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 \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) because the solution will be quite basic.)

Short 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.

Step-by-step solution

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: \[K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]^2\]

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: \[4 \times 10^{-10} = (x)(2x)^2\]

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: \[[\mathrm{Ni}^{2+}] = 4.64 \times 10^{-4} \mathrm{M}\] \[[\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] = 2(4.64 \times 10^{-4}) \mathrm{M} = 9.28 \times 10^{-4} \mathrm{M}\]

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 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:

• \[K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]^2\]
• 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.

By employing the proper balance and expression, chemists deduce essential insights on the extent to which salts dissolve in variable conditions.

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:

• \[\mathrm{NiC}_{2}\mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ni}^{2+}(aq) + 2\mathrm{C}_{2}\mathrm{O}_{4}^{2-}(aq)\]

Given an initial amount (0.020 mol in a 1.0 L solution), you can calculate the concentrations of these ions.

If we assume \(x\) is the amount of nickel oxalate that dissociates:

• \([\mathrm{Ni}^{2+}] = x\)
• \([\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] = 2x\)

Putting these values in the Ksp expression: \[4 \times 10^{-10} = (x)(2x)^2 = 4x^3\]
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.

• 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.

By comprehending ionic equilibrium, one can predict concentrations of ions in solutions and understand the chemical behavior in cases like nickel oxalate solubility without significant internal complexity, benefiting processes like purification and analysis in chemistry.