Given Molarity and Ka, Solve For PH
Posted 5:08 AM

In water, when [H+] and [A-] are produced, [H+] and [A-] are approx equal (call it x). If Ka is very small compared to M (which it usually is), then you can usually reduce it to sqrt(M*Ka).

x^2/(M-x) = Ka
x = sqrt(M*Ka-Ka^2/4)-Ka/2
ph = -log(x)

Labels: biotech

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