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Find the pH of the equivalance point and the volume (mL) of 0.0884 M KOH needed to reach the equivalance point in the titration of 23.4 mL of 0.0390 M HNO2. Ka HNO2 = 4.5 x 10^(-4) Kb NO2- = 2.2 x 10^(-11) (* I already solved the volume of KOH at equivalance point. It ended up being 10.3237106 mL. I am just having trouble calculating the pH of the final solution at equivalance point. Please show a step-by-step process :)

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