Calculate the pressure of \(\mathrm{H}_{2}\) (in atm) required to maintain equilibrium with respect to the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pb}(s)+2 \mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{H}_{2}(g) $$ Given that \(\left[\mathrm{Pb}^{2+}\right]=0.035 M\) and the solution is buffered at \(\mathrm{pH} 1.60\).

Short Answer

Expert verified
The pressure of H2 required to maintain equilibrium is 1.23 atm.

Step by step solution

01

Find \([\mathrm{H}^{+}]\)

Given that the pH of the solution is 1.60, we can use the definition of pH to find the concentration of \(\mathrm{H}^{+}\) ions. \(\mathrm{pH}=-\log [\mathrm{H}^+]\). Solving this gives us \( [\mathrm{H}^+]= 10^{-\mathrm{pH}} = 10^{-1.60} = 0.025 \, \mathrm{M}\).
02

Find the equilibrium constant (K)

At equilibrium, the concentration of solid Pb is constant and is not included in the equilibrium constant expression. Therefore, the expression becomes \(K = \frac{[\mathrm{Pb}^{2+}]}{[\mathrm{H}^{+}]^2}\). We know the equilibrium constant for this reaction at \(25^{\circ} \mathrm{C}\) is \(K = 0.80\). So, we can solve for \([\mathrm{Pb}^{2+}]\): \(0.80 = \frac{[\mathrm{Pb}^{2+}]}{(0.025\,\mathrm{M})^2}\) which gives us \([\mathrm{Pb}^{2+}]\) is equal to `0.05 M'.
03

Calculate the pressure of \(\mathrm{H}_{2}\)

Finally, using the ideal gas law, we have \(\mathrm{P}_{\mathrm{H}_{2}} = \frac{\mathrm{nRt}}{\mathrm{V}} = \frac{[\mathrm{H}_{2}] \times \mathrm{RT}}{1 \mathrm{M}}\). The ideal gas constant \(\mathrm{R} = 0.0821 \, \mathrm{L \cdot atm/mol \cdot K}\), and \(\mathrm{T} = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \, \mathrm{K}\). \(\mathrm{n} = [\mathrm{H}_{2}]\). We can rearrange this to find \([\mathrm{H}_{2}]\) to get \([\mathrm{H}_{2}] = \mathrm{P} \times \frac{1 \mathrm{M}}{\mathrm{RT}} = \frac{\mathrm{P_{H_{2}}}}{0.0821 \times 298.15}\). Substituting our known values, we get \(\mathrm{P}_{\mathrm{H}_{2}} = [\mathrm{H}_{2}] \times 0.0821 \times 298.15 = 0.05 \times 0.0821 \times 298.15 = 1.23 \, \mathrm{atm}\).

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Key Concepts

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

Equilibrium Constant
Understanding the equilibrium constant (K) is key to mastering chemical equilibrium calculations. To imagine what equilibrium means, picture a scale in perfect balance—that is essentially what's happening at the molecular level in a chemical reaction at equilibrium.

The equilibrium constant is a ratio of the concentration of the products to the concentration of the reactants, raised to the power of their stoichiometric coefficients. In the provided exercise, the equilibrium constant expression for the reaction \(\mathrm{Pb}(s) + 2 \mathrm{H}^{+}(aq) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{H}_{2}(g)\) would exclude the solid lead (\(\mathrm{Pb}\) ) because its concentration remains constant. So, the expression simplifies to \(K = \frac{[\mathrm{Pb}^{2+}]}{[\mathrm{H}^{+}]^2}\).

By calculating K, you can predict the direction of the reaction—whether it will go forward or reverse to reach equilibrium. A higher K value indicates that at equilibrium, the reaction favors the formation of products, while a lower K value suggests that reactants are favored.
pH and Hydrogen Ion Concentration
The pH of a solution is a measure of how acidic or basic that solution is and is defined as the negative logarithm of the hydrogen ion concentration: \(\mathrm{pH} = -\log [\mathrm{H}^{+}]\).

In our exercise, the pH is given as 1.60; a lower pH value indicates a higher concentration of hydrogen ions. This relationship is logarithmic, not linear—meaning each unit change in pH equals a tenfold change in hydrogen ion concentration. For example, a pH of 3 is ten times more acidic than a pH of 4.

Understanding the link between pH and \([\mathrm{H}^{+}]\) is not just important for solving equilibrium problems, but is also essential in fields such as biochemistry, environmental science, and medicine, where precise pH control is vital.
Ideal Gas Law
The ideal gas law is a critical equation in chemistry and physics which describes the behavior of an ideal gas. It is often written as \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

From manipulating the ideal gas law, we can calculate the pressure exerted by a gas if the number of moles and the temperature are known. In the given solution, the number of moles of \(\mathrm{H}_{2}\) was derived from its equilibrium concentration, which was found using the equilibrium constant and the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{H}^{+}\).

It's crucial to always use Kelvin for the temperature in gas law calculations, as it is the standard scientific unit for thermodynamic temperature scale, and remember that the values for R may vary depending on the units of pressure and volume used.

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Most popular questions from this chapter

A \(9.00 \times 10^{2}-\mathrm{mL} 0.200 \mathrm{M} \mathrm{MgI}_{2}\) was electrolyzed. As a result, hydrogen gas was generated at the cathode and iodine was formed at the anode. The volume of hydrogen collected at \(26^{\circ} \mathrm{C}\) and \(779 \mathrm{mmHg}\) was \(1.22 \times 10^{3} \mathrm{~mL}\). (a) Calculate the charge in coulombs consumed in the process. (b) How long (in min) did the electrolysis last if a current of \(7.55 \mathrm{~A}\) was used? (c) A white precipitate was formed in the process. What was it and what was its mass in grams? Assume the volume of the solution was constant.

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