Problem 86 - Chemistry

Question

What are the concentrations of \(\mathrm{HSO}_{4}^{-}, \mathrm{SO}_{4}^{2-}\), and \(\mathrm{H}_{3} \mathrm{O}^{+}\) in a \(0.20 \mathrm{M} \mathrm{KHSO}_{4}\) solution? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a strong acid; \(K_{\mathrm{a}}\) for \(\mathrm{HSO}_{4}^{-}=1.3 \times 10^{-2}\).)

Step-by-Step Solution

Answer

[HSO4-] = 0.149 M, [SO4^2-] = 0.051 M, [H3O+] = 0.051 M

Step 1: Understand the Dissociation

The compound \( \mathrm{KHSO}_{4} \) dissociates in water as follows: \( \mathrm{KHSO}_{4} \rightarrow \mathrm{K}^{+} + \mathrm{HSO}_{4}^{-} \). Since \( \mathrm{KHSO}_{4} \) completely dissociates, the initial concentration of \( \mathrm{HSO}_{4}^{-} \) is \( 0.20 \, \mathrm{M} \).

Step 2: Analyze the Reaction of HSO4-

The bisulfate ion \( \mathrm{HSO}_{4}^{-} \) further dissociates as a weak acid: \( \mathrm{HSO}_{4}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{SO}_{4}^{2-} \). The equilibrium constant \( K_{a} = 1.3 \times 10^{-2} \) governs this dissociation.

Step 3: Set Up the Equilibrium Expression

Let the concentration of \( \mathrm{H}^{+} \) and \( \mathrm{SO}_{4}^{2-} \) produced be \( x \). At equilibrium, we have:- \([\mathrm{HSO}_{4}^{-}] = 0.20 - x\)- \([\mathrm{H}^{+}] = x\)- \([\mathrm{SO}_{4}^{2-}] = x\)The equilibrium expression is \( K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{HSO}_{4}^{-}]} \).

Step 4: Substitute and Solve for x

Substitute into the equilibrium expression:\[1.3 \times 10^{-2} = \frac{x^2}{0.20 - x}\]Assume \( x \) is small compared to \( 0.20 \), so \( 0.20 - x \approx 0.20 \), resulting in:\[1.3 \times 10^{-2} \approx \frac{x^2}{0.20}\]Solve for \( x \):\[x^2 = 2.6 \times 10^{-3}\]\[x = \sqrt{2.6 \times 10^{-3}} \approx 0.051\]

Step 5: Calculate Final Concentrations

Using the calculated \( x \), find the concentrations:- \([\mathrm{H}^{+}] = x = 0.051 \, \mathrm{M}\)- \([\mathrm{SO}_{4}^{2-}] = x = 0.051 \, \mathrm{M}\)- \([\mathrm{HSO}_{4}^{-}] = 0.20 - x = 0.20 - 0.051 = 0.149 \, \mathrm{M}\)

Key Concepts

Dissociation of Strong Acids

Strong acids dissociate completely in solution to release their ions. For example, when you place potassium bisulfate (\( \mathrm{KHSO}_{4} \)) in water, it dissociates into \( \mathrm{K}^{+} \) and \( \mathrm{HSO}_{4}^{-} \). This dissociation is important because it sets the stage for the subsequent steps in chemical equilibrium. Since \( \mathrm{KHSO}_{4} \) is a strong acid, it's safe to assume that all of it dissociates at first, making the initial concentration of \( \mathrm{HSO}_{4}^{-} \) equal to the initial concentration of \( \mathrm{KHSO}_{4} \) (in this case, 0.20 M).
Understanding the initial dissociation helps in determining how the further dissociation occurs. When strong acids dissociate fully, they significantly affect the ion concentrations, leading to the next equilibrium stage for weak acids, like \( \mathrm{HSO}_{4}^{-} \).

Equilibrium Constant \( (K_a) \)

The equilibrium constant, \( K_a \), is a key player in understanding how weak acids like \( \mathrm{HSO}_{4}^{-} \) further dissociate in solution. It tells us how far the dissociation reaction goes. For this example, the reaction is:\[\mathrm{HSO}_{4}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{SO}_{4}^{2-}\]The given \( K_a \) value for \( \mathrm{HSO}_{4}^{-} \) is \( 1.3 \times 10^{-2} \). This value indicates that the bisulfate ions only partly dissociate, as opposed to completely like strong acids.
The \( K_a \) expression is:\[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{SO}_4^{2-}]}{[\mathrm{HSO}_4^{-}]}\]This expression allows us to calculate how many hydrogen ions \( \mathrm{H}^{+} \) and sulfate ions \( \mathrm{SO}_{4}^{2-} \) are present at equilibrium. Knowing \( K_a \) helps us figure out the balance of our solution.

Equilibrium Concentrations

Determining equilibrium concentrations involves knowing the change in concentrations from the initial state to the equilibrium state.
In our example, let the concentration of \( \mathrm{H}^{+} \) and \( \mathrm{SO}_{4}^{2-} \) produced be \( x \).
At equilibrium:

• \([\mathrm{HSO}_{4}^{-}] = 0.20 - x\)
• \([\mathrm{H}^{+}] = x\)
• \([\mathrm{SO}_{4}^{2-}] = x\)

Using the approximation that \( x \) is small compared to \( 0.20 \), we can simplify our calculations by considering that \( 0.20 - x \approx 0.20 \).
This simplifies the equilibrium expression to:\[1.3 \times 10^{-2} \approx \frac{x^2}{0.20}\]Solving gives \( x \approx 0.051 \).
Thus, the concentrations are:

• \([\mathrm{H}^{+}] = 0.051 \mathrm{M}\)
• \([\mathrm{SO}_{4}^{2-}] = 0.051 \mathrm{M}\)
• \([\mathrm{HSO}_{4}^{-}] = 0.149 \mathrm{M}\)

These equilibrium concentrations give insight into the balance of ions in the solution.