Suppose, in the saturated solution of Ca(OH)2 prepared by the TA, some of the dissolved Ca2+ jons re-precipitate as Caco3. This removes Ca2+ ions from the solution. What will the solid Ca(OH)2 present in the bottom of the beaker do in order to minimize the effect of this shift away from equilibrium? What would the effect be on the [OH -] and the calculated Ksp value? O The removal of Ca2+ ions by the precipitation of CaCO3 from solution results in the dissolution of more Ca2+ from Ca(OH)2. As a result, the [OH-] will increase and the estimate of Kep will be too high. O Solid substances are not part of equilibrium equations, so their presence would leave the value of Ksp unaffected. O The removal of Ca2+ ions by the precipitation of CaCo3 from solution results in the co-precipitation of Ca(OH)2 as it attempts to maintain its equilibrium concentrations. As a result, the [OH ] will decrease and the estimate of Ksp will be too low.

27. For PbCl2 (Ksp = 2.4 x 10-4), will a precipitate of PbCl, form when 0.10 L of 3.0 x 10- M Pb(NO,), is added to 400 mL of 9.0 x 10-2 M NaCl? * Yes. because Q> Ksp. O No. because Q< Ksp O No. because Q = Ksp O Yes. because Q< Ksp

The solubility product constant for Al(OH)3 is 1.90x10-33, and the formation constant for Al(OH)4– is 7.70x1033 at 25°C. Calculate the value of the equilibrium constant*, K, under standard conditions for: Al(OH)3 ​(s) + OH- (aq) ------> Al (OH)4 - (aq)Another way to write the equation for the equilibrium includes the waters of hydration: Al(H2​O)3(OH)3 (s) + OH- (aq) ------> Al(H2O)2​(OH)4​- (aq)

Copper (II) Phosphate is extremely insoluble, but forms a complex ion with CN- that can drive the Copper (II) solubility through Le Chatlier’s principle. Given the two reactions below, determine KT. Cu3(PO4)2 (s)  -->3 Cu2+ (aq) + 2 PO43-(aq)              K1=2.05x10-35 Cu(CN)42- (aq)  -->Cu2+ (aq) + 4 CN- (aq)       K2=1.0x10-25 Cu3(PO4)2 (s)  +  12 CN- (aq)  -->  3 Cu(CN)42- (aq)  +   2 PO43- (aq)                 KT=???

2. Calcium Benzoate (CaBz2): a. A saturated CaBz2 sample solution is at equilibrium and the initially ion concentrations are [Ca²+]ia and [Bz];=b, CaBz₂(s) Ca²+ (aq) + 2Bz¯(aq). This sample is allowed to saturate, and the remaining solids will be filtered out. The final benzoate ion concentration in solution is [BZ];= c as determined from titration. What is the solubility product of CaBz2 in terms of a, b and c? b. The lab experiment uses the saturated CaBz2 solution at equilibrium before titration. Why must the CaBz2 solution be saturated?

How mug barium dissolves? (milligrams)

What concentration of SO? is in equilibrium with Ag,SO,(8) and 3.20 × 10-³ M Ag*? The Ksp of Ag, SO, can be found in this table. (sof] = M

1. General concept and calculations: a. The solubility product of AgCl at T = 298 K is 1.8 x 10-¹0. For a given silver solution with concentration [Ag] 3 10 M, what will be the maximum [Cl] concentration before AgCl(s) precipitation? - b. Given the saturated Ag CrO4 solution a concentration of 6.627 10 M in water, determine its solubility product of the equilibrium Ag₂ CrO42Ag* + Cro

Copper (II) Phosphate is extremely insoluble, but forms a complex ion with CN- that can drive the Copper (II) solubility through Le Chatlier’s principle. Given the two reactions below, determine KT.   Cu3(PO4)2 (s)  -->    3 Cu2+ (aq) +  2 PO43-           (aq)             K1=2.05x10-35 Cu(CN)42- (aq)    -->    Cu2+ (aq)   +   4 CN- (aq)                    K2=1.0x10-25   Cu3(PO4)2 (s)  +  12 CN- (aq)  -->  3 Cu(CN)42- (aq)  +   2 PO43- (aq)                 KT=??? 3b. Of the 4 special types of reaction, which one is depicted by K1 and which one is depicted by the reverse of K2?

In this experiment, Ag+(aq) reacts with SCN- (aq) to form a precipitate.  Adding AgNO3 (aq), which contains Ag+, will essentially increase, not change, or decrease  the concentration of SCN- and drive the equilibrium reaction to the right, left or neither way When this happens, the value of Kc will be higher, lower or stay constant