Calculate the pH of a 0.340 M NaO₂CCO₂H solution.pK₁ = 1.250 and pK₂ = 4.266.

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First, let's write the balanced equation for the dissociation of NaO₂CCO₂H: NaO₂CCO₂H ⇌ H+ + O−₂CCO₂SNa The acid dissociation constant (Ka) for this acid is just the reciprocal of the base dissociation constant (Kb) for its conjugate base: Ka = Kw/Kb = 10^-14/4.52x10^-10 = 2.21x10^-5 The equilibrium expression for this reaction is: Ka = [H+][O−₂CCO₂SNa]/[NaO₂CCO₂H] We can make the following assumptions to simplify the calculation: - The initial concentration of NaO₂CCO₂H is equal to its equilibrium concentration, since it is the limiting reactant. - The concentration of O−₂CCO₂SNa (the conjugate base of the acid) is very high due to the complete dissociation of the salt in water, so we can assume that its concentration is constant. With these assumptions, we can write: Ka = [H+]^2/[NaO₂CCO₂H] [H+]^2 = Ka[NaO₂CCO₂H] [H+]^2 = (2.21x10^-5)(0.340) [H+] = 1.73x10^-3 M pH = -log[H+] = 2.79

Calculate the PH of a 0