A 0.0100 M aqueous solution is prepared from h2so4; treating the first dissociation as complete and using \(K_{a2}=1.2\times 10^{-2}\) for \(\mathrm{HSO_4^-}\), what are \([\mathrm{H^+}]\) and the pH?

Using \(K_{a2}\) gives \([\mathrm{H^+}]\approx 1.45\times 10^{-2}\,\mathrm{M}\) and \(\mathrm{pH}\approx 1.84\), lower than the \(2.00\) predicted by assuming only the first dissociation.

h2so4 as a Diprotic Acid: pH of Sulfuric Acid Solution

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Problem

The formula h2so4 represents sulfuric acid, a diprotic (polyprotic) acid. A \(0.0100\,\mathrm{M}\) aqueous solution is prepared. Treat the first dissociation as complete and use \(K_{a2}=1.2\times 10^{-2}\) at \(25^\circ\mathrm{C}\) for the second dissociation:

\[ \mathrm{H_2SO_4(aq)} \rightarrow \mathrm{H^+(aq)} + \mathrm{HSO_4^-(aq)} \] \[ \mathrm{HSO_4^-(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{SO_4^{2-}(aq)} \]

Determine \([\mathrm{H^+}]\) and the pH.

Solution

1) Chemical meaning of “diprotic” for h2so4

A diprotic acid can donate two protons per formula unit in water. For sulfuric acid, the first proton is released essentially completely in dilute solution (strong-acid behavior), while the second proton comes from \(\mathrm{HSO_4^-}\) and is governed by an equilibrium constant \(K_{a2}\).

Strategy: compute the “baseline” \([\mathrm{H^+}]\) after the first dissociation, then add the extra \([\mathrm{H^+}]\) generated by the second dissociation using an ICE table.

2) Step 1: first dissociation (treated as complete)

Let the formal concentration be \(C_0 = 0.0100\,\mathrm{M}\). After complete first dissociation:

\[ [\mathrm{H^+}]_{\text{initial}} = C_0,\qquad [\mathrm{HSO_4^-}]_{\text{initial}} = C_0,\qquad [\mathrm{SO_4^{2-}}]_{\text{initial}} = 0. \]

3) Step 2: second dissociation equilibrium for \(\mathrm{HSO_4^-}\)

For \(\mathrm{HSO_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{SO_4^{2-}}\), define the additional amount that dissociates as \(x\) (in \(\mathrm{M}\)).

Species	Initial (M)	Change (M)	Equilibrium (M)
\(\mathrm{HSO_4^-}\)	\(C_0\)	\(-x\)	\(C_0 - x\)
\(\mathrm{H^+}\)	\(C_0\)	\(+x\)	\(C_0 + x\)
\(\mathrm{SO_4^{2-}}\)	\(0\)	\(+x\)	\(x\)

Write the equilibrium expression:

\[ K_{a2}=\frac{[\mathrm{H^+}][\mathrm{SO_4^{2-}}]}{[\mathrm{HSO_4^-}]} = \frac{(C_0 + x)\,x}{C_0 - x}. \]

4) Solve for \(x\) (quadratic form)

Substitute \(K_{a2}=1.2\times 10^{-2}\) and \(C_0=0.0100\):

\[ 1.2\times 10^{-2} = \frac{(0.0100 + x)\,x}{0.0100 - x}. \]

Rearrange to a quadratic equation:

\[ K_{a2}(C_0 - x) = x(C_0 + x) \] \[ K_{a2}C_0 - K_{a2}x = C_0x + x^2 \] \[ x^2 + (K_{a2}+C_0)x - K_{a2}C_0 = 0. \]

Insert numerical values:

\[ x^2 + (0.012 + 0.0100)x - (0.012)(0.0100) = 0 \] \[ x^2 + 0.0220x - 1.20\times 10^{-4} = 0. \]

Use the quadratic formula (positive root):

\[ x=\frac{-0.0220 + \sqrt{(0.0220)^2 + 4\cdot(1.20\times 10^{-4})}}{2} = \frac{-0.0220 + \sqrt{9.64\times 10^{-4}}}{2}. \]

\[ \sqrt{9.64\times 10^{-4}} \approx 0.0310 \quad \Rightarrow \quad x \approx \frac{-0.0220 + 0.0310}{2} \approx 4.52\times 10^{-3}\,\mathrm{M}. \]

5) Compute \([\mathrm{H^+}]\) and pH

Total hydrogen ion concentration is the initial \(C_0\) from the first dissociation plus the additional \(x\) from the second:

\[ [\mathrm{H^+}] = C_0 + x = 0.0100 + 0.00452 = 0.01452\,\mathrm{M}. \]

Then:

\[ \mathrm{pH} = -\log_{10}([\mathrm{H^+}]) = -\log_{10}(0.01452) \approx 1.84. \]

Comparison (common shortcut): ignoring the second dissociation would give \([\mathrm{H^+}]\approx 0.0100\,\mathrm{M}\) and \(\mathrm{pH}=2.00\). Accounting for \(K_{a2}\) lowers the pH to about \(1.84\).

Visualization

The first dissociation of h2so4 is treated as complete in dilute solution, producing \(\mathrm{H^+}\) and \(\mathrm{HSO_4^-}\). The second dissociation is an equilibrium that increases \([\mathrm{H^+}]\) beyond \(C_0\).

Final result

For \(0.0100\,\mathrm{M}\) h2so4 with the first dissociation complete and \(K_{a2}=1.2\times 10^{-2}\), the second dissociation contributes \(x\approx 4.52\times 10^{-3}\,\mathrm{M}\), giving \([\mathrm{H^+}]\approx 1.45\times 10^{-2}\,\mathrm{M}\) and \(\mathrm{pH}\approx 1.84\).

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Problem

Solution

1) Chemical meaning of “diprotic” for h2so4

2) Step 1: first dissociation (treated as complete)

3) Step 2: second dissociation equilibrium for \(\mathrm{HSO_4^-}\)

4) Solve for \(x\) (quadratic form)

5) Compute \([\mathrm{H^+}]\) and pH

Visualization

Final result

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