved solid" in a saturated solution. Calculating Molar Solubility. She has over 10 years of biology research experience in academia. Take a look at the reaction that occurs when silver (I) chloride dissolves in water: When the solution is saturated, the dissolving of AgCl and formation of solid AgCl are occurring at the same time, or the reaction is at equilibrium: The equilibrium constant can be written where the power on the components is the coefficient in the balanced equation: Since there is no concentration for the solid AgCl, this can be removed from the equation. Then Ks = [La3+][IO3–]3 = S(3S)3 = 27S4, 27S4 = 6.2 × 10–12, S = ( ( 6.2 ÷ 27) × 10–12 )¼ = 6.92 × 10–4 M. Cadmium is a highly toxic environmental pollutant that enters wastewaters associated with zinc smelting (Cd and Zn commonly occur together in ZnS ores) and in some electroplating processes. The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH)2 is formed — that is, all of the Cd2+ gets precipitated. University-level students should be able to derive these relations for ion-derived solids of any stoichiometry. Notice how a much wider a range of values can display on a logarithmic plot. Watch the recordings here on Youtube! In most practical cases, 17.1: The Solubility of Slightly Soluble Salts, Relating Solubilities to Solubility Constants. The key to solving solubility problems is to properly set up your dissociation reactions and define solubility. Say that the Ksp for AgCl is 1.7 x 10-10. But for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag+ concentration. The plots shown below illustrate the common ion effect for silver chromate as the chromate ion concentration is increased by addition of a soluble chromate such as Na2CrO4. In like manner, there is a 1:1 molar ratio between dissolved AgBr and Br¯ in solution. This is found by the equilibrium constant for the reaction. Solution for Calculate the molar solubility of Pbl2(s) in a 0.100 mol/L solution of Nal(aa) at SATP. For example, if some quantity x of fluoride ion is added to a solution initially in equilibrium with solid CaF2, we have, \[K_{sp} = [Ca^{2+}][ F^–]^2 = S (2S + x)^2 . \[La(IO_3)_3 \rightleftharpoons La^{3+ }+ 3 IO_3^–\], If the solubility is S, then the equilibrium concentrations of the ions will be, [La3+] = S and [IO3–] = 3S. Copyright 2020 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. 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