Sulphurous acid (H2SO3) has Ka1 = 1.7 x 10^-2 and Ka2 = 6.4 x 10^-8. The pH of 0.588 M is …… (Round

Q: Why do we completely ignore the second dissociation constant ($K_{a_2}$)?

Because $K_{a_1}$ is about a million times larger than $K_{a_2}$ ($10^{-2}$ vs $10^{-8}$). The amount of $H^+$ produced by the second step is so minuscule that it doesn't affect the overall pH.

Q: Is the approximation $1-\alpha \approx 1$ always valid?

Not always! It's generally safe if $\alpha$ is less than 5% (or $0.05$). Here, $\alpha$ is around 17%, which is a bit high, but using the approximation still leads us to the correct nearest integer for the pH. If you need extreme precision, you must solve the full quadratic equation.

Q: What if I solved the full quadratic equation without the approximation?

You would get $\alpha \approx 0.156$, leading to $[H^+] \approx 0.0918 \text{ M}$. The pH would be $-\log(0.0918) \approx 1.037$. Rounding to the nearest integer still gives you exactly 1!

Q: Why is the concentration of $HSO_3^-$ equal to $H^+$?

Because in the first dissociation step, one molecule of $H_2SO_3$ breaks apart to form exactly one $H^+$ ion and one $HSO_3^-$ ion. They are produced in a 1:1 stoichiometric ratio.

Q: How do I quickly calculate the square root of $0.0289$ without a calculator?

Notice that $289$ is a perfect square! $17 \times 17 = 289$. Therefore, the square root of $0.0289$ (which is $289 \times 10^{-4}$) is simply $17 \times 10^{-2}$, or $0.17$.

Question

Sulphurous acid ($	ext{H}_2	ext{SO}_3$) has $K_{a_1} = 1.7 	imes 10^{-2}$ and $K_{a_2} = 6.4 	imes 10^{-8}$. The pH of $0.588 	ext{ M}$ is ……
(Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a beaker filled with a $0.588 	ext{ M}$ solution of sulphurous acid ($	ext{H}_2	ext{SO}_3$). Sulphurous acid is a diprotic acid, meaning it has the potential to release two protons ($	ext{H}^+$) into the solution. However, it doesn't release them all at once; it does so in two distinct steps, each governed by its own equilibrium constant.

We are given two dissociation constants: $K_{a_1} = 1.7 	imes 10^{-2}$ and $K_{a_2} = 6.4 	imes 10^{-8}$. Notice how $K_{a_1}$ is significantly larger than $K_{a_2}$ ($K_{a_1} \gg K_{a_2}$). This massive difference tells us a crucial physical reality: almost all the hydrogen ions in the solution will come from the first dissociation. The second dissociation is so weak that its contribution to the total $[	ext{H}^+]$ is negligible. Therefore, we can safely ignore the second step and focus entirely on the first ionization.

### The Master Equation

Let's write down the chemical equation for the first ionization:
$$ 	ext{H}_2	ext{SO}_3 ightleftharpoons 	ext{H}^+ + 	ext{HSO}_3^- $$

Initially, before any dissociation happens, the concentration of $	ext{H}_2	ext{SO}_3$ is $0.588 	ext{ M}$, and the concentrations of the products are zero. As the system reaches equilibrium, a certain fraction of the acid, let's call it $\alpha$ (the degree of dissociation), breaks apart.

At equilibrium, the concentrations will be:
- $[	ext{H}_2	ext{SO}_3] = 0.588(1 - \alpha)$
- $[	ext{H}^+] = 0.588\alpha$
- $[	ext{HSO}_3^-] = 0.588\alpha$

Now, we plug these equilibrium concentrations into the expression for the first acid dissociation constant, $K_{a_1}$:
$$ K_{a_1} = \frac{[	ext{H}^+][	ext{HSO}_3^-]}{[	ext{H}_2	ext{SO}_3]} $$
$$ 1.7 	imes 10^{-2} = \frac{(0.588\alpha)(0.588\alpha)}{0.588(1 - \alpha)} = \frac{0.588\alpha^2}{1 - \alpha} $$

### The Crucial Approximation

Solving a quadratic equation can be tedious. However, because sulphurous acid is a weak acid, the degree of dissociation $\alpha$ is relatively small compared to $1$. This allows us to make a powerful approximation: $1 - \alpha \approx 1$.

Applying this approximation simplifies our equation beautifully:
$$ 1.7 	imes 10^{-2} \approx 0.588\alpha^2 $$

### Final Calculation

Now, let's solve for $\alpha$:
$$ \alpha^2 = \frac{1.7 	imes 10^{-2}}{0.588} \approx 0.0289 $$
Taking the square root of both sides gives us:
$$ \alpha = \sqrt{0.0289} = 0.17 $$

With $\alpha$ in hand, we can easily find the concentration of hydrogen ions:
$$ [	ext{H}^+] = 0.588\alpha = 0.588 	imes 0.17 = 0.09996 	ext{ M} $$
This value is incredibly close to $0.1 	ext{ M}$, which can be written as $10^{-1} 	ext{ M}$.

Finally, to find the pH, we take the negative logarithm of the hydrogen ion concentration:
$$ 	ext{pH} = -\log[	ext{H}^+] $$
$$ 	ext{pH} = -\log(10^{-1}) = 1 $$

**The pH of the solution is exactly 1.**

Animated Solution for Chemistry - Ionic Equilibrium: Sulphurous acid ( $H_{2} SO_{3}$ ) has $K_{a_{1}} = 1.7 \times 1 0^{- 2}$ and $K_{a_{2}} = 6.4 \times 1 0^{- 8}$ . The pH of $0.588 M$ is …… (Round off to the nearest integer)

Enter Numerical Value:

Visualized Solution

$H_{2} S O_{3}$ Solution

Comparing $K_{a}$ Values

First Ionization Setup

Equilibrium Expression

Approximation

Solving for $α$

Calculating $[H^{+}]$

Final pH Calculation

The Way Forward

The Sigma Insight: pH, Buffer and Indicator

Analyzing the Setup

The Master Equation

The Crucial Approximation

Final Calculation

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