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Step-by-Step Solution
Step 1: Identify the relevant dissociation constants
Sulphurous acid ( H_{2}SO_{3} ) has two dissociation constants:
K_{a_{1}} = 1.7 \times 10^{-2} and K_{a_{2}} = 6.4 \times 10^{-8}.
Because K_{a_{1}} \gg K_{a_{2}}, the contribution of H^{+} ions from the second dissociation is negligible compared to the first.
Step 2: Write the expression for the first dissociation constant
Let the initial concentration of H_{2}SO_{3} be c. The dissociation can be represented as:
H_{2}SO_{3} \rightleftharpoons H^{+} + HSO_{3}^{-}.
Define \alpha as the degree of dissociation for the first step. Then, at equilibrium:
[H_{2}SO_{3}] = c(1 - \alpha)
[H^{+}] = c \alpha
[HSO_{3}^{-}] = c \alpha
The acid dissociation constant K_{a_{1}} is:
K_{a_{1}} = \dfrac{[H^{+}][HSO_{3}^{-}]}{[H_{2}SO_{3}]}
= \dfrac{(c \alpha)(c \alpha)}{c (1 - \alpha)}
= \dfrac{c \alpha^{2}}{(1 - \alpha)}.
Step 3: Substitute the given values
Given c = 0.588\,\text{M} and K_{a_{1}} = 1.7 \times 10^{-2}. Thus,
\dfrac{0.588\, \alpha^{2}}{1 - \alpha} = 1.7 \times 10^{-2}.
Rewriting,
0.588\, \alpha^{2} = (1 - \alpha) \times 1.7 \times 10^{-2}.
Multiply both sides by 100 to simplify numerical handling:
58.8\, \alpha^{2} = 1.7 - 1.7\, \alpha.
Arrange terms into a standard quadratic form:
58.8\, \alpha^{2} + 1.7\, \alpha - 1.7 = 0.
Step 4: Solve the quadratic equation for α
Compute \alpha using the quadratic formula:
\alpha = \dfrac{-b \pm \sqrt{b^{2} + 4ac}}{2a},
where a = 58.8, b = 1.7, and c = -1.7. Substituting:
\alpha = \dfrac{-1.7 \pm \sqrt{(1.7)^{2} + 4 \times 58.8 \times 1.7}}{2 \times 58.8}.
Upon calculation,
\alpha \approx 0.156.
Step 5: Determine the hydrogen ion concentration
Since \alpha is the degree of dissociation,
[H^{+}] = c\, \alpha = 0.588 \times 0.156 \approx 0.092\,\text{M}.
Step 6: Calculate pH
The \text{pH} is given by:
\text{pH} = -\log[H^{+}] = -\log(0.092).
Numerically,
\text{pH} \approx 1.036.
Rounding off to the nearest integer:
\boxed{1}.
