The pH of 10.0 mL of 0.400 M hypochlorous acid solution, before the addition of 0.500 M NaOH, is 3.96.
The acid-base reaction between hypochlorous acid and NaOH is the following:
HClO(aq) + NaOH(aq) ⇄ H₂O(l) + NaClO(aq) (1)
Before the addition of NaOH, we have the following equilibrium dissociation of the acid in water
HClO(aq) + H₂O(l) ⇄ H₃O⁺(aq) + ClO⁻(aq) (2)
0.400 - x x x
The acid constant for the reaction (2) is:
[tex] K_{a} = \frac{[H_{3}O^{+}][ClO^{-}]}{[HClO]} [/tex]
[tex] 3.0 \cdot 10^{-8} = \frac{x*x}{(0.400 - x)} [/tex]
[tex] 3.0 \cdot 10^{-8}(0.400 - x) - x^{2} = 0 [/tex]
After solving the above quadratic equation, we have:
[tex] x_{1} = 1.095 \cdot 10^{-4} M = [H_{3}O^{+}] = [ClO^{-}] [/tex]
[tex] x_{2} = -1.096 \cdot 10^{-4} M = [H_{3}O^{+}] = [ClO^{-}] [/tex]
Now, taking the positive value of concentrations (it cannot be negative), we can calculate the pH of the solution
[tex] pH = -log([H_{3}O^{+}]) = -log(1.095 \cdot 10^{-4}) = 3.96 [/tex]
Therefore, the pH of the hypochlorous acid solution before the addition of NaOH is 3.96.
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