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Chapter 7: Problem 22

The first proton of sulfuric acid is completely ionized, but the second proton is only partially dissociated, with an acidity constant \(K_{a 2}\) of \(1.2 \times 10^{-2} .\) Calculate the hydrogen ion concentration in a \(0.0100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution.

Short Answer

Expert verified
The hydrogen ion concentration is approximately 0.0101 M.

Step by step solution

01

Understand the problem

We have 0.0100 M sulfuric acid, which dissociates in two steps: first as a strong acid to give all its protons, and second as a weak acid with \( K_{a2} = 1.2 \times 10^{-2} \). We need to calculate the hydrogen ion concentration \([H^+])\) in the solution.
02

Initial Dissociation of Sulfuric Acid

Considering \(H_2SO_4\) is a strong acid, the initial dissociation fully converges to \(HSO_4^-\) and \(H^+\). This results in an initial \([H^+])\) of 0.0100 M due to complete dissociation.
03

Partial Dissociation of the Second Proton

In the second step, \(HSO_4^-\) partially dissociates: \(HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}\). We use the equilibrium constant \((K_{a2})\) to determine the contribution of \(HSO_4^-\) dissociation to the \([H^+])\).
04

Set Up Expression for Equilibrium

At equilibrium, the concentration of \(HSO_4^-\) is reduced by \(x\) and that of \(H^+\) and \(SO_4^{2-}\) are increased by \(x\). The expression for \(K_{a2}\) is: \[ K_{a2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} = \frac{(0.0100 + x)x}{(0.0100 - x)} \approx \frac{(0.0100)x}{0.0100} = 1.2 \times 10^{-2} \]
05

Solve for x

Since \(x\) is small compared to 0.0100, we assume \( (0.0100 - x) \approx 0.0100 \).So, \[ x = \frac{1.2 \times 10^{-4}}{1} = 1.2 \times 10^{-4} \] which is the concentration of \( H^+ \) contributed by partial dissociation.
06

Calculate Total Hydrogen Ion Concentration

The total \([H^+])\) is the sum of fully dissociated \([H^+])\) from the first proton and \([H^+])\) from the second step: \[0.0100 + 1.2 \times 10^{-4} = 0.0101 M\].

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Key Concepts

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

Strong Acid
Sulfuric acid (\(H_2SO_4\)) is a classic example of a strong acid. In chemistry, a strong acid is defined by its ability to completely dissociate in water.
Unlike weak acids, which only partially dissociate, strong acids like \(H_2SO_4\) release all available hydrogen ions into the solution. This results in a high concentration of \([H^+]\) ions almost instantly.
  • The first dissociation of sulfuric acid is complete, meaning \(H_2SO_4\) converts entirely into \(HSO_4^-\) and \(H^+\).
  • This ensures the initial concentration of \([H^+]\) is the same as the concentration of the acid itself.
This behavior is crucial when calculating the hydrogen ion concentration in solutions involving strong acids.h4 Example:In a \(0.0100 \, M\) sulfuric acid solution, the initial hydrogen ion concentration is \(0.0100 \, M\) due to complete dissociation.
Acidity Constant
When discussing acids, the acidity constant, or \(K_a\), is a key factor, especially for the second dissociation of sulfuric acid. The \(K_a\) value helps us understand how easily an acid releases its protons in solution.
Sulfuric acid's second dissociation has a \(K_{a2}\) of \(1.2 \times 10^{-2}\).
  • This indicates that the second proton dissociates much less readily than the first.
  • The larger the \(K_a\) value, the weaker the bond holding the proton, and hence, the stronger the acid in that dissociation step.
The knowledge of \(K_{a2}\) allows us to calculate the additional \([H^+]\) formed from this partial dissociation and plays a vital role in acid-base equilibrium calculations.
Hydrogen Ion Concentration
The hydrogen ion concentration \([H^+]\) in a solution is key to understanding the solution's acidity. For sulfuric acid:
  • The initial \([H^+]\) is from the complete dissociation of the first proton, equaling the molarity of the acid (\(0.0100 \, M\)).
  • Additional \([H^+]\) results from the lesser, partial dissociation of the second proton.
Combining these contributions, the solution's total \([H^+]\) is \(0.0101 \, M\).
Remember that even slight changes in \([H^+]\) can significantly affect a solution's pH and reactivity.
Equilibrium Expression
The equilibrium expression is a mathematical depiction of a chemical equilibrium state. For partial dissociation of \(HSO_4^-\) in sulfuric acid:
  • The formula represents the relationship between reactants and products at equilibrium: \[ K_{a2} = \frac{[H^+][SO_4^{2-}]}{[HSO_4^-]} \]
  • At equilibrium, the remove of \(HSO_4^-\) produces more \([H^+]\) and \([SO_4^{2-}]\)
In the problem, it simplifies to \(0.0100x/0.0100 = 1.2 \times 10^{-2}\). This exemplifies how to determine unknown concentrations, such as \(x\), informing us of the precise contribution from the second dissociation to the overall acidity.

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