Asked by steven

How would you prepare 500 ml of a 1.8 M phosphate buffer (pH 6.8) containing 4.5 x 10-6 M CaCl2 and 52 mM sucrose?

Answers

Answered by DrBob222
pH = pK + log(base)/(acid)
6.8 = pK2 + log(base)/(acid)
Substitute pK2 and solve for b/a ratio.
That's two unknowns. The second equation is
base + acid = 1.8M
Solve the two equations simultaneously for acid and base. Along the route, and before you make it up to the final volume, add enough solid CaCl2 to make3 1.8M and enough sucrose to make it 52 mM.
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