Question

at must be the equilibrium concentration of Question 1 (5 pts). A. If the equilibrium concentration of Ce is 0.00010 M, what must be the oqu Cod in the solution? K (Ce (CO)) -59% 10%. (Answer: 3.9 x 10M] B. If the concentration of Ca?" of the solution in 1A is in will some CaCO, begin to precipitate? K (Caco) -1.3 x 10". (Hint: What is the the association of Caland Co, at these concentrations and how does that compare K for this association reaction?) would make C. Using your answers above, in a solution containing 0,010 M Сe and 0,010 M C possible to use C.O. (i.e., add it to the solution) to precipitate 99.0% of the Ce" in a solution of (i.e., reduce the concentration to 0.00010 M) without precipitating any Cam?! Question 2. (10 pts). For each of the following compounds, state whether a solution of nure water will be acidic, neutral, or basic after the dissolution of the compound. Explain your answer for each compound. stantaneously brought from zero to 0.010 M, le? k.(CaCO.) - 1.3 x 10". (Hint: What is the value of Q for concentrations and how does that compare to the value of A. NaBr B. Fe(NO) C. K PO D. NH4Cl Question 3 (4 pts). Given that HCN is a weak acid (that reacts with water to produce CN and H,0") with a K, value of 6.2 x 10", write the reaction for CN reacting with water and calculate the K, this reaction (i.e., for CN reacting with water). Question 4 (2 pts). The equilibrium constant for the reaction H20(1)=H* (aq) + OH(aq) is 1.0 10" at 25°C. What is the value of the equilibrium constant for the reaction 4H2O(l) = 4H+ (aq) + 40H(aq) at the same temperature? Question 5 (7 pts). An aqueous system is made by dissolving solid calcium oxalate (CaCZ02) in pure water. Write down two mathematically independent (ie., different) mass balance equations and the charge balance equation for this system. Assume (1) all species are aqueous unless otherwise specified and (2) there is enough of each species to allow equilibrium to be reached. The pertinent chemical mequations are:

Answer 1

Question 2.

A. Neutral

Explanation: NaBr + H2O ----> NaOH + HBr, it releases neither H⁺ nor OH^- in the aqueous solution.

B. Acidic

Explanation: Fe³⁺ + 3H₂O <----> Fe(OH)3 + 3H⁺, it releases H⁺ ions in the aqueous solution.

C. Basic

Explanation: PO₄^3- + 3H₂O <----> H₃PO₄ + 3OH^-, it releases OH^- ions in the aqueous solution.

D. Acidic

Explanation: NH₄⁺ + H₂O <----> NH₄OH + H⁺, it releases H⁺ ions in the aqueous solution.