Problem 114 - Chemistry and Chemical Reactivity

Question

The cancer drug cisplatin, \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2},\) can be made by reacting \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtCl}_{4}\) with ammonia in aqueous solution. Besides cisplatin, the other product is \(\mathrm{NH}_{4} \mathrm{Cl}\). (a) Write a balanced equation for this reaction. (b) To obtain \(12.50 \mathrm{g}\) of cisplatin, what mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{PtCl}_{4}\) is required? What volume of \(0.125 \mathrm{M} \mathrm{NH}_{3}\) is required? (c) Cisplatin can react with the organic compound pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N},\) to form a new compound. $$\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(\mathrm{aq})+x \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq}) \rightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{x}(\mathrm{s})$$ Suppose you treat 0.150 g of cisplatin with what you believe is an excess of liquid pyridine \((1.50 \mathrm{mL} ; d=0.979 \mathrm{g} / \mathrm{mL}) .\) When the reaction is complete, you can find out how much pyridine was not used by titrating the solution with standardized HCl. If 37.0 mL of 0.475 M HCl is required to titrate the excess pyridine, $$\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})$$ what is the formula of the unknown compound \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{x} ?\)

Step-by-Step Solution

Answer

The formula of the unknown compound is \(\mathrm{Pt(NH}_3)_2\mathrm{Cl}_2(\mathrm{C}_5\mathrm{H}_5\mathrm{N})_2\). 15.55 g of \((\mathrm{NH}_4)_2 \mathrm{PtCl}_4\) and 0.667 L of 0.125 M \(\mathrm{NH}_3\) are needed.

Step 1: Write Balanced Equation for Reaction

The balanced equation for the formation of cisplatin is:\[ \text{\((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4} + 2 \mathrm{NH}_3 \rightarrow \mathrm{Pt(NH}_{3})_{2}\mathrm{Cl}_{2} + 2 \mathrm{NH}_4\mathrm{Cl}\)} \]

Step 2: Calculate Molar Masses

First, calculate the molar masses of the relevant compounds:- \(\mathrm{Pt(NH}_{3})_{2}\mathrm{Cl}_{2}:\) \(195.08 + 2(14.01 + 3 \times 1.01) + 2 \times 35.45 = 300.01 \, \text{g/mol}\)- \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}:\) \(2(14.01 + 4 \times 1.01) + 195.08 + 4 \times 35.45 = 372.97 \, \text{g/mol}\)

Step 3: Calculate Moles of Cisplatin Needed

To find the number of moles of cisplatin from a mass of \(12.50 \, \mathrm{g}\):\[ \text{moles of cisplatin} = \frac{12.50 \, \mathrm{g}}{300.01 \, \mathrm{g/mol}} = 0.0417 \, \text{mol} \]

Step 4: Use Stoichiometry to Find Moles of \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}\) Needed

According to the balanced equation, the molar ratio between \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}\) and cisplatin is 1:1. Therefore, the moles of \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}\) needed are:\[ 0.0417 \, \text{mol} \]

Step 5: Calculate Mass of \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}\) Needed

Using the molar mass calculated earlier, the mass of \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}\) required is:\[ 0.0417 \, \text{mol} \times 372.97 \, \text{g/mol} = 15.55 \, \text{g} \]

Step 6: Calculate Volume of \(\mathrm{NH}_3\) Solution Required

From the balanced equation, 2 moles of \(\mathrm{NH}_3\) are required per mole of cisplatin. Thus:\[ \text{moles of } \mathrm{NH}_3 = 0.0417 \, \text{mol} \times 2 = 0.0834 \, \text{mol} \]Using the molarity of \(0.125 \, \text{M}\):\[ \text{Volume of } \mathrm{NH}_3 = \frac{0.0834 \, \text{mol}}{0.125 \, \text{M}} = 0.667 \, \text{L} \]

Step 7: Calculate Excess Pyridine Used

The initial mass of pyridine is calculated by its volume and density.\[ \text{mass of pyridine} = 1.50 \, \text{mL} \times 0.979 \, \text{g/mL} = 1.4685 \, \text{g} \]Convert to moles using pyridine molar mass \((C_5H_5N = 79.10 \, \text{g/mol})\):\[ 1.4685 \, \text{g} / 79.10 \, \text{g/mol} = 0.0186 \, \text{mol} \]

Step 8: Calculate Excess Pyridine After Reaction

From the titration, we calculate the moles of excess pyridine:\[ 0.475 \, \text{M} \times 0.037 \, \text{L} = 0.017575 \, \text{mol} \]

Step 9: Determine Moles of Pyridine Reacted

Moles of pyridine that reacted:\[ 0.0186 \, \text{mol} - 0.017575 \, \text{mol} = 0.001025 \, \text{mol} \]

Step 10: Determine Formula of Complex

0.150 g of cisplatin corresponds to:\[ \text{moles of cisplatin} = \frac{0.150 \, \text{g}}{300.01 \, \text{g/mol}} = 0.0005 \, \text{mol} \]The ratio of pyridine to cisplatin is 2:1, thus the formula is:\[ \mathrm{Pt(NH}_{3})_{2}\mathrm{Cl}_{2}(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N})_2 \]

Key Concepts

Chemical Reactions

In the world of chemistry, chemical reactions are transformations where substances, known as reactants, change into different substances, called products. To describe these transformations precisely, chemists use chemical equations. A balanced chemical equation ensures that the same amount of each element is present on both sides of the reaction. This is crucial because atoms take part in reactions without being created or destroyed. In our case, forming cisplatin involves the reaction of \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}\) with \(\mathrm{NH}_3\). The balanced equation is:

• \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4} + 2 \mathrm{NH}_3 \rightarrow \mathrm{Pt(NH}_{3})_{2}\mathrm{Cl}_{2} + 2 \mathrm{NH}_4\mathrm{Cl}\)

This equation means every molecule of the platinum complex reacts with two molecules of ammonia to form a molecule of cisplatin and two ammonium chloride molecules. When writing and balancing reactions, keep in mind:

• Matter is conserved; the number of atoms for each element must be the same on both sides.
• Balanced reactions reflect stoichiometry, showing the relation between reactants and products in a specific ratio.

Understanding these fundamentals helps us carry out precise laboratory syntheses and analyses.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. Using stoichiometry, chemists can predict the amounts of substances required or produced in a reaction. For example, if we need to produce 12.50 grams of cisplatin, we can use stoichiometry to find the necessary amounts of other reactants. First, calculate the moles of cisplatin needed:

• Molar mass of cisplatin is 300.01 g/mol.
• Moles of cisplatin = \( \frac{12.50\, \text{g}}{300.01\, \text{g/mol}} = 0.0417\, \text{mol} \).

Now, apply stoichiometry from the balanced equation:

• The molar ratio of \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}\) to cisplatin is 1:1.
• Moles of \((\mathrm{NH}_{4})_{2} \mathrm{PtCl}_{4}\) needed = 0.0417 mol.
• Mass required = \( 0.0417\, \text{mol} \times 372.97\, \text{g/mol} = 15.55\, \text{g} \).

In essence, stoichiometry lets us bridge theoretical reactions with real-world laboratory practices.

Coordination Compounds

Coordination compounds are unique chemical structures where a metal atom is bonded to a group of surrounding molecules or ions, called ligands. Cisplatin is an example of such a compound where platinum is bound to two ammonia molecules and two chloride ions. These compounds are characterized by their ability to form stable bonds and resist dissociation.

• Coordination compounds play significant roles in biological systems and industry.
• Cisplatin, specifically, is widely used in cancer treatment.

When cisplatin reacts with an organic molecule like pyridine, a new coordination complex is formed. Pyridine acts as a ligand that replaces the possible water molecules in the coordination sphere around platinum. Understanding these formations helps in designing new drugs and materials with essential properties for medical and industrial applications.

Titration Analysis

Titration is a laboratory procedure used to determine the concentration of a reactant in solution by adding a titrant of known concentration until the reaction reaches a known endpoint. In this exercise, pyridine's reaction with HCl demonstrates titration's utility in determining excess reactant quantities.

• The reaction of pyridine with HCl is given by: \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} + \mathrm{HCl} \rightarrow \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} + \mathrm{Cl}^{-} \).
• You need 37.0 mL of 0.475 M HCl to titrate the leftover pyridine, revealing how much did not react with cisplatin.

To find the formula of the new complex, understand how much pyridine reacted. Knowing the initial amount of pyridine and the moles that reacted allows us to determine the coordination numbers, vital for inferring the unknown complex formula. This systematic approach in titration not only aids in confirming reaction completeness but is essential for verifying compounds in various scientific fields.