Problem 39 - Chemistry: An Atoms-Focused Approach

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: Calculate the pH and pOH values for the following solutions: a. Stomach acid in which [HCl] = 0.155 M 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 = 2.20, pOH = 11.80 d. pH = 1.90, pOH = 12.10

Step 1: Determine the concentration of \(\mathrm{H^+}\) ions

Since \(\mathrm{HCl}\) is a strong acid, it will completely dissociate: \(\mathrm{HCl} \rightarrow \mathrm{H^+} + \mathrm{Cl^{-}}\). Therefore, \([\mathrm{H^+}] = 0.155 \mathrm{M}\).

Step 2: Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) values

Using the formula \(\mathrm{pH} = -\log([\mathrm{H^{+}}])\), we get \(\mathrm{pH} = -\log(0.155) = 0.81\). Then, \(\mathrm{pOH} = 14 - \mathrm{pH} = 14 - 0.81 = 13.19\). #b. \(0.00500 M \mathrm{HNO}_{3}\)#

Step 1: Determine the concentration of \(\mathrm{H^+}\) ions

Since \(\mathrm{HNO_3}\) is a strong acid and completely dissociates, we have \([\mathrm{H^+}] = 0.00500 M\).

Step 2: Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) values

Using the same formula as before, we get \(\mathrm{pH} = -\log(0.00500) = 2.30\), and \(\mathrm{pOH} = 14 - 2.30 = 11.70\). #c. A 2: 1 mixture of \(0.0125 M \mathrm{HCl}\) and \(0.0125 M \mathrm{NaOH}\)#

Step 1: Determine the concentration of \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) ions after the reaction

Since \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) are both strong, they will react with each other: \(\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}\). Due to the 2:1 mixture, we have 2 moles of \(\mathrm{HCl}\) reacting with 1 mole of \(\mathrm{NaOH}\). The concentration of \(\mathrm{H^+}\) ions left in the solution is: \([\mathrm{H^+}] = ([\mathrm{HCl}] - [\mathrm{NaOH}]/2) = (0.0125 - 0.0125/2) = 0.00625 M\).

Step 2: Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) values

Using the formula, we get \(\mathrm{pH} = -\log(0.00625) = 2.20\), and \(\mathrm{pOH} = 14 - 2.20 = 11.80\). #d. A 3: 1 mixture of \(0.0125 M \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(0.0125 \mathrm{MKOH}\)#

Step 1: Determine the concentration of \(\mathrm{H^+}\) and \(\mathrm{OH^-}\) ions after the reaction

In this case, we have \(\mathrm{H_{2}SO_{4}}\) reacting with \(\mathrm{KOH}\): \(\mathrm{H_{2}SO_{4}} + 2\mathrm{KOH} \rightarrow \mathrm{K_2SO_{4}} + 2\mathrm{H_2O}\). Since this is a 3:1 mixture, we will use up all the \(\mathrm{KOH}\), and \([\mathrm{H^+}]_{\text{leftover}} = ([\mathrm{H_2SO_4}] - [\mathrm{KOH}]/2) = (0.0125 - 0.0125/2) = 0.00625 M\). However, \(\mathrm{H_{2}SO_{4}}\) is a strong diprotic acid, meaning it dissociates twice providing 2 moles of H+ ions. So, the effective concentration of \(\mathrm{H^+}\) ions: \([\mathrm{H^+}] = 2 \times 0.00625 = 0.0125 M\).

Step 2: Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) values

Using the formula, we get \(\mathrm{pH} = -\log(0.0125) = 1.90\), and \(\mathrm{pOH} = 14 - 1.90 = 12.10\).

Key Concepts

pH Calculation

The calculation of \(\mathrm{pH}\) is a fundamental concept in acid-base chemistry and gives us an understanding of how acidic or basic a solution is. \(\mathrm{pH}\) is calculated using the formula \(\mathrm{pH} = -\log([\mathrm{H^+}])\), where \(\mathrm{[H^+]}\) represents the concentration of hydrogen ions in a solution.
For strong acids, like \(\mathrm{HCl}\) or \(\mathrm{HNO_3}\), which dissociate completely in water, the concentration of \(\mathrm{H^+}\) ions is equal to the acid concentration. This makes the calculation straightforward:
* Calculate \(\mathrm{pH}\) directly from the concentration of the acid.
* Use logarithmic concepts to compute the exact \(\mathrm{pH}\) value.
Recognizing this formula helps us quickly assess the nature of the solution. A lower \(\mathrm{pH}\) value indicates higher acidity.

pOH Calculation

The \(\mathrm{pOH}\) represents the basicity of the solution and works hand in hand with \(\mathrm{pH}\). To determine \(\mathrm{pOH}\), we use the formula \(\mathrm{pOH} = 14 - \mathrm{pH}\). This calculation is rooted in the relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\), which always adds up to 14 at 25°C, a condition known as the ion product of water balance.
In solutions containing strong acids, where the \(\mathrm{pH}\) is easily computed, \(\mathrm{pOH}\) offers another angle from which to understand the solution's acidity or basicity:
* \(\mathrm{pOH}\) indicates the remaining basicity in the presence of acids.
* It provides insights into how much hydroxide ion \(\mathrm{(OH^-)}\) concentration is there in the solution.
This calculation helps visualize a complete picture of the aqueous system.

Strong Acids

Strong acids, such as \(\mathrm{HCl}\) and \(\mathrm{HNO_3}\), are characterized by their ability to completely dissociate in water. This means that when introduced into the solution, they fully break apart into \(\mathrm{H^+}\) ions and their corresponding anions. Understanding the behavior of strong acids is crucial for accurately predicting the acid strength and its impact on solutions:
* They lead to a direct correlation between acid concentration and \(\mathrm{H^+}\) ion concentration.
* Strong acids are excellent choices for \(\mathrm{pH}\) calculation practice because they simplify the process.
* They always produce a consistent level of \(\mathrm{H^+}\), critical for reaction predictability.
This complete dissociation is key in applications ranging from titrations to various industrial processes.

Acid-Base Reactions

Acid-base reactions are essential chemical processes where acids and bases neutralize each other. Understanding these reactions involves recognizing how acids release \(\mathrm{H^+}\) ions while bases capture them, often resulting in the formation of water and salts.
In mixtures such as \(\mathrm{HCl}\) with \(\mathrm{NaOH}\), acids react with bases to undergo a neutralization reaction:

• The balanced reaction equation is \(\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}\).
• These reactions lead to the formation of neutral water and salt.
• Calculations involve determining leftover concentrations of ions after reactions.

In cases involving strong acids and bases, these reactions help determine leftover \(\mathrm{H^+}\) or \(\mathrm{OH^-}\) concentrations, crucial for accurately calculating the resulting \(\mathrm{pH}\) or \(\mathrm{pOH}\). This makes them vital for a range of chemical processes, helping drive numerous applications in laboratory and industrial settings.