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Chapter 15: Problem 60

A solution contains \(0.25\) \(M\) \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.25 M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) Can the metal ions be separated by slowly adding \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) Assume that for successful separation \(99 \%\) of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\).

Short Answer

Expert verified
Yes, the metal ions can be separated by slowly adding Na2CO3 to the mixture. At the concentration of CO鈧兟测伝 required for 99% of Ni虏鈦� to precipitate, only 0.144% of Cu虏鈦� has precipitated, meeting the condition for successful separation.

Step by step solution

01

Find the Ksp values for nickel and copper carbonate

Consult a table of solubility product constants (Ksp) to find the Ksp values for nickel carbonate (NiCO3) and copper carbonate (CuCO3). The Ksp values are: NiCO3: \(K_{sp} = 1.6 \times 10^{-7}\) CuCO3: \(K_{sp} = 2.3 \times 10^{-10}\)
02

Write balanced chemical equations for the precipitation reactions

Write balanced chemical equations for the precipitation of nickel carbonate and copper carbonate: Ni虏鈦�(aq) + CO鈧兟测伝(aq) 鈫� NiCO鈧�(s) Cu虏鈦�(aq) + CO鈧兟测伝(aq) 鈫� CuCO鈧�(s)
03

Calculate concentration of CO鈧兟测伝 required for precipitation

Use the Ksp expressions and given concentrations of the metal ions to determine the concentration of carbonate ions (CO鈧兟测伝) required for precipitation. The Ksp expressions are: For NiCO3: \(K_{sp} = [Ni^{2+}][CO_{3}^{2-}]\) For CuCO3: \(K_{sp} = [Cu^{2+}][CO_{3}^{2-}]\) Since the concentrations of Ni虏鈦� and Cu虏鈦� are both 0.25 M, we can plug in these values and solve for the carbonate ion concentrations: For NiCO3: \(1.6 \times 10^{-7} = (0.25)[CO_{3}^{2-}]\) \(CO_{3}^{2-} = 6.4 \times 10^{-7}\, M\) For CuCO3: \(2.3 \times 10^{-10} = (0.25)[CO_{3}^{2-}]\) \(CO_{3}^{2-} = 9.2 \times 10^{-10}\, M\)
04

Compare the carbonate ion concentrations and determine if separation is possible

Now, we must compare the carbonate ion concentrations needed for precipitation to determine if the metal ions can be separated. We want 99% of one metal ion to precipitate before the other begins to precipitate, so we need to check if the first to precipitate has reached 99% at the concentration required for the other to start. For Ni虏鈦� to precipitate, the concentration of CO鈧兟测伝 must reach \(6.4 \times 10^{-7}\, M\). At this concentration, the fractional precipitation of Cu虏鈦� is: \(\frac{9.2 \times 10^{-10}}{6.4 \times 10^{-7}} = 1.44 \times 10^{-3}\) This means that only about 0.144% of Cu虏鈦� has precipitated when 99% of Ni虏鈦� has precipitated, which meets the condition for successful separation. Therefore, the metal ions can be separated by slowly adding Na2CO3 to the mixture.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Solubility Product Constant
The solubility product constant, often denoted as \( K_{sp} \), is a crucial concept in understanding how compounds dissolve or precipitate in a solution. It tells us the maximum product of the ion concentrations that can exist in a solution without forming a precipitate.
For any sparingly soluble ionic compound like nickel carbonate (\( ext{NiCO}_3 \)) or copper carbonate (\( ext{CuCO}_3 \)), the \( K_{sp} \) expression is written as:\[K_{sp} = [ ext{Metal}^{2+}][ ext{CO}_3^{2-}]\]
  • For \( ext{NiCO}_3 \), the \( K_{sp} = 1.6 \times 10^{-7} \).
  • For \( ext{CuCO}_3 \), the \( K_{sp} = 2.3 \times 10^{-10} \).
The lower the \( K_{sp} \), the less soluble the compound is. This implies that \( ext{CuCO}_3 \) precipitates easier than \( ext{NiCO}_3 \) under the same conditions.
Exploring Precipitation Reactions
Precipitation reactions occur when two solutions are mixed, leading to the formation of an insoluble solid called a precipitate. Understanding these reactions helps us predict whether a particular compound will remain dissolved or form a precipitate.
In the case of nickel and copper ions, adding carbonate ions \(( ext{CO}_3^{2-})\) results in two potential reactions:
  • \( ext{Ni}^{2+}( ext{aq}) + ext{CO}_3^{2-}( ext{aq}) \rightarrow ext{NiCO}_3( ext{s}) \)
  • \( ext{Cu}^{2+}( ext{aq}) + ext{CO}_3^{2-}( ext{aq}) \rightarrow ext{CuCO}_3( ext{s}) \)
Precipitation is influenced by the concentration of ions and their respective \( K_{sp} \) values. A compound starts to precipitate once the product of its ion concentrations reaches its \( K_{sp} \). The higher the concentration of \( ext{CO}_3^{2-} \), the more likely a precipitate will form.
Calculating Concentration for Precipitation
Calculating the concentrations of ions needed for precipitation is essential in separating metal ions from a solution. By using the \( K_{sp} \) expressions, we can determine the point at which a solid will start to form.
For \( ext{NiCO}_3 \) and \( ext{CuCO}_3 \), we know:\[K_{sp} = [ ext{Metal}^{2+}][ ext{CO}_3^{2-}]\]Given the conditions:
  • \([ ext{Ni}^{2+} ] = 0.25 \, M\)
  • \( 1.6 \times 10^{-7} = (0.25)[ ext{CO}_3^{2-}] \)
  • \([ ext{CO}_3^{2-}] = 6.4 \times 10^{-7} \, M\)
For \( ext{CuCO}_3 \):
  • \([ ext{Cu}^{2+} ] = 0.25 \, M\)
  • \( 2.3 \times 10^{-10} = (0.25)[ ext{CO}_3^{2-}] \)
  • \([ ext{CO}_3^{2-}] = 9.2 \times 10^{-10} \, M\)
By calculating when these ion products meet their \( K_{sp} \), we can see if \( 99\% \) of one can precipitate without affecting the other, thereby achieving successful separation.

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Most popular questions from this chapter

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) b. \(\mathrm{V}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}\)

Order the following solids (a-d) from least soluble to most soluble. Ignore any potential reactions of the ions with water. a. \(\mathrm{AgCl} \quad K_{\mathrm{sp}}=1.6 \times 10^{-10}\) b. \(\mathrm{Ag}_{2} \mathrm{S} \quad K_{\mathrm{sp}}=1.6 \times 10^{-49}\) c. \(\mathrm{CaF}_{2} \quad K_{\mathrm{sp}}=4.0 \times 10^{-11}\) d. CuS \(\quad K_{\mathrm{sp}}=8.5 \times 10^{-45}\)

The common ion effect for ionic solids (salts) is to significantly decrease the solubility of the ionic compound in water. Explain the common ion effect.

A solution is prepared by adding 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to \(0.50 \mathrm{L}\) of \(3.0 M \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {overall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$ \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) $$

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{mol} / \mathrm{L}\).
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