What is the pH of a .01100 M solution of NH3?
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Use NH3 + HOH --> (NH4+) + (OH-)
Kb = 1.8 x 10^-5 (given in a textbook - rate constant)
Kb = (NH4+)(OH-)/(NH3) = (x^2)/0.01100
1.8 x 10^-5 = (x^2)/0.01100
x = 0.000445 = OH
p(OH) = -log(OH)
p(OH) = -log(0.000445)
p(OH) = 3.35
pH + pOH = 14
pH + 3.35 = 14
pH = 10.7 (approximately)
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