pH of Polyprotic Acids: H2S, H3PO4 & H2SO4 Guide | ChemCa

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pH of Polyprotic Acids

Step-by-step strategies to calculate pH and ion concentrations for Diprotic and Triprotic acids like $H_2S$, $H_3PO_4$, and $H_2SO_4$.

Ionic Equilibrium pH Calculation

The Dominant Step Principle

Polyprotic acids dissociate in steps. For most weak polyprotic acids (like $H_2CO_3, H_2S, H_3PO_4$), the first dissociation constant ($K_{a1}$) is significantly larger than the second ($K_{a2}$).

Typically: $K_{a1} \gg K_{a2} \gg K_{a3}$ (often by factors of $10^4$ to $10^6$).

Key Approximation: Since the first step produces the vast majority of $H^+$, we can calculate the pH by treating the acid as monoprotic using only $K_{a1}$. The subsequent steps contribute negligible $H^+$.

Stepwise Dissociation ($H_2A$)

$$ \text{Step 1: } H_2A \rightleftharpoons H^+ + HA^- \quad (K_{a1}) $$

$$ \text{Step 2: } HA^- \rightleftharpoons H^+ + A^{2-} \quad (K_{a2}) $$

*Step 2 is suppressed by the Common Ion Effect ($H^+$ from Step 1).

Case 1: Weak Diprotic Acid ($K_{a1} \gg K_{a2}$)

Solved Example: Carbonic Acid

Calculate the pH and $[CO_3^{2-}]$ concentration in a $0.05 M$ solution of $H_2CO_3$.
Given: $K_{a1} = 4 \times 10^{-7}$, $K_{a2} = 5 \times 10^{-11}$.

Step 1: Determine pH (First Dissociation)

Since $K_{a1} \gg K_{a2}$, we treat it as a monoprotic acid. Check if we can neglect $\alpha$ ($C/K_{a1} > 100$).

$$ [H^+] \approx \sqrt{K_{a1} \cdot C} = \sqrt{4 \times 10^{-7} \times 0.05} $$ $$ [H^+] = \sqrt{20 \times 10^{-9}} = \sqrt{2 \times 10^{-8}} \approx 1.41 \times 10^{-4} M $$

$pH = -\log(1.41 \times 10^{-4}) = 4 - 0.15 = \mathbf{3.85}$

Step 2: Concentration of Second Ion ($A^{2-}$)

This is a standard result for weak polyprotic acids. From Step 1, $[H^+] \approx [HA^-]$. Now write the Step 2 equilibrium:

$$ K_{a2} = \frac{[H^+][CO_3^{2-}]}{[HCO_3^-]} $$

Substituting $[H^+] \approx [HCO_3^-]$ (from Step 1):

$$ K_{a2} \approx \frac{[H^+][CO_3^{2-}]}{[H^+]} \implies [CO_3^{2-}] \approx K_{a2} $$ $$ \therefore [CO_3^{2-}] = 5 \times 10^{-11} M $$

*Note: The concentration of the divalent ion equals the second dissociation constant.

Case 2: The $H_2SO_4$ Exception

Sulfuric Acid ($H_2SO_4$) is unique because the first step is Strong (Complete dissociation) while the second step is Weak.

Step 1 (Strong)

$$ H_2SO_4 \longrightarrow H^+ + HSO_4^- $$ $$ [H^+]_1 = C $$

Step 2 (Weak)

$$ HSO_4^- \rightleftharpoons H^+ + SO_4^{2-} $$ $$ \text{Initial: } C \quad C \quad 0 $$ $$ \text{Eq: } C-x \quad C+x \quad x $$

Example: 0.01M $H_2SO_4$ ($K_{a2} = 1.2 \times 10^{-2}$)

$$ K_{a2} = \frac{(C+x)(x)}{C-x} $$ $$ 1.2 \times 10^{-2} = \frac{(0.01+x)x}{0.01-x} $$

Since $K_{a2}$ is comparable to $C$, we cannot ignore x. We must solve the quadratic equation.
Solving gives $x \approx 0.0045 M$.

Total $[H^+]$:

$$ [H^+]_{total} = C + x = 0.01 + 0.0045 = 0.0145 M $$ $$ pH = -\log(0.0145) \approx 1.84 $$

Summary Formulas

Case Type	Condition	Calculation
Weak Diprotic ($H_2S$)	$K_{a1} \gg K_{a2}$	$[H^+] \approx \sqrt{K_{a1}C}$ $[A^{2-}] \approx K_{a2}$
Strong-Weak ($H_2SO_4$)	Step 1 Complete	$[H^+] = C + x$ Solve Quad for x
Triprotic ($H_3PO_4$)	$K_{a1} \gg K_{a2} \gg K_{a3}$	$[H^+] \approx \sqrt{K_{a1}C}$ $[HPO_4^{2-}] \approx K_{a2}$

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