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mols of HNO3 = 87.5 millimol.

mols of KOH = 112.5 millimol

Moles of KOH left over = 25 millimol

Vol = 800 ml

Molarity of OH- = 0.03125

Thus pH = 14+log(0.03125) = 12.49

2007-05-02 20:54:38 · answer #1 · answered by ag_iitkgp 7 · 1 0

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