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A metal ion buffer was prepared from \(0.030 \mathrm{M} \mathrm{ML}\) and \(0.020 \mathrm{M} \mathrm{L}\), where \(\mathrm{ML}\) is a metal-ligand complex and \(\mathrm{L}\) is free ligand. $$ \mathrm{M}+\mathrm{L} \rightleftharpoons \mathrm{ML} \quad K_{\mathrm{f}}=4.0 \times 10^{8} $$ Calculate the concentration of free metal ion, M, in this buffer.

Short Answer

Expert verified
The concentration of free metal ion, M, is approximately \(3.75 \times 10^{-9} \mathrm{M}\).

Step by step solution

01

Write the equilibrium expression

The equilibrium expression for the metal-ligand complex formation is given by: \[K_f = \frac{[ML]}{[M][L]}\]where \([ML]\), \([M]\), and \([L]\) are the concentrations of the metal-ligand complex, free metal ion, and free ligand, respectively. The formation constant, \(K_f\), is \(4.0 \times 10^8\).
02

Substitute known values and variables

We know that \([ML] = 0.030 \mathrm{M}\) and \([L] = 0.020 \mathrm{M}\). Let \([M]\) be the concentration of the free metal ion we want to determine. Substitute these into the equilibrium expression:\[4.0 \times 10^8 = \frac{0.030}{[M] \times 0.020}\]
03

Solve for the concentration of metal ion [M]

Rearrange the equation to solve for \([M]\):\[[M] = \frac{0.030}{4.0 \times 10^8 \times 0.020}\]Calculate:\[[M] = \frac{0.030}{8.0 \times 10^6} = 3.75 \times 10^{-9} \mathrm{M}\]

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Key Concepts

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

Equilibrium Expression
An equilibrium expression is a mathematical formula that explains the relationship between the concentrations of reactants and products when a chemical reaction reaches equilibrium. In the context of a metal-ligand complex, it helps determine how a metal ion (M) combines with a ligand (L) to form a complex (ML).
For the reaction \( M + L \rightleftharpoons ML \), the equilibrium expression is given by the formula:
  • \( K_f = \frac{[ML]}{[M][L]} \)
This expression shows that the formation constant \( K_f \) is equal to the concentration of the metal-ligand complex \([ML]\) divided by the product of the concentrations of the free metal ion \([M]\) and the free ligand \([L]\).
This concept is crucial for understanding how changes in concentration affect the formation of the complex, and it allows us to calculate unknown concentrations given certain constants.
Formation Constant
The formation constant, denoted by \( K_f \), is a specific type of equilibrium constant. It measures the strength or stability of a metal-ligand complex in a solution. A high \( K_f \) value means that the complex is very stable, and the formation of the complex is highly favored.
For the given reaction \( M + L \rightleftharpoons ML \), the formation constant \( K_f \) is \( 4.0 \times 10^8 \). This indicates that once the metal ion and the ligand come together, they tend to remain bonded, forming the complex ML rather than existing separately.
Understanding \( K_f \) is essential when predicting how easily a metal ion will form a complex with a ligand. This is especially useful in metal ion buffer systems where controlling metal ion concentration is critical.
Metal-Ligand Complex
A metal-ligand complex is a chemical species formed when a metal ion binds with one or more ligands. The binding usually involves the sharing or donating of electrons, resulting in a more stable arrangement. For the equation \( M + L \rightleftharpoons ML \), \( ML \) represents this complex.
Such complexes are prevalent in fields such as coordination chemistry and biochemistry. They play vital roles in many biological processes and industrial applications. Factors like the type of ligand, charge on the metal, and the formation constant can influence the stability of the complex.
Comprehension of metal-ligand complexes is crucial for fields dealing with metal ion buffers, where the stability of complexes must be manipulated to achieve desired concentrations of metal ions.
Concentration Calculation
Calculating the concentration of species in a reaction involves substituting known values into the equilibrium expression and solving for the unknowns. In the scenario discussed, we need to find the concentration of free metal ion, \([M]\), in a solution where \([ML] = 0.030 \mathrm{M}\) and \([L] = 0.020 \mathrm{M}\).
Using the equilibrium expression \( K_f = \frac{[ML]}{[M][L]} \), and the given \( K_f \) of \( 4.0 \times 10^8 \), substitute the known values:
  • \( 4.0 \times 10^8 = \frac{0.030}{[M] \times 0.020} \)
Rearrange and solve for \([M]\):
  • \([M] = \frac{0.030}{4.0 \times 10^8 \times 0.020}\)
  • \([M] = \frac{0.030}{8.0 \times 10^6} = 3.75 \times 10^{-9} \mathrm{M}\)
This calculation is vital in determining how much free metal ion remains in the solution after forming some of the complex.

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Most popular questions from this chapter

Standard addition with confidence interval. Ammonia in seawater was measured with an ammonia-selective electrode. A \(100.0 \mathrm{~mL}\) aliquot of seawater was treated with \(1.00 \mathrm{~mL}\) of \(10 \mathrm{M} \mathrm{NaOH}\) to convert \(\mathrm{NH}_{4}^{+}\)to \(\mathrm{NH}_{3}\). Therefore, \(V_{0}=101.0 \mathrm{~mL}\). A reading was then taken with the electrode. Then a series of \(10.00 \mathrm{~mL}\) aliquots of standard \(\mathrm{NH}_{4}^{+} \mathrm{Cl}^{-}\)were added with results in the table. \begin{tabular}{ll|ll} \(V_{\mathrm{S}}(\mathrm{mL})\) & \(E(\mathrm{~V})\) & \(V_{\mathrm{S}}(\mathrm{mL})\) & \(E(\mathrm{~V})\) \\ \hline 0 & \(-0.0844\) & \(30.00\) & \(-0.0394\) \\ \(10.00\) & \(-0.0581\) & \(40.00\) & \(-0.0347\) \\ \(20.00\) & \(-0.0469\) & & \\ \hline \end{tabular} sousce: Data derived from \(H\). Van \(R_{y m y y k, E}\) W. Hall, S. J. Peffesch, and A. E. Wiedeman, "Extending the Marine Microcosm Laboratory," J. Chem. Ed. 2007, 84, 306 . The standard contains \(100.0 \mathrm{ppm}\) ( \(\mathrm{mg} / \mathrm{L}\) ) of nitrogen in the form of \(\mathrm{NH}_{4}^{+} \mathrm{Cl}^{-}\). A separate experiment determined that the electrode slope \(\beta R T(\ln 10) / F\) is \(0.0566 \mathrm{~V}\). (a) Prepare a standard addition graph. Find the concentration and \(95 \%\) confidence interval for ammonia nitrogen \((\mathrm{ppm})\) in the \(100.0 \mathrm{~mL}\) of seawater. (b) Standard addition is best if the additions increase analyte to \(1.5\) to 3 times its original concentration. Does this experiment fall in that range? A criticism of this experiment is that too much added standard creates error because the standards contribute too much to the computed result and the reading from the initial solution is not weighted heavily enough.

Solutions with a wide range of \(\mathrm{Hg}^{2+}\) concentrations were prepared to calibrate an experimental \(\mathrm{Hg}^{2+}\) ion-selective electrode. For the range \(10^{-5}<\left[\mathrm{Hg}^{2+}\right]<10^{-1} \mathrm{M}, \mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) was used directly. The range \(10^{-11}<\left[\mathrm{Hg}^{2+}\right]<10^{-6} \mathrm{M}\) could be covered by the buffer system \(\mathrm{HgCl}_{2}(s)+\mathrm{KCl}(a q)\) (based on \(\mathrm{p} K_{\mathrm{sp}}\) for \(\mathrm{HgCl}_{2}=13.16\) ). The range \(10^{-15}<\left[\mathrm{Hg}^{2+}\right]<10^{-11} \mathrm{M}\) was obtained with \(\mathrm{HgBr}_{2}(s)+\) \(\mathrm{KBr}(a q)\) (based on \(\mathrm{p} K_{\mathrm{sp}}\) for \(\mathrm{HgBr}_{2}=17.43\) ). The resulting calibration curve is shown in the figure. Calibration points for the \(\mathrm{HgCl}_{2}\) \(/ \mathrm{KCl}\) buffer are not in line with the other data. Suggest a possible explanation. \(\mathrm{Hg}^{2+}\) ion-selective electrode calibration curve from J. A. Shatkin, H. S. Brown, and S. Licht, "Composite Graphite lon Selective Electrode Array Potentiometry for the Detection of Mercury and Other Relevant Ions in Aquatic Systems," Anal. Chem. \(1995,67,1147\). It was not stated in the paper, but we presume that all solutions had the same ionic strength.

The \(\mathrm{H}^{+}\)ion-selective electrode on the Phoenix Mars Lander has selectivity coefficients \(K_{\mathrm{H}^{+}}^{\mathrm{Pot}} . \mathrm{Na}^{+}=10^{-8.6}\) and \(K_{\mathrm{H}^{+} . \mathrm{Ca}^{2+}}^{\mathrm{Pot}}=10^{-7.8}\). Let A be the primary ion sensed by the electrode and let its charge be \(z_{A} .\) Let \(X\) be an interfering ion with charge \(z_{X}\). The relative error in primary ion activity due to an interfering ion is \({ }^{46}\) $$ \text { Error in } \mathcal{A}_{\Lambda}(\%)=\frac{\left(K_{\Lambda, X}^{P_{0}}\right)^{\nabla / 2_{\Lambda}} \mathcal{A}_{X}}{\mathcal{A}_{\Lambda}^{z_{X} / z_{\Lambda}}} \times 100 $$ This expression applies for errors less than \(\sim 10 \%\). If the \(\mathrm{pH}\) is \(8.0\) \(\left(\mathcal{A}_{\mathrm{H}^{+}}=10^{-8.0}\right)\) and \(\mathcal{A}_{\mathrm{Na}^{+}}=10^{-2.0}\), what is the relative error in measuring \(\mathcal{A}_{\mathrm{H}^{+}}\)? If the \(\mathrm{pH}\) is \(8.0\) and \(\mathcal{A}_{\mathrm{Ca}^{2+}}=10^{-2.0}\), what is the relative error in measuring \(\mathcal{A}_{\mathrm{H}^{+}}\)?

The selectivity coefficient, \(K_{\mathrm{Li}^{+}, \mathrm{H}^{+}}^{\mathrm{Pot}}\), for a \(\mathrm{Li}^{+}\)ion-selective electrode is \(4 \times 10^{-4}\). When this electrode is placed in \(3.44 \times 10^{-4} \mathrm{M}\) \(\mathrm{Li}^{+}\)solution at \(\mathrm{pH} 7.2\), the potential is \(-0.333 \mathrm{~V}\) versus S.C.E. What would be the potential if the \(\mathrm{pH}\) were lowered to \(1.1\) and the ionic strength were kept constant?

A cyanide ion-selective electrode obeys the equation $$ E=\text { constant }-0.05916 \log \left[\mathrm{CN}^{-}\right] $$ The potential was \(-0.230 \mathrm{~V}\) when the electrode was immersed in \(1.00 \mathrm{mM} \mathrm{NaCN}\). (a) Evaluate the constant in the equation. (b) Using the result from part (a), find \(\left[\mathrm{CN}^{-}\right]\)if \(E=-0.300 \mathrm{~V}\). (c) Without using the constant from part (a), find \(\left[\mathrm{CN}^{-}\right]\)if \(E=\) \(-0.300 \mathrm{~V}\).

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