Question

A sample of \(0.96 \mathrm{~L}\) of \(\mathrm{HCl}\) gas at \(372 \mathrm{mmHg}\) and \(22^{\circ} \mathrm{C}\) is bubbled into \(0.034 \mathrm{~L}\) of \(0.57 \mathrm{M} \mathrm{NH}_{3} .\) What is the \(\mathrm{pH}\) of the resulting solution? Assume the volume of solution remains constant and that the \(\mathrm{HCl}\) is totally dissolved in the solution.

Short answer

The pH of the resulting solution is approximately 2.19.

Step-by-step solution

Convert Pressure to Atmospheres

First, we need to convert the pressure from mmHg to atmospheres (atm). We use the conversion factor where 1 atm is equivalent to 760 mmHg. \[372 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 0.4895 \text{ atm}\]

Calculate Moles of HCl

With the pressure in atmospheres, apply the ideal gas law to find the moles of \(\mathrm{HCl}\):\[PV = nRT\]Where:- \(P = 0.4895 \text{ atm}\)- \(V = 0.96 \text{ L}\)- \(R = 0.0821 \; \text{L atm K}^{-1} \text{ mol}^{-1}\)- \(T = 22^{\circ} \mathrm{C} + 273.15 = 295.15 \; \text{K}\)Solve for \(n\) (moles of \(\mathrm{HCl}\)):\[n = \frac{PV}{RT} = \frac{(0.4895)(0.96)}{0.0821 \times 295.15} = 0.0196 \; \text{mol}\]

Calculate Moles of NH3

Calculate the moles of \(\mathrm{NH}_3\) in the solution using its molarity and volume:\[\text{Moles of } \mathrm{NH}_3 = 0.57 \text{ M} \times 0.034 \text{ L} = 0.01938 \; \text{mol}\]

Determine Limiting Reactant and Remaining \\(\mathrm{NH}_3\\)

Compare the moles of \(\mathrm{HCl}\) and \(\mathrm{NH}_3\) to determine the limiting reactant.- Moles of \(\mathrm{HCl}\) \((0.0196)\) are slightly more than moles of \(\mathrm{NH}_3\) \((0.01938)\), so \(\mathrm{NH}_3\) is the limiting reactant.Reacting these, the moles of \(\mathrm{NH}_3\) will be used up completely, and an excess of \(0.0196 - 0.01938 = 0.00022 \) moles of \(\mathrm{HCl}\) will remain.

Calculate pH of the Solution

The remaining moles of \(\mathrm{HCl}\) (0.00022 mol) determine the pH of the solution since \(\mathrm{HCl}\) is a strong acid.The concentration of \(\mathrm{HCl}\) in the final solution is:\[\text{Concentration of HCl} = \frac{0.00022 \text{ mol}}{0.034 \text{ L}} = 0.00647 \text{ M}\]The pH is calculated using the formula:\[\text{pH} = -\log[H^+] = -\log[0.00647]\]\[\text{pH} \approx 2.19\]

Key concepts

Ideal Gas Law

The ideal gas law is crucial in calculating the moles of a gas under specific conditions of pressure, volume, and temperature. It is expressed as \(PV = nRT\), where:

• P is the pressure in atmospheres.
• V is the volume in liters.
• n is the number of moles.
• R is the ideal gas constant \(0.0821 \, \text{L atm K}^{-1} \text{ mol}^{-1}\).
• T is the temperature in Kelvin.

To use the ideal gas law effectively, ensure that all units are consistent. In the given problem, we used this formula to convert the conditions of HCl gas into moles by rearranging it to \(n = \frac{PV}{RT}\). This calculation allows us to proceed with further chemical calculations crucial for determining pH.

Limiting Reactant

The concept of a limiting reactant is vital when dealing with chemical reactions, especially when two reactants are involved. A limiting reactant is the substance that is completely consumed when the chemical reaction is complete.
In this context, we compare the moles of HCl and NH\(_3\). Though the numbers might seem close, NH\(_3\) is slightly less than HCl in moles, making it the limiting reactant. This means all NH\(_3\) will react with HCl, and there will be a leftover amount of HCl. Understanding the limiting reactant helps in knowing which reactant determines the extent of the reaction.

Strong Acid

Strong acids completely dissociate in water, meaning they release their hydrogen ions entirely. HCl is a classic example, dissolving in a solution to produce H\(^+\) ions.
This characteristic is important for pH calculations because the concentration of HCl directly informs us of the H\(^+\) concentration. In the aforementioned exercise, after identifying the excess HCl, we calculated its concentration, leading directly to the pH calculation since pH = \(-\log[H^+]\).
Understanding strong acids is key in various chemistry problems, especially when calculating the acidity of solutions.

Moles Calculation

Calculating moles is a fundamental step in solving chemistry problems, as it helps quantify the chemical substances involved. In our solution, moles of both HCl and NH\(_3\) were calculated.

• For gases like HCl, we utilized the ideal gas law.
• For NH\(_3\), we used the formula \(\text{moles} = \text{molarity} \times \text{volume}\).

Both calculations were essential for understanding how much of each reactant was present, which further informed us about the limiting reactant and excess chemicals. This is a frequent requirement in chemistry, ensuring accurate reactions and results.