Question

A sample of \(7.5 \mathrm{~L}\) of \(\mathrm{NH}_{3}\) gas at \(22^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50-L solution of \(0.40 \mathrm{M} \mathrm{HCl}\). Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{~L}\), calculate the \(\mathrm{pH}\) of the resulting solution.

Short answer

The final pH of the solution after bubbling the NH3 gas into the HCl solution is approximately 9.15.

Step-by-step solution

Find the moles of NH3 using the Ideal Gas Law

From the given information, we have the volume (V) = 7.5 L, pressure (P) = 735 torr, and temperature (T) = 22°C = 295 K. We also know the gas constant (R) = 62.36 (L·torr)/(K·mol). We can now use the Ideal Gas Law (PV=nRT) to find the moles of NH3 gas: \( PV = nRT \) \( n = \frac{PV}{RT} = \frac{(735 \,\mathrm{torr})(7.5 \,\mathrm{L})}{(62.36 \,\mathrm{(L \cdot torr)/(K \cdot mol)})(295 \,\mathrm{K})} = 0.542 \, \mathrm{mol} \) So, the number of moles of \(\mathrm{NH_3}\) is 0.542 mol.

Find the moles of HCl in the solution

Given information: the solution's volume (V) = 0.50 L, and the concentration (C) = 0.40 M. To find the moles of HCl, we use the equation: \( n = CV = (0.40 \, \mathrm{M})(0.50 \, \mathrm{L}) = 0.20 \, \mathrm{mol} \) So, the number of moles of \(\mathrm{HCl}\) is 0.20 mol.

Find the stoichiometry of the NH3 and HCl reaction

When \(\mathrm{NH_3}\) and \(\mathrm{HCl}\) react, they neutralize each other, forming \(\mathrm{NH_4^+}\) and \(\mathrm{Cl^-}\): \( \mathrm{NH_3} + \mathrm{HCl} \rightarrow \mathrm{NH_4^+} + \mathrm{Cl^-} \) Since the reaction ratio is 1:1, the amount of produced \(\mathrm{NH_4^+}\) and \(\mathrm{Cl^-}\) will be determined by the limiting reagent. As we have 0.542 mol of \(\mathrm{NH_3}\) and 0.20 mol of \(\mathrm{HCl}\), the limiting reagent is \(\mathrm{HCl}\) since we have a smaller amount of it.

Calculate the equilibrium concentrations

After the reaction, we have: Remaining moles of NH3 = \( 0.542 \,\mathrm{mol}\) - \(0.20 \,\mathrm{mol}\) = 0.342 mol Moles of \( \mathrm{NH_4^+} \) = 0.20 mol (as this is formed from the reaction) The final volume is still 0.50 L, so we can find the concentrations of NH3 and \( \mathrm{NH_4^+} \): \[ [NH_3] = \frac{0.342\,\mathrm{mol}}{0.50\,\mathrm{L}} = 0.684 \,\mathrm{M} \] \[ [NH_4^+] = \frac{0.20\,\mathrm{mol}}{0.50\,\mathrm{L}} = 0.40 \,\mathrm{M} \]

Calculate the pH using the Henderson-Hasselbalch equation

We know the equilibrium concentrations of NH3 and \( \mathrm{NH_4^+} \). We also know that the acid dissociation constant for the conjugate acid of NH3 (\( \mathrm{NH_4^+} \)) is \( K_a = 5.6 \times 10^{-10} \). We can now use the Henderson-Hasselbalch equation to calculate the pH of the solution: \( pH = pK_a + \log \Bigg(\frac{[base]}{[acid]}\Bigg) = -\log(K_a) + \log \Bigg(\frac{[NH_3]}{[NH_4^+]}\Bigg) \) Plug in the values: \( pH = -\log(5.6 \times 10^{-10}) + \log \Bigg(\frac{0.684}{0.40}\Bigg) \) \( pH \approx 9.15 \) So, the final pH of the solution after bubbling the NH3 gas into the HCl solution is approximately 9.15.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental concept in chemistry that allows you to relate the pressure, volume, temperature, and amount of a gas. This law is expressed by the formula \( PV = nRT \), where:

• \( P \) represents the pressure of the gas.
• \( V \) is the volume of the gas.
• \( n \) stands for the number of moles of the gas.
• \( R \) is the ideal gas constant.
• \( T \) is the temperature measured in Kelvin.

In the original problem, you encounter a situation where you have 7.5 liters of ammonia gas at a specific temperature and pressure. By rearranging the Ideal Gas Law to solve for \( n \) (the moles of gas), you can determine the amount of ammonia present. This understanding allows you to predict how much of the gas will participate in the subsequent chemical reactions.

Henderson-Hasselbalch Equation

To determine the pH of a buffer solution, the Henderson-Hasselbalch equation proves incredibly useful. It provides a direct relationship between the [base]/[acid] ratio in a solution and its pH:\[ pH = pK_a + \log\left(\frac{[base]}{[acid]}\right) \]Here:

• \( pH \) is the measure of acidity or basicity.
• \( pK_a \) is the negative logarithm of the acid dissociation constant.
• [base] and [acid] refer to the concentrations of the basic and acidic components in the solution.

In our exercise, after determining the equilibrium concentrations of ammonia (\([NH_3]\)) and ammonium ion (\([NH_4^+]\)), one can apply the Henderson-Hasselbalch equation by plugging in these values along with \( K_a \) of the ammonium ion. This provides a simplified method to accurately calculate the pH, which came out to be approximately 9.15 in the scenario given.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantities of reactants and products in a chemical reaction. It allows chemists to calculate how much of each substance is consumed or produced. When given a balanced chemical equation, the stoichiometry helps to determine the proportions in which the reactants combine.In the reaction between ammonia (\(\mathrm{NH_3}\)) and hydrochloric acid (\(\mathrm{HCl}\)), the stoichiometry is quite straightforward because they react in a 1:1 mole ratio:\[ \mathrm{NH_3} + \mathrm{HCl} \longrightarrow \mathrm{NH_4^+} + \mathrm{Cl^-} \]Given that you have calculated the moles of both \(\mathrm{NH_3}\) and \(\mathrm{HCl}\), you can use stoichiometry to determine how much \(\mathrm{NH_3}\) and \(\mathrm{HCl}\) are consumed and how much \(\mathrm{NH_4^+}\) and \(\mathrm{Cl^-}\) are created. This understanding is crucial for figuring out which reactant is the limiting reagent, as this dictates the extent of the reaction.

Limiting Reagent

The concept of a limiting reagent is foundational to understanding chemical reactions. In any chemical reaction, the limiting reagent is the substance that is completely used up first, thereby stopping the reaction and limiting the amount of product formed.In our example, there are 0.542 moles of \(\mathrm{NH_3}\) and 0.20 moles of \(\mathrm{HCl}\). Since they react in a 1:1 ratio, \(\mathrm{HCl}\) is the limiting reagent because you have less of it available compared to the \(\mathrm{NH_3}\). A clear identification of the limiting reagent helps to predict the amount of product formed and the reactant left in excess. After \(\mathrm{HCl}\) is exhausted, you are left with excess \(\mathrm{NH_3}\), which plays a key role in determining the final concentrations of substances in the solution and the resulting pH calculation.