Calculate the ph of a 0.10 m solution of hydrazine, n2h4. kb for hydrazine is 1.3×10−6.

N2H4 + H2O = N2H5(+) + OH(-) 

Kb = 1.3 E-06 = [N2H5(+)][OH(-)] / [N2H4] = [OH(-)]^2 / 0.1 

Thus [OH(-)] = 0.000361 

pOH = -log[OH(-)] = 3.44 

pH = 14 - pOH = 10.6

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The first digit is in the middle school 100s place