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In calculating the \(\mathrm{pH}\) of a \(1.0 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{3},\) we can ignore the \(\mathrm{H}^{+}\) ions produced by the ionization of the bisulfate \(\left(\mathrm{HSO}_{3}^{-}\right)\) ion; however, in calculating the \(\mathrm{pH}\) of a \(1.0 M\) solution of sulfuric acid, we cannot ignore the \(\mathrm{H}^{+}\) ions produced by the ionization of the bisulfate ion. Why?

Short Answer

Expert verified
Answer: We can ignore the H鈦 ions produced by the ionization of the bisulfate ion (HSO鈧冣伝) in the case of a 1.0 M H鈧係O鈧 solution because H鈧係O鈧 is a weak acid, and the second dissociation has a much smaller K鈧 value compared to its K鈧 value. In the case of a 1.0 M H鈧係O鈧 solution, we cannot ignore the H鈦 ions from the ionization of the bisulfate ion (HSO鈧勨伝) due to the higher K鈧 value when compared to its K鈧 value, making it significant for calculating the pH.

Step by step solution

01

Understand the dissociation of the acids and their constants

Both H鈧係O鈧 and H鈧係O鈧 are diprotic acids, meaning they can donate two protons (H鈦 ions) to the solution. The dissociation and the corresponding equilibrium constant (K鈧) can be represented as follows for both acids: H鈧係O鈧 鈬 H鈦 + HSO鈧冣伝 (K鈧) HSO鈧冣伝 鈬 H鈦 + SO鈧兟测伝 (K鈧) H鈧係O鈧 鈬 H鈦 + HSO鈧勨伝 (K鈧) HSO鈧勨伝 鈬 H鈦 + SO鈧劼测伝 (K鈧) The value of K鈧 is much greater than K鈧 for both acids. However, the K鈧 value for H鈧係O鈧 is much larger than that for H鈧係O鈧, making H鈧係O鈧 a stronger acid whereas H鈧係O鈧 is a weak acid.
02

Calculate the pH of the H鈧係O鈧 solution

Since H鈧係O鈧 is a weak acid, the primary dissociation process will contribute most to the H鈦 ions in the solution. As K鈧 is much smaller than K鈧, the equilibrium process of HSO鈧冣伝 will hardly contribute to the H鈦 concentration. Therefore, we can simply consider the K鈧 value for H鈧係O鈧 in calculating the pH of a 1.0 M solution and ignore the H鈦 ions contributed by HSO鈧冣伝 ions.
03

Calculate the pH of the H鈧係O鈧 solution

On the other hand, H鈧係O鈧 is a strong acid with a very large K鈧 value, which means it will almost completely dissociate into H鈦 ions and HSO鈧勨伝 ions. However, the K鈧 value for H鈧係O鈧 is also significant in comparison to its K鈧 value when compared to H鈧係O鈧. While calculating the pH of a 1.0 M H鈧係O鈧 solution, we cannot ignore the H鈦 ions contributed by the HSO鈧勨伝 ions as they also affect the overall concentration of H鈦 ions in the solution. We need to consider both K鈧 and K鈧 values in calculating the pH of the H鈧係O鈧 solution.
04

Conclusion

In summary, the difference in calculating the pH of a H鈧係O鈧 solution and a H鈧係O鈧 solution is due to the significant difference in their K鈧 and K鈧 values. For the weak acid H鈧係O鈧, we can ignore the H鈦 ions produced by the ionization of the bisulfate ion (HSO鈧冣伝) given the K鈧 value is much smaller than the K鈧 value. However, for the strong acid H鈧係O鈧, we cannot ignore the H鈦 ions contributed by the ionization of the bisulfate ion (HSO鈧勨伝) due to the higher K鈧 value when compared to its K鈧 value.

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Key Concepts

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

Diprotic Acids
Diprotic acids are a fascinating group of acids that have the ability to donate two hydrogen ions (H鈦) per molecule when they dissolve in water. This characteristic is what sets them apart from monoprotic acids, which can donate only one proton. Common examples of diprotic acids include sulfurous acid (\(\text{H}_2\text{SO}_3\) and sulfuric acid (\(\text{H}_2\text{SO}_4\).
When diprotic acids dissolve in water, they dissociate in two steps. First, they lose one H鈦 ion to form a hydrogen sulfate or bisulfate ion. In the second step, this ion can lose another H鈦 ion, resulting in a sulfate or sulfite ion. However, each dissociation step has its own equilibrium constant and affects the \(\text{pH}\) calculation and ionization of the solution differently.
Equilibrium Constant
The equilibrium constant, represented as \(K_a\), is a crucial concept in understanding acids and their behavior in solutions. For any acid dissociation reaction, the equilibrium constant is a measure of the strength of the acid: the larger the \(K_a\), the stronger the acid.
In the case of diprotic acids like \(\text{H}_2\text{SO}_3\) and \(\text{H}_2\text{SO}_4\), each dissociation step has its own equilibrium constant: \(K_1\) for the first dissociation and \(K_2\) for the second. Generally, \(K_1\) is greater than \(K_2\), as the first dissociation step is always more favorable than the second. This is because the first proton is more readily lost than the second.
The substantial difference between the \(K_1\) and \(K_2\) values can determine whether the second dissociation step significantly affects the hydrogen ion concentration in a solution, which is critical for calculating the overall \(\text{pH}\).
pH Calculation
Calculating the \(\text{pH}\) of a solution reflects how acidic or basic the solution is, primarily based on the concentration of H鈦 ions in the solution. The formula for calculating \(\text{pH}\) is \(\text{pH} = -\log[\text{H}^+]\).
In a diprotic acid solution, the \(\text{pH}\) calculation varies depending on the values of \(K_1\) and \(K_2\). For a strong acid like \(\text{H}_2\text{SO}_4\) where \(K_1\) is very large, it nearly completely dissociates, adding a significant amount of H鈦 ions to the solution. This means that both dissociations must be considered.
Conversely, a weaker acid like \(\text{H}_2\text{SO}_3\) primarily dissociates in just the first step because its \(K_2\) is considerably smaller, thus its contribution to H鈦 ion concentration is negligible in pH calculations.
Ionization
Ionization is the process by which an acid breaks apart to release H鈦 ions into the solution. For diprotic acids, ionization happens in two stages, resulting in different ionic species being formed in the solution.
Upon dissolving, an acid like \(\text{H}_2\text{SO}_4\) first releases an H鈦 ion, forming a bisulfate ion (HSO鈧勨伝), which can further ionize to produce another H鈦 ion and sulfate ion (SO鈧劼测伝).
  • The initial ionization is typically easier, contributing more significantly to the water's acidity.
  • The secondary ionization can still be relevant, especially in strong acids, where this step significantly impacts the solution's total \(\text{pH}\).
In weak acids where \(K_2\) is much smaller, the second ionization is less significant and often disregarded in \(\text{pH}\) calculations when the accuracy permits.

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