Problem 35 - Chemistry The Science in Context

Question

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the following solutions: a. stomach acid in which \([\mathrm{HCl}]=0.155 \mathrm{M}\) b. \(0.00500 M \mathrm{HNO}_{3}\) c. a 2: 1 mixture of \(0.0125 M \mathrm{HCl}\) and \(0.0125 M \mathrm{NaOH}\) d. a 3: 1 mixture of \(0.0125 M \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(0.0125 \mathrm{MKOH}\)

Step-by-Step Solution

Answer

Question: Determine the pH and pOH values for the following solutions: a. Stomach acid with a concentration of 0.155 M HCl b. 0.00500 M HNO₃ c. A 2:1 mixture of 0.0125 M HCl and 0.0125 M NaOH d. A 3:1 mixture of 0.0125 M H₂SO₄ and 0.0125 M KOH Answer: a. pH ≈ 0.81, pOH ≈ 13.19 b. pH ≈ 2.30, pOH ≈ 11.70 c. pH = 7, pOH = 7 d. pH ≈ 2.20, pOH ≈ 11.80

Step 1: a. Stomach acid with \([\mathrm{HCl}] = 0.155 \mathrm{M}\)

As HCl is a strong acid, it will ionize completely into \(\mathrm{H}^{+}\) and \(\mathrm{Cl}^{-}\) ions. So, \([\mathrm{H}^{+}] = 0.155 \mathrm{M}\). Now, let's calculate the pH: \(pH = -\log_{10} [\mathrm{H}^{+}] = -\log_{10}(0.155) \approx 0.81\) Using the relationship between pH and pOH, we can calculate pOH: \(pOH = 14 - pH \approx 14 - 0.81 \approx 13.19\)

Step 2: b. \(0.00500 M \mathrm{HNO}_{3}\)

Since HNO\(_{3}\) is a strong acid, it will completely ionize, giving a concentration of \([\mathrm{H}^{+}] = 0.00500 \mathrm{M}\). Now, let's calculate the pH: \(pH = -\log_{10} [\mathrm{H}^{+}] = -\log_{10}(0.00500) \approx 2.30\) Using the relationship between pH and pOH, we can calculate pOH: \(pOH = 14 - pH \approx 14 - 2.30 \approx 11.70\)

Step 3: c. A 2: 1 mixture of \(0.0125 M \mathrm{HCl}\) and \(0.0125 M \mathrm{NaOH}\)

As the molarities of the strong acid HCl and strong base NaOH are equal, they will neutralize each other and form water. The resulting solution will be neutral with a pH of 7, and thus, pOH will also be 7. pH = 7 pOH = 7

Step 4: d. A 3: 1 mixture of \(0.0125 M \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(0.0125 M \mathrm{KOH}\)

The molarities of H\(_2\)SO\(_4\) and KOH are both 0.0125M. Consider the reaction between \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{KOH}\). Only one proton from H\(_2\)SO\(_4\) will react with one molecule of KOH, and we can write the equation as: \(\mathrm{H}_{2} \mathrm{SO}_{4} + \mathrm{KOH} \rightarrow \mathrm{KHSO}_{4} + \mathrm{H}_{2} \mathrm{O}\) After the reaction, all the KOH will be consumed, and half of the H\(_2\)SO\(_4\) will remain in the solution. So, the final concentration of \([\mathrm{H}^{+}]\) in the solution will be half the initial concentration of the acid, which is \(0.0125/2 = 0.00625 \mathrm{M}\). Now, let's calculate the pH: \(pH = -\log_{10} [\mathrm{H}^{+}] = -\log_{10}(0.00625) \approx 2.20\) Using the relationship between pH and pOH, we can calculate pOH: \(pOH = 14 - pH \approx 14 - 2.20 \approx 11.80\)

Key Concepts

pH Calculation

Calculating the pH of a solution involves understanding the concentration of hydrogen ions \([\mathrm{H}^+]\). pH is a measure of how acidic or basic a solution is on a scale from 0 to 14. A low pH means the solution is acidic, while a high pH means it's basic. Here's how to calculate pH:

• Identify the concentration of hydrogen ions, \([\mathrm{H}^+]\), in the solution. For strong acids like \(\mathrm{HCl}\) and \(\mathrm{HNO}_3\), this is equal to the acid concentration because they ionize completely.
• Use the formula \(pH = -\log_{10}[\mathrm{H}^+]\) to find the pH. This formula converts the hydrogen ion concentration into a simple numeric scale.

If you know the pH and need to find the pOH, use the relationship \(pH + pOH = 14\). This helps determine the basicity of the solution. For example, if the pH is calculated as 2.30, then \(pOH = 14 - 2.30 = 11.70\). Doing these calculations helps you understand the chemical nature of the solution.

Strong Acids

Strong acids are unique in that they completely dissociate in water, meaning every molecule releases a hydrogen ion \([\mathrm{H}^+]\). Because of this property, calculating the hydrogen ion concentration becomes straightforward: it's equal to the acid's molarity.
Well-known strong acids include \(\mathrm{HCl}\), \(\mathrm{HNO}_3\), and \(\mathrm{H}_2\mathrm{SO}_4\). Understanding strong acids is crucial for accurate pH calculations. In solutions of strong acids, the pH is directly related to the acid concentration because of the complete dissociation. For instance:

• If you have a \(0.155 \, M \, \mathrm{HCl}\) solution, then \([\mathrm{H}^+] = 0.155 \, M\).
• Applying the pH formula, \(pH = -\log_{10}(0.155)\), gives you a pH of approximately 0.81.

This property distinguishes strong acids from weak acids, which do not dissociate fully and require equilibrium calculations to determine pH.

Neutralization Reactions

Neutralization reactions occur when an acid and a base react to form water and a salt. This reaction is significant because it can alter the pH of a solution. For example, when equal amounts of a strong acid like \(\mathrm{HCl}\) and a strong base like \(\mathrm{NaOH}\) are mixed, they neutralize each other and the solution becomes neutral, having a pH of 7. In some cases, the mixing of acids and bases does not perfectly balance:

• Consider a 3:1 mixture of \(\mathrm{H}_2\mathrm{SO}_4\) and \(\mathrm{KOH}\). In this case, only one of the two protons from \(\mathrm{H}_2\mathrm{SO}_4\) reacts with \(\mathrm{KOH}\), resulting in a partially neutralized solution.
• Leftover hydrogen ions from the incomplete neutralization will affect the pH.

Practically, this means you'll need to consider the remaining acid when calculating pH, as was done in determining the concentration of \([\mathrm{H}^+]\) and subsequent pH in the problem. Neutralization reactions are a key part of understanding how solutions react and balance naturally.