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Calculate the pH of a 0.028 m NACN solution Ka 4.9 x 10 -10. the answer is 10.88 please help?
Wiki User
∙ 15y ago
This is a weak base problem.
Assume
F = 0.028M
Ka = 4.9 10^-10
Kb = Kw / Ka
Kw - 10^-14 therefor pKw=14
x = [OH-]
Kw / Ka = Kb = x^2 / (F-x)
solve for "x" by use of quadratic formula to get the [OH-]
pOH = -log [OH-]
then plug pOH into the pH equation
pH =pKw - pOH
Wiki User
∙ 15y ago
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