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If the pKa is 8.3, when Ka = 10^-pKa = 5.012*10^-9.

Then, [H+]= sq. rt(Ka*C0). {C0 is the intial concentration of the acid}.

[H+] = sq. rt[(5.012*10^-9)(0.05M)}

= 1.58*10^-5 M

Then, pH = -log[H+]

pH = -log (1.58*10^-5M)

= 4.8

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Q: What is the pH of a 0.05M solution of TRIS acid with a pKa of 8.3?
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