What is the molar solubility of AgCl in 0.10 M NH3 solution? 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. \[La(IO_3)_3 \rightleftharpoons La^{3+ }+ 3 IO_3^–\], If the solubility is S, then the equilibrium concentrations of the ions will be, [La3+] = S and [IO3–] = 3S. 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 . Silver chloride is fairly insoluble. Relevance. Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. 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. © copyright 2003-2020 Study.com. Legal. The molar solubility of AgCl in 0.10 M NH3 is: Answer Save. K sp = [Ag +] 2 = 1.6 x 10 -10. Describe the visible change that occurs if H3O+ is... A 45-mL sample of 0.015 M calcium chloride, CaCl2,... Titration of a Strong Acid or a Strong Base, LeChatelier's Principle: Disruption and Re-Establishment of Equilibrium, Collision Theory: Definition & Significance, Gibbs Free Energy: Definition & Significance, The Common Ion Effect and Selective Precipitation, Acid-Base Buffers: Calculating the pH of a Buffered Solution, The pH Scale: Calculating the pH of a Solution, Buffer System in Chemistry: Definition & Overview, Ionic Equilibrium: Definition & Calculations, Dalton's Law of Partial Pressures: Calculating Partial & Total Pressures, What is Salt Hydrolysis? That corresponds to a molar solubility of 1.3 x 10-5 moles/L. The solubility product (k s p ) of C a (O H) 2 at 2 5 0 C is 4. The molar solubility of AgCl is equal to the concentration of silver ion and chloride ion at equilibrium. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. 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. For very soluble substances (like sodium nitrate, NaNO 3), this value can be quite high, exceeding 10.0 moles per liter of solution in some cases. The overall reaction has an equilibrium constant that is large enough that the answer obtained using this technique will deviate somewhat from the correct answer. 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. All rights reserved. We can express this quantitatively by noting that the solubility product expression, \[[Ca^{2+}][F^–]^2 = 1.7 \times 10^{–10} \label{8}\], must always hold, even if some of the ionic species involved come from sources other than CaF2(s). Since {eq}[Ag^+]=[Cl^-] {/eq}, they can be both represented by a single variable, x. The solubility product, Ksp, for AgCl in water is 1.77 × 10−10 at room temperature, which indicates that only 1.9 mg (that is, {\displaystyle {\sqrt {1.77\times 10^ { … For very soluble substances (like sodium nitrate, NaNO 3), this value can be quite high, exceeding 10.0 moles per liter of solution in some cases.. For insoluble substances like silver bromide (AgBr), the molar solubility can be quite small. K sp for AgCl is 1.6 × 10 −10.Calculate the molar solubility of AgCl in 1.0 M NH 3. The solubility of manganese(II) hydroxide is 2.2 x... What is the molar solubility of CaC2O4-H2O? University-level students should be able to derive these relations for ion-derived solids of any stoichiometry. 7.2 × 10−10D. 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In the case of AgBr, the value is 5.71 x 10¯ 7 moles per liter. Say that the K sp for AgCl is 1.7 x 10-10. 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? [Ag +] = (1.6 x 10 -10) ½. 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.