To calculate the pH of a 0.009 mol/dm³ solution of tetraoxosulphate (VI) acid, also known as sulfuric acid (H₂SO₄), we need to consider how sulfuric acid dissociates in water.
Sulfuric acid is a strong acid and dissociates completely in its first step:
\[ \text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^- \]
The first dissociation produces 1 mole of \( \text{H}^+ \) ions for every mole of \( \text{H}_2\text{SO}_4 \).
However, the bisulfate ion (\( \text{HSO}_4^- \)) can also dissociate further, but to a lesser extent:
\[ \text{HSO}_4^- \rightarrow \text{H}^+ + \text{SO}_4^{2-} \]
In dilute solutions, the second dissociation is usually not complete, and we can assume it contributes additional \( \text{H}^+ \).
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From the first dissociation, if we start with 0.009 mol/dm³ of \( \text{H}_2\text{SO}_4 \), we will produce: \[ [\text{H}^+] = 0.009 , \text{mol/dm}^3 \]
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The second dissociation of \( \text{HSO}_4^- \) can also contribute \( \text{H}^+ \). For a rough estimate, we'll use the dissociation constant \( K_a \) for \( \text{HSO}_4^- \). The \( K_a \) for bisulfate is approximately \( 1.2 \times 10^{-2} \).
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We can set up an expression for the concentration of \( \text{H}^+ \) from the second dissociation: \[ K_a = \frac{[\text{H}^+][\text{SO}_4^{2-}]}{[\text{HSO}_4^-]} \]
Let's denote \( x \) as the concentration of \( \text{H}^+ \) from this second dissociation. Thus: \[ K_a \approx \frac{(0.009 + x)(x)}{(0.009 - x)} \]
For most calculations we can assume \( x \) will be small compared to 0.009, and simplify to: \[ K_a \approx \frac{(0.009)(x)}{0.009} \] \[ x \approx K_a \] \[ x \approx 1.2 \times 10^{-2} \text{ mol/dm}^3 \]
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The total \( [\text{H}^+] \) becomes: \[ [\text{H}^+] \approx 0.009 + 0.012 = 0.021 \text{ mol/dm}^3 \]
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Finally, to find the pH: \[ \text{pH} = -\log[\text{H}^+] = -\log(0.021) \]
Calculating the logarithm: \[ \text{pH} \approx 1.68 \]
So the pH of a 0.009 mol/dm³ solution of tetraoxosulphate (VI) acid is approximately 1.68.
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