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3). I need to make a 0.50M Tris buffer with a pH of 8.75. I have access to Tris (a weak acid) and the sodium salt of Tris (thThe Ka of Tris is 8.5x10-9 and the pKa of Tris is 8.07. What are the [HA] and [A-] concentrations of this 0.50M buffer at pH 8.75?

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Answer #1

Solution-

pH= pKa + log(base/acid)

=>8.75= 8.07 + log(base/acid)

=> log(A-/HA) = 0.68

=> (A-/HA) 10^0.68 = 4 .786301

=> (A-)=  4 .786301*(HA)

=>concentration of buffer = 0.5M

So (A-)+(HA)= 0.5

4 .786301*(A-)+(HA)= 0.5

5.786301*(HA)= 0.5

(HA)= 0.5/5.786301

=0.086411

=(A-)= 4.786301*(HA)

=4.786301*0.086411

=0.413589

(HA)= 0.0864

(A-)=0.414 M

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