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solubility equilibrium is meaningful when both sides (solids and dissolved ions) are simultaneously present. If you keep adding salt, the salt has reached its solubility limit, and the solution is saturated in NaCl. The situation is now described by NaCl(s) = Na+(aq)+ Cl–(aq) lead chloride (PbCl2), silver chloride (AgCl) are sparingly soluble. When these sparingly soluble salts are dissolved in water, equilibrium is established between the undissolved solid salt and ions of the dissolved salt. For example: AgCl (s) = Ag+ (aq) + Cl– (aq)

The equilibrium constants for saturated solution and solid formation (precipitate) are called solubility product, Ksp. For unsaturated and supersaturated solutions, the system is not at equilibrium, and ion products, Qsp, which have the same expression as Ksp is used. Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. The resulting solution is called a saturated solution.

is only slightly soluble- negligible http://www.thebigger.com/chemistry/ionic-equilibria/what-are-the-differences-between-solubility-product-andionic-product/ -->Ksp is equilibrium constant for sparingly soluble salts. The symbols (aq) indicate that these ions are surrounded by water molecules in the solution. If the solution is not saturated, no precipitate will form In this case, the product is called the ion product, Qsp.

The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound combining one of the ions of the precipitate is added to the solution in equilibrium with the precipitate.

Anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH.

Ksp = [Ca++] [F-]² For every mole of CaF2 that dissolves, 1 mole of Ca++ and 2 moles of F- are in solution. Therefore, if molar solubility = x, then Ksp = (x) (2x)² = 4x³ Given x = 1.24x10^-3 M, Ksp = (4) (1.24x10^-3 M)³ = 7.63x10^-9.

CaCO3 Ca2+ + CO32at equilibrium [Ca2+]= x [CO32-] = x + 0.20 2.8 x 10^-9 = (x)( x+0.20) x = molar solubility = 1.4 x 10^-8 M

If Qsp is greater than Ksp, then a precipitate will form. If Ksp is greater than Qsp, a precipitate will not form experimentally.

diluted, so half concentration Qsp = [Ag+] [Cl-]= 0.020 x 0.015= 0.0003

Molar solubility is solubility in units of mole per unit volume

Ksp = [Pb2+][I-]2 let amount of PbI2 dissolve per unit volume of solution be x => [Pb2+] = x => [I-] = 0.2 + 2x => Ksp = x(0.1 + 2x)^2 Solve for x

The Ksp equals the product of the concentrations of the ions created when the substance is dissolved, individually raised to the power of their coefficients. Their concentrations are synonymous and equal to their solubility, both measured in moles/Liter Ex: To find the Ksp of Calcium Phosphate you would first write to ionic equation. Ca3(PO4)2 3Ca 2+ + 2PO4 3Then we write an equilibrium expression Ksp = [Ca 2+]^3 * [PO4 3-]^2 Now for you problem we have to do the opposite. We are given the Ksp and need to find the solubility. PbI2 Pb 2+ + 2IKsp = [Pb 2+] * [I-]^2 The solubility will be represented by the letter "x" in this case. 7.9 x 10^-9 = (x) * (2x)^2 (Notice that I substituted 2x for the concentration of the iodide ion because for every one lead ion, there are two iodide.) This simplifies to: 7.9 x 10^-9 = 4x^3 1.25 x 10^-3 mols/L...