Problem 66
Question
To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\left(K_{u p}=4 \times 10^{-10}\right)\) in 1.0 L of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}{\underline{\phantom{xx}}}^{2-}\) because the solution will be quite basic.)
Step-by-Step Solution
Verified Answer
The final concentration of \(\mathrm{NH}_{3}\) needed to dissolve 0.020 moles of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\) in 1.0 L of solution is approximately \(9.22 \times 10^{-4}\) M.
1Step 1: Write the balanced dissociation equation for Nickel Oxalate and Ksp expression.
The balanced dissociation equation for Nickel Oxalate can be written as:
\[\mathrm{NiC}_{2}\mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ni}^{2+}(aq) + 2\mathrm{C}_{2}\mathrm{O}_{4}^{2-}(aq)\]
The Ksp expression for the dissociation of Nickel Oxalate is given by:
\[K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]^2\]
2Step 2: Calculate the concentration of Nickel and Oxalate ions in the solution.
Initially, we have 0.020 moles of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\) in 1.0 L of solution. If x moles of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\) dissolve, the concentration of \(\mathrm{Ni}^{2+}\) is x and the concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) is 2x. Now we can plug these values into the Ksp expression and solve for x.
\[K_{sp} = (x)(2x)^2\]
Given the Ksp value of \(4 \times 10^{-10}\), we can now solve for x:
\[4 \times 10^{-10} = (x)(2x)^2\]
3Step 3: Solve for x to determine the concentrations of Nickel and Oxalate ions.
To find the value of x, we could simplify the equation from previous step:
\[4 \times 10^{-10} = 4x^3\]
Divide both sides by 4:
\[1 \times 10^{-10} = x^3\]
Now, take the cube root of both sides:
\[x = \sqrt[3]{1 \times 10^{-10}}\]
\[x \approx 4.64 \times 10^{-4}\]
Now we know the concentrations of Nickel (x) and Oxalate ions (2x) in the solution:
\[[\mathrm{Ni}^{2+}] = 4.64 \times 10^{-4} \mathrm{M}\]
\[[\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] = 2(4.64 \times 10^{-4}) \mathrm{M} = 9.28 \times 10^{-4} \mathrm{M}\]
4Step 4: Find the required concentration of \(\mathrm{NH}_{3}\).
To find the required concentration of \(\mathrm{NH}_{3}\), we can use the concentration of \(\mathrm{Ni}^{2+}\) and \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions. Since the total concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions remains constant, we can assume that any increase in ammonia concentration would lead to a decrease in the concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) ions. We can now set up an equation using the Ksp expression:
\[K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]^2\]
Plugging in the known concentration values and defining the required \(\mathrm{NH}_{3}\) concentration as y, we have:
\[4 \times 10^{-10} = (4.64 \times 10^{-4})(9.28 \times 10^{-4} - y)^2\]
Now, we can solve for y:
\[y = 9.28 \times 10^{-4} - \sqrt{\frac{4 \times 10^{-10}}{4.64 \times 10^{-4}}}\]
Calculating, we have the result:
\[y \approx 9.22 \times 10^{-4} \mathrm{M}\]
Therefore, the final concentration of \(\mathrm{NH}_{3}\) needed to dissolve 0.020 moles of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\) in 1.0 L of solution is approximately \(9.22 \times 10^{-4}\) M.
Key Concepts
Ksp calculationnickel oxalate dissolutionionic equilibrium
Ksp calculation
The Ksp, or solubility product constant, is a measure used to describe the solubility of a dissolved substance under equilibrium conditions. It gives insights into how much of a solid can dissolve in a solution to form a saturated solution. The Ksp is particular to the salt being dissolved and depends on the temperature and the ionic strength of the solution.
To calculate the Ksp, it's crucial to write the balanced dissociation equation for the salt. For nickel oxalate, \[\mathrm{NiC}_{2}\mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ni}^{2+}(aq) + 2\mathrm{C}_{2}\mathrm{O}_{4}^{2-}(aq)\]
The equilibrium expression for the Ksp is formulated as:
To calculate the Ksp, it's crucial to write the balanced dissociation equation for the salt. For nickel oxalate, \[\mathrm{NiC}_{2}\mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ni}^{2+}(aq) + 2\mathrm{C}_{2}\mathrm{O}_{4}^{2-}(aq)\]
The equilibrium expression for the Ksp is formulated as:
- \[K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]^2\]
- For nickel oxalate, with a known Ksp value of \(4 \times 10^{-10}\), we substitute the ion concentrations into this expression to solve for an unknown concentration or extent of solubility.
nickel oxalate dissolution
Nickel oxalate, like many salts, dissolves in water by separating into its constituent ions. This process is described by the dissolution equation:
If we assume \(x\) is the amount of nickel oxalate that dissociates:
Solve the equation for \(x\) to determine the concentrations, showing how dissolution directly ties to solubility concepts. Understanding this process reveals how salts interact in aqueous solutions and their concentration dynamics.
- \[\mathrm{NiC}_{2}\mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ni}^{2+}(aq) + 2\mathrm{C}_{2}\mathrm{O}_{4}^{2-}(aq)\]
If we assume \(x\) is the amount of nickel oxalate that dissociates:
- \([\mathrm{Ni}^{2+}] = x\)
- \([\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] = 2x\)
Solve the equation for \(x\) to determine the concentrations, showing how dissolution directly ties to solubility concepts. Understanding this process reveals how salts interact in aqueous solutions and their concentration dynamics.
ionic equilibrium
Ionic equilibrium in solutions refers to the state where the rate of dissolution of a salt equals the rate at which it precipitates out of solution. This balance involves the intricate interaction of ions and the equilibrium constant, Ksp, especially for sparingly soluble salts like nickel oxalate.
When nickel oxalate dissolves, it reaches a point where the concentrations of its ions no longer increase; instead, they are constant, forming an equilibrium.
When nickel oxalate dissolves, it reaches a point where the concentrations of its ions no longer increase; instead, they are constant, forming an equilibrium.
- At this point, even if more salt were added, it would not dissolve further unless conditions change (temperature, pressure, or the presence of other ions).
- This concept is crucial in predicting whether precipitates will form or dissolve when ions are mixed.
Other exercises in this chapter
Problem 64
For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c)
View solution Problem 65
From the value of \(K_{f}\) listed in Table \(17.1,\) calculate the concentration of \(\mathrm{Ni}^{2}(a q)\) and \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^
View solution Problem 69
(a) Will \(\mathrm{Ca}(\mathrm{OH})_{2}\) precipitate from solution if the pH of a \(0.050 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) is adjusted to \(8.0 ?
View solution Problem 70
(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the pH of a \(0.020 \mathrm{M}\) solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2
View solution