blossie73021 blossie73021
  • 27-10-2017
  • Chemistry
contestada

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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