8.58e-2 mol A - 1.2e-1 = -0.0342 mol
This means that there are 0.0342 mol of excess NaOH
[OH-] = 0.0342 mol/ 1L =0.0342 M
pOH = -log(0.0342) = 1.466
pH = 14 - 1.466
pH = 12.534
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