Question

Cisplatin, \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), is a drug widely used in chemotherapy. It can react with the weak base pyridine, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\). Suppose \(3.11 \mathrm{~g}\) of cisplatin are treated with \(2.00 \mathrm{~mL}\) of pyridine \((d=0.980 \mathrm{~g} / \mathrm{mL})\). The unreacted pyridine is then titrated with \(\mathrm{HCl}\) according to the following reaction: $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}(l)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}^{+}(a q) $$ The complete reaction requires \(31.2 \mathrm{~mL}\) of \(0.0245 \mathrm{M} \mathrm{HCl}\). (a) How many moles of pyridine were unused in the cisplatin reaction? (b) How many moles of pyridine would react with one mole of cisplatin?

Short answer

a) The amount of unused pyridine in moles is 0.0007644 moles. b) The number of pyridine moles that would react with one mole of cisplatin is approximately 2 moles.

Step-by-step solution

Calculate the initial moles of pyridine

Given, volume of pyridine = 2.00 mL and density = 0.980 g/mL. We can first find the mass of pyridine. Mass of pyridine = Volume × Density = 2.00 mL × 0.980 g/mL = 1.96 g Now, we can find the moles of pyridine. The molecular weight of pyridine (\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\)) is \(12(6)+1(5)+14 = 93\mathrm{~g/mol}\). Moles of pyridine = \(\frac{Mass}{Molar\, mass}\) = \(\frac{1.96}{93}\) = 0.02108 moles

Calculate the moles of HCl used in the titration

The complete reaction requires 31.2 mL of 0.0245 M HCl. We can calculate the moles of HCl as follows: Moles of HCl = Molarity × Volume (in Liters) = 0.0245 M × 0.0312 L = 0.0007644 moles

Find the moles of unused pyridine

Since both HCl and unused pyridine quantities are stoichiometrically equivalent, their molar amounts are equal. Moles of unused pyridine = Moles of HCl = 0.0007644 moles

Calculate the moles of cisplatin

Given, 3.11 g of cisplatin is treated. The molecular weight of cisplatin (\(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\)) is \(195.08 + 2(14+3)+2(35.45) = 300.05\mathrm{~g/mol}\). We can find the moles of cisplatin as: Moles of cisplatin = \(\frac{Mass}{Molar\, mass}\) = \(\frac{3.11}{300.05}\) = 0.01037 moles

Calculate the moles of pyridine that reacted with cisplatin

Moles of pyridine that reacted = Initial moles of pyridine - Moles of unused pyridine Reacted pyridine moles = 0.02108 - 0.0007644 = 0.02032 moles

Determine the moles of pyridine that would react with one mole of cisplatin

We will now find the ratio of moles of reacted pyridine to moles of cisplatin. Ratio = \(\frac{moles\,of\,reacted\,pyridine}{moles\,of\,cisplatin}\) = \(\frac{0.02032}{0.01037}\) = 1.96 Since 1.96 is close to 2, we can assume that approximately 2 moles of pyridine react with one mole of cisplatin.