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Chemistry
12 July, 07:04
Calculate the ph of 0.63 m nh3 (kb=1.8*10-5).
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Trey
12 July, 07:28
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[OH-] = √1.8*10^-5 * 0.15
[OH-] = 1.64*10^-3M
pOH = - log (1.64*10^-3)
pOH = 2.79
pH = 14 - 2.79
pH = 11.21
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