7.00 grams of Cobalt (II) nitrate are dissolved in 1.00 liter of water. What is the initial concentration of CO (2+) in moles per liter? Then, 0.22 moles of solid KOH are added to this solution. The solution volume remains 1.00 liter. Given that Kf = 5.0 x 10^9 for the complex ion, Co(OH)4 (2-) find the final concentrations of Co (2+), Co(OH)4 (2-), and OH-. Also find the pH of this final solution.
The Correct Answer and Explanation is :
To solve this problem, we’ll need to go through several steps:
- Calculate initial concentration of Co(2+):
- Molecular weight of Cobalt(II) nitrate ([Co(NO_3)_2]) = 58.93 (Co) + 2(14.01 (N) + 316.00 (O)) = 58.93 + 2*(14.01 + 48.00) = 182.95 g/mol.
- Amount of Cobalt(II) nitrate = 7.00 g.
- Moles of Cobalt(II) nitrate = ( \frac{7.00 \text{ g}}{182.95 \text{ g/mol}} \approx 0.038 \text{ mol} ).
- Since Cobalt(II) nitrate dissociates completely into Co(2+) and NO3(-) in solution, the initial concentration of Co(2+) = 0.038 M.
- Reaction with KOH:
- KOH also dissociates completely in solution providing OH(-) ions and K(+) ions.
- Moles of KOH = 0.22 mol, thus moles of OH(-) = 0.22 mol.
- Initially, OH(-) concentration = 0.22 M.
- Formation of complex ion Co(OH)4(2-):
- The reaction can be represented as: ( Co^{2+} + 4 OH^- \rightarrow Co(OH)_4^{2-} ).
- The reaction’s equilibrium expression is: ( K_f = \frac{[Co(OH)_4^{2-}]}{[Co^{2+}][OH^-]^4} ).
- Given ( K_f = 5.0 \times 10^9 ), if x mol/L of ( Co(OH)_4^{2-} ) forms, then ( [Co^{2+}] = 0.038 – x ) and ( [OH^-] = 0.22 – 4x ).
- Solve for x using Kf:
- ( 5.0 \times 10^9 = \frac{x}{(0.038-x)(0.22-4x)^4} ).
- Solving this equation (we will use a computational approach for accuracy), we find x.
- Calculate pH:
- ( [OH^-] ) remaining can be used to find pOH: ( pOH = -\log [OH^-] ).
- pH = 14 – pOH.
Let’s calculate x for the complex ion concentration, determine remaining concentrations, and calculate the pH.
The calculated concentration of the complex ion ( Co(OH)_4^{2-} ) is approximately ( 0.038 ) M, almost completely consuming the initial Co(2+).
Updated concentrations:
- ( [Co(OH)_4^{2-}] \approx 0.038 ) M.
- ( [Co^{2+}] \approx 0.038 – 0.038 \approx 0 ) M (practically zero).
- ( [OH^-] = 0.22 – 4 \times 0.038 \approx 0.068 ) M.
Calculating pH:
To find the pH, we first calculate the pOH from the remaining OH- concentration and then use ( pH + pOH = 14 ) to find the pH.
[ pOH = -\log[OH^-] ]
[ pH = 14 – pOH ]
Let’s calculate the pOH and pH from the remaining ( [OH^-] ).
The pOH of the solution is approximately 1.17, and the corresponding pH is approximately 12.83.
Explanation
Step-by-Step Breakdown:
- Initial Cobalt Concentration: We started by calculating the initial molarity of Cobalt(II) ions in the solution, which was 0.038 M, based on the amount of Cobalt(II) nitrate dissolved.
- Reaction with KOH: KOH dissociated fully into K+ and OH-, introducing 0.22 M of hydroxide ions into the solution. This high concentration of OH- was set to react with Cobalt(II) ions.
- Formation of Complex Ion: The reaction between Cobalt(II) ions and hydroxide ions formed the complex ion ( Co(OH)_4^{2-} ). Given the high stability constant ( K_f ) of ( 5 \times 10^9 ), this reaction was heavily favored, driving the conversion of nearly all available Cobalt(II) ions into the complex form.
- Equilibrium Calculations: Using the equilibrium expression and the provided formation constant, we calculated the extent of the reaction, finding that nearly all Co(2+) was converted into ( Co(OH)_4^{2-} ) (0.038 M), leaving virtually none of the original Co(2+) ions free in the solution.
- Hydroxide Ion Concentration and pH: After the reaction, the remaining concentration of hydroxide ions was about 0.068 M. This allowed us to calculate the pOH and, subsequently, the pH of the solution, resulting in a final pH of approximately 12.83. This high pH value is consistent with the presence of excess hydroxide ions in the solution.
This detailed stepwise approach demonstrates the comprehensive use of stoichiometry, equilibrium constants, and acid-base concepts to analyze and predict the outcome of chemical reactions in solution.
Now, let’s create a diagram illustrating these concepts and reactions visually.
Here is the educational diagram illustrating the chemical reactions involved in your scenario. This diagram visually represents the dissolution of Cobalt(II) nitrate, its reaction with KOH, the formation of the complex ion ( Co(OH)_4^{2-} ), and the final pH of the solution. The molecular structures, reaction arrows, and pH scale are clearly labeled for better understanding. You can view and refer to this diagram to get a clearer picture of the entire process and outcomes.