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- Al(OH)3 (s) ⇌ Al3+ (aq) + 3 OH− (aq) Ksp = 1.3 × 10-33 (a) What is the minimum pH at which Al(OH)3(s) will precipitate from a solution that is 0.075 M in Al3+? (b) A solution has [Al]3+ = 0.075 M and [CH3COOH] = 1.00 M. What is the maximum quantity of NaCH3COO that can be added to 500 mL of this solution before precipitation of Al(OH)3(s) begins? Given Ka[CH3COOH] = 1.8 × 10-5 .CoCl42-(aq) + 6H2O(l) >< Co(H2O)62+(aq) + 4Cl-(aq) 1) What was the original color of the cobalt solution? 2) What color did you observe after adding the concentrated HCl solution? 3) What color did you observe after adding the DI water? 4) Assume that the dynamic equilibrium represented in the equation is present in the original solution you placed in the test tube. Based on LeChatelier’s Principle, as stated above, you will interpret your observations in 2, above. a) When hydrochloric acid was added, which reaction (forward or reverse) was favored? What evidence supports this? Why would this reaction be favored? b) Assume that the system rapidly returns to equilibrium. At this new position of equilibrium, is there more or less Co(H2O)62+present after the addition of the acid? Based on this, what happened to the amount of CoCl42-? ANSWER ALLL PARTS!Given the equation: Ag+(aq)+2NH3(aq)⟶[Ag(NH3)2]+(aq) ?f=2.00×10^7 determine the concentration of NH3(aq) that is required to dissolve 295 mg of AgCl(s) in 100.0 mL of solution. The ?sp of AgCl is 1.77×10^−10. I've worked this through several times, but I am just not coming up with the right answer! Thanks :)
- In Reaction 3, you change the temperature of your solution of 1.0 M CoCl2. Using your lab data to help you, answer the following questions below. Co(H2O)6^2+ (aq) + 4 Cl- <---> CoCl4^2- (aq) + 6H20(I) (Pink) (blue) At warmer temperatures, more of the (reactants/prodcuts) exist at equilibrium as compared to the reaction at room temperature. At warmer temperatures, the color of the solution will (be more pink/be more blue/ same color) At colder temperatures, more of the (reactants/prodcuts) exist at equilibrium as compared to the reaction at room temperature. At colder temperatures, the color of the solution will (be more pink/be more blue/no change) The value of Kc will (decrease/increse) when the temperature of the solution is increased, in comparison to the Kc value at room temperature. (Hint: Think about the equation Kc = products/reactants) The reaction studied in this lab is…Co(H2O)62+(aq) + 4 Cl–(aq)⇄CoCl42–(aq) + 6 H2O What happens to the equilibrium using Le Chateliers principle when DI water is added to this? What about when silver nitrate was added? Any precipitate formed? If so, what is the precipitate?2. Calculate the molar solubility of lead(II) phosphate in a pH 10 solution. Pb3(PO4)2(s) 3 Pb²+ (aq) + 2 PO4³- (aq) Pb²+ (aq) + 3 OH(aq) Pb(OH)3(aq) Ksp = 1.0 x 10-54 Kf = 8.0 x 1013 Pb3(PO4)2(s) + 9 OH(aq) 3 Pb(OH)3 (aq) + 2 PO4³- (aq)
- You dissolve 0.194 g of NaOH in 2.00 L of a buffer solution that has [H2PO4] = |HPO, | = 0.188 M. What is the pH of the solution before adding NaOH? (K, for dihydrogen phosphate ion is 6.2 x 10-8.) pH = What is the pH of the solution after adding NaOH? pH =What change will be caused by addition of a small amount of Ba(OH)2 to a buffer solution containing nitrous acid, HNO2, and potassium nitrite, KNO₂? O The concentration of hydronium ions will increase significantly. O The concentration of nitrous acid will increase as will the concentration of hydronium ions. O The concentration of nitrite ion will decrease and the concentration of nitrous acid will increase. The concentration of nitrous acid will decrease and the concentration of nitrite ions will increase.Bromothymol blue is yellow in acid solutions. HBB is acidic form of the indicator. The indicator is blue in basic solutions. BB- is basic form of the indicator. HBB Н20 H30+ BB- (a) Predict what will happen if NaOH is added to the reaction system. And why ? (b). Predict what will happen if HCl solution is added to the reaction system. And why ?
- The initial in the unknown analysis is the addition of 6. M HCl. This results in the formation of a precipitate. The solid precipitate is separate from the liquid solution. The solution could potentially contain K+, Ca2+, and Al3+. Next, 3M NH3 is added to the solution to separate the Al3+. This is a result of the fact that the Al3+ forms the hydroxide, Al(OH)3. But Ca2+ also forms the hydroxide, Ca(OH)2. Why does the Ca(OH)2 not separate out with the Al(OH)3?Identify the spectator ion(s) in the following reaction.Zn(OH)2(s) + 2K+(aq) + 2OH–(aq) → 2K+(aq) + Zn(OH)4–(aq) Zn(OH)2 K+ and OH- K+ K+ and Zn(OH)4 -2 Zn(OH)4– Provide explanation for answer.Iron concentrations greater than 8.30×10-6 M in water used for laundry purposes can cause staining. What [OH-] is required to reduce Fe2+ to this level by precipitation of Fe(OH)2? (Ksp = 8.0×10-16)