December 04, 2020, 03:55:06 PM
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Topic: Ph question, I'm stuck  (Read 379 times)

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Offline jmxwell

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Re: Ph question, I'm stuck
« Reply #15 on: October 29, 2020, 12:48:49 PM »
[H3O+] = Ka(Ca - [H3O+])/(Cb +[H3O+]
[H3O+]^2 + (Cb +Ka)[H3O+] - KaCa = 0
x^2 + (0.2+3.85)x -(3.85.0.1) = 0
x^2 + 4.05x - 0.385=0
x=0.09
ph = -log[H+] = -log(0.09) = 1.04
ph = 3.85+1.04 =4.89

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