Problem 64
Question
To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) because the solution will be quite basic.)
Step-by-Step Solution
Verified Answer
The concentration of \(\mathrm{NH}_{3}\) needed to dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\) in \(1.0 \mathrm{~L}\) of solution is approximately 0.475 M.
1Step 1: Write the chemical equations for the formation of the Nickel-ammonia complex and its dissociation reactions
First, let's write the balanced equation for the dissolution of NiC2O4 in an NH3 solution:
\[ \mathrm{NiC}_{2}\mathrm{O}_{4(s)} + 6 \mathrm{NH}_{3(aq)} \rightleftharpoons \mathrm{Ni(NH_{3})_{6}^{2+}(aq)} + \mathrm{C}_{2}\mathrm{O}_{4^{2-}(aq)} \]
The Ni(NH3)62+ complex can dissociate into Ni2+ and 6 molecules of ammonia by the following reaction:
\[ \mathrm{Ni(NH_{3})_{6}^{2+}(aq)} \rightleftharpoons \mathrm{Ni^{2+}(aq)} + 6 \mathrm{NH}_{3(aq)} \]
2Step 2: Write the Ksp expression and the stability constant expression
Next, we can write the Ksp expression for Nickel Oxalate and the stability constant (Kf) expression for the formation of the Nickel-ammonia complex.
Ksp expression for NiC2O4:
\[ K_{sp} = [\mathrm{Ni}^{2+}] [\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] \]
Stability constant (Kf) expression for Ni(NH3)62+:
\[ K_{f} = \frac{[\mathrm{Ni(NH_{3})_{6}^{2+}]}{[\mathrm{Ni}^{2+}][\mathrm{NH}_{3}]^{6}} \]
3Step 3: Calculate the concentration of the Nickel-ammonia complex and NH3
Let x be the concentration of \(\mathrm{Ni^{2+}}\) and \(\mathrm{C}_{2}\mathrm{O}_{4^{2-}}\) ions, since they will both have the same concentration. Thus, from the Ksp expression, we can write:
\[ K_{sp} = (4 \times 10^{-10}) = [\mathrm{Ni}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4^{2-}}] = x^2 \]
Now, we have 0.020 mol of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\), so the concentration of \(\mathrm{Ni(NH_{3})_{6}^{2+}}\) will be 0.020 mol/L. Let y be the concentration of \(\mathrm{NH}_{3}\) required to dissolve the given amount of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\). From the stability constant expression, we can write:
\[ K_{f} = \frac{[\mathrm{Ni(NH_{3})_{6}^{2+}]}{[\mathrm{Ni}^{2+}][\mathrm{NH}_{3}]^{6}} = \frac{0.020}{x (y - x)^6} \]
Now, we need to eliminate x from this equation. From Ksp, we know that \(x^2 = 4 \times 10^{-10}\). Therefore, \(x = \sqrt{4 \times 10^{-10}}\) and we can substitute x into the stability constant expression.
\[ K_{f} = \frac{0.020}{\sqrt{4 \times 10^{-10}} (y - \sqrt{4 \times 10^{-10}})^6} \]
Finally, we can solve for the concentration of NH3 needed (y) by substituting the Kf value for Nickel-ammonia complex. This value can be found in literature or tables, for example: Kf = 5.0 × 10^8.
\[ 5.0 \times 10^8 = \frac{0.020}{\sqrt{4 \times 10^{-10}} (y - \sqrt{4 \times 10^{-10}})^6} \]
Solving for y, we obtain:
\[ y \approx 0.475 \, M \]
Thus, the concentration of \(\mathrm{NH}_{3}\) required to dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\) in \(1.0 \mathrm{~L}\) of solution is approximately 0.475 M.
Key Concepts
Solubility Product Constant (Ksp)Complex Ion FormationStability Constant (Kf)
Solubility Product Constant (Ksp)
The solubility product constant, or Ksp, is a crucial concept in chemical equilibrium, particularly when dealing with sparingly soluble salts. It represents the extent to which a compound will dissolve in water. In other words, Ksp reflects the solubility of the compound by defining the equilibrium between the solid salt and its dissociated ions in solution. The expression for Ksp is given by the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation.
For instance, the Ksp for nickel(II) oxalate (NiC2O4) at equilibrium can be expressed as:
For instance, the Ksp for nickel(II) oxalate (NiC2O4) at equilibrium can be expressed as:
- \[ K_{sp} = [\mathrm{Ni}^{2+}] [\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] \]
Complex Ion Formation
Complex ion formation occurs when a metal ion binds with one or more ligands, creating charged entities known as complexes. In the context of our exercise, nickel forms a complex ion with ammonia:
Complex ions can significantly increase the solubility of metal salts in solution. In this scenario, the formation of \[\mathrm{Ni(NH_3)_6^{2+}}\] allows otherwise insoluble NiC2O4 to dissolve in the presence of ammonia. The balance in this equilibrium is crucial, as too little ammonia would not be able to stabilize enough nickel ions, leading to precipitation.
- \[ \mathrm{Ni(NH_{3})_{6}^{2+}}\]
Complex ions can significantly increase the solubility of metal salts in solution. In this scenario, the formation of \[\mathrm{Ni(NH_3)_6^{2+}}\] allows otherwise insoluble NiC2O4 to dissolve in the presence of ammonia. The balance in this equilibrium is crucial, as too little ammonia would not be able to stabilize enough nickel ions, leading to precipitation.
Stability Constant (Kf)
The stability constant, denoted as Kf, measures the stability of a complex ion formed in solution. Its expression outlines the equilibrium between the complex ion and the free metal ions plus the ligands. A higher Kf value suggests that the complex ion is highly stable and predominant in the equilibrium mixture.
For the complex ion
In the exercise, knowing the Kf value for the nickel-ammonia complex helps us calculate the needed concentration of ammonia in the solution to ensure the dissolution of 0.020 mol of NiC2O4.
For the complex ion
- \[\mathrm{Ni(NH_3)_6^{2+}}\]
- \[ K_{f} = \frac{[\mathrm{Ni(NH_{3})_{6}^{2+}]}{[\mathrm{Ni}^{2+}][\mathrm{NH}_{3}]^{6}} \]
In the exercise, knowing the Kf value for the nickel-ammonia complex helps us calculate the needed concentration of ammonia in the solution to ensure the dissolution of 0.020 mol of NiC2O4.
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