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Some \(\mathrm{K}_{2} \mathrm{SO}_{3}\) and \(\mathrm{KHSO}_{3}\) are dissolved in 250.0 \(\mathrm{mL}\) of solution and the resulting \(\mathrm{pH}\) is \(7.25 .\) Which is greater in this buffer solution, the concentration of \(\mathrm{SO}_{3}^{2-}\) or the concentration of \(\mathrm{HSO}_{3}-7\) If \(\left[\mathrm{SO}_{3}^{2-}\right]=1.0 \mathrm{M}\) in this solution, calculate the concentration of \(\mathrm{HSO}_{3}\) .

Short Answer

Expert verified
In conclusion, the concentration of SO3^2- is greater than the concentration of HSO3- in the given buffer solution, and the concentration of HSO3- is 0.316 M.

Step by step solution

01

Write down the Henderson-Hasselbalch equation for a buffer solution

The Henderson-Hasselbalch equation for a buffer solution is given by: pH = pKa + log\(\frac{[\text{A}^-]}{[\text{HA}]}\) Where pH is the solution's pH, pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid, and [\text{A}^-] and [\text{HA}] are the molar concentrations of the conjugate acid/base pair. In our case, the weak acid is HSO3- and its conjugate base is SO3^2-.
02

Calculate pKa using the given pH value

We are given that the pH of the solution is 7.25. To find the pKa of HSO3-, we can use the relationship between pH, pOH, pKw, pKa, and pKb: pOH = 14 - pH = 14 - 7.25 = 6.75 We can say \(pK_a = p_{OH}\), That gives, \(pK_a = 6.75\)
03

Use the Henderson-Hasselbalch equation to find the concentration of HSO3-

Now we can plug the values of pH, pKa, and [\(\text{SO}_{3}^{2-}\)] into the Henderson-Hasselbalch equation: 7.25 = 6.75 + log\(\frac{[\text{SO}_{3}^{2-}]}{[\text{HSO}_{3}^{-}]}\) Rearrange the equation to solve for [\(\text{HSO}_{3}^{-}\)]: [\(\text{HSO}_{3}^{-}\)] = [\(\text{SO}_{3}^{2-}\)] / 10^(\(pH - pK_a\)) [\(\text{HSO}_{3}^{-}\)] = \(1.0 M\) / 10^(\(7.25 - 6.75\)) [\(\text{HSO}_{3}^{-}\)] = \(1.0 M\) / 10^(\(0.5\)) [\(\text{HSO}_{3}^{-}\)] = 0.316 M
04

Compare the concentrations and determine which ion has a higher concentration

Now that we have the concentration of HSO3- (0.316 M), we can compare it to the given concentration of SO3^2- (1.0 M): [\(\text{SO}_{3}^{2-}\)] = 1.0 M [\(\text{HSO}_{3}^{-}\)] = 0.316 M Since the concentration of SO3^2- (1.0 M) is greater than the concentration of HSO3- (0.316 M), the concentration of SO3^2- is greater in this buffer solution. In conclusion, the concentration of SO3^2- is greater than the concentration of HSO3- in the given buffer solution, and the concentration of HSO3- is 0.316 M.

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Key Concepts

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

Buffer Solution
A buffer solution is a special kind of solution that resists drastic changes in pH when small amounts of acids or bases are added. This amazing property helps maintain a stable environment, which is crucial in many chemical and biological processes. Buffer solutions are typically composed of a weak acid and its conjugate base or a weak base and its conjugate acid.

For example, in our context, the buffer solution consists of the weak acid \( ext{HSO}_3^-\) and its conjugate base \( ext{SO}_3^{2-}\). This pair helps the solution maintain a relatively constant pH even when minor amounts of other substances are introduced. When an acid \(\text{H}^+\) is added, it reacts with the \(\text{SO}_3^{2-}\) ion forming \(\text{HSO}_3^-\), and when a base \(\text{OH}^-\) is added, it will react with \(\text{HSO}_3^-\) forming \(\text{SO}_3^{2-}\). This ability to neutralize small amounts of added acid or base is what defines a buffer solution.

The behavior of buffer solutions can be explained and calculated using the Henderson-Hasselbalch equation, a widely used mathematical relationship that connects pH, the concentration of acid-base pairs, and acid dissociation constant.
Acid Dissociation Constant
The acid dissociation constant, represented as \(K_a\), is a measure of the strength of an acid in a solution. It reflects the tendency of an acid to donate protons to the base. The larger the \(K_a\) value, the stronger the acid, implying it dissociates more in solution.

In this particular exercise, the value \(pK_a\) is more commonly used, which is simply the negative logarithm of the \(K_a\) value: \(pK_a = -\log(K_a)\). This transformation is useful because it makes it easier to work with the numbers, especially when they are very large or very small. In our buffer solution context, \(pK_a\) tells us about the strength of the weak acid \(\text{HSO}_3^-\).

Using the Henderson-Hasselbalch equation, we were able to deduce the \(pK_a\) from the given pH of the solution, which is 7.25. Essentially, it shows how the \(pH\) of a buffer is related to the relative concentrations of the acid and base pair, which is crucial in determining how the buffer functions.
Conjugate Acid-Base Pair
A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton \((\text{H}^+)\). When an acid donates its proton, it forms its conjugate base, and when a base accepts a proton, it forms its conjugate acid. This dynamic is central to understanding buffer solutions and acid-base chemistry.

In this exercise, the conjugate acid-base pair is \(\text{HSO}_3^-\) and \(\text{SO}_3^{2-}\). The weak acid \(\text{HSO}_3^-\) loses a proton to become its conjugate base \(\text{SO}_3^{2-}\). Conversely, \(\text{SO}_3^{2-}\) can gain a proton to become \(\text{HSO}_3^-\). This reversible transformation is what makes them a conjugate pair.

The ratio of the concentrations of the components of the conjugate acid-base pair determines the pH of the solution, as expressed through the Henderson-Hasselbalch equation. Their interaction ensures that the solution can resist changes in pH, maintaining stability, which is vital in many biochemical and industrial processes where pH stability is required.

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Most popular questions from this chapter

Consider the titration of 80.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) by 0.400 \(\mathrm{M} \mathrm{HCl}\) . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of HCl have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}} \\ {\text { c. } 30.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}}\end{array} $$

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Two drops of indicator HIn \(\left(K_{2}=1.0 \times 10^{-9}\right),\) where HIn is yellow and In - is blue, are placed in 100.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{MCl}\) . a. What color is the solution initially? b. The solution is titrated with 0.10\(M\) NaOH. At what pH will the color change (yellow to greenish yellow) occur? c. What color will the solution be after 200.0 \(\mathrm{mL}\) NaOH has been added?

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