Answer:
pH = 2.10
Explanation:
We name an acid as diprotic because it can release two protons:
H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
We propose the mass balance:
Analytical concentration = [H₂A] + [HA⁻] + [A⁻²]
As Ka₂ is so small, we avoid the [A⁻²] so:
0.18 M = [H₂A] + [HA⁻]
But we can not avoid the HA⁻, because the Ka₁. Ka₁'s expression is:
Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]
We propose the charge balance:
[H₃O⁺] = [HA⁻] + [A⁻²] + [OH⁻]
As we did not consider the A⁻², we can miss the term and if
Kw = H⁺ . OH⁻
We replace Kw/H⁺ = OH⁻. So the new equation is:
[H₃O⁺] = [HA⁻] + Kw / [H₃O⁺]
The acid is so concentrated, so we can avoid the term with the Kw, so:
[H₃O⁺] = [HA⁻]
In the mass balance we would have:
0.18 M = [H₂A]
We replace at Ka₁
Ka₁ = [H₃O⁺] . [HA⁻] / [H₂A]
Ka1 . 0.18 / [H₃O⁺] = [HA⁻]
We replace at the charge balance:
[H₃O⁺] = Ka1 . 0.18 / [H₃O⁺]
[H₃O⁺]² = 3.4×10⁻⁴ . 0.18
[H₃O⁺] = √(3.4×10⁻⁴ . 0.18)
[H₃O⁺] = 7.82×10⁻³
- log [H₃O⁺] = pH → - log 7.82×10⁻³
pH = 2.10
Following are the calculation to the pH:
For First ionization:
[tex][H^+] = (K_{a1} \times C)^{\frac{1}{2}}[/tex]
where
C = initial concentration of acid [tex]= 0.18\ M[/tex]
[tex][H^{+}] = (3.4 \times 10^{-4} \times 0.18 \ M)^{\frac{1}{2}}\\[/tex]
[tex][H^{+}] = 0.007\ M[/tex]
For Second ionization:
[tex][H^{+}] = K_{a2} \\\\[/tex]
[tex][ H^{+} ] = 6.7 \times 10^{-9}\ M \\[/tex]
[tex]Total [H^{+}] = 0.007\ M + 6.7 \times 10^{-9}\ M\\\\Total [H^{+}] = 0.007 \ M[/tex]
[tex]pH = -\log[H^+] \\\\pH = -\log(0.007 \ M)\\\\pH = 2.15[/tex]
Therefore, the pH is "2.15".
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