Salt of a Strong Base and a Weak Acid. pH of Solution.?
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Calculate the pH of a 5.00×10-1 M aqueous solution of sodium hypobromite (NaOBr). (For hypobromous acid, HOBr, Ka = 2.00×10-9).
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Answer:
The pH will be basic. Why? Because of the partial reaction: BrO(-) + H2O = HBrO + OH(-). The constant is Kb , with Kb = Ke: Ka. Ke = 10^(-14) at 25°C. Kb = 5x10^(-6). You resolve a second degree equation as follows: Kb = w^2: (C-w), with w = (OH-) and C = 0.5 mol per liter. So: 5x10^(-6) = w^2 : (0.5 -w). Or: w^2 +5x10^(-6).w - 2.5x10^(-) = 0. w = 1.578x10^(-3) mol per liter. pH = 11.20 at 25°C.
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