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How many millimeters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride to give a pH of 7.60 in a final volume of 0.250 mL?
Useful information: Tris hyrochloride fw = 157.596 g/ml
The Kb for tris is 1.2 x 10^-6
You can represent tris hydrochloride as BH+ and its conjugate base, tris, as B.
10.0 g of trisHCl = 0.06345 moles
Kb = [OH-] x (BH+/B) is the equation I will use for a buffer based on a base (tris)
Now we sub in all our known values.
Kb we knoe 1.2 x 10^-6
[OH-] = 10^-6.4
BH+ = 0.06345 moles
But BH+ + OH ---> B
Initial 0.06345 x 0
Change -x -x +x
-----------------------------------------------------------
Final 0.06345-x 0 x
so
1.2 x 10^-6 = [10^-6.4] x {(0.06345 -x)/x}
solve for x, which will be moles of NaOH added, then use the 0.500 M concentration value to calculate volume needed.