blossie73021 blossie73021

27-10-2017
Chemistry

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Calculate the ph of 0.63 m nh3 (kb=1.8×10−5).

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destinymcbrydeoyh0xi destinymcbrydeoyh0xi

27-10-2017

[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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