Question
what is the [CH3COO^-]/[CH3COOH] ration in an acetate buffer (Ka=1.76 x 10^-5) at pH 3.8? show calculations
Answers
DrBob222
Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base)(acid)]
3.8 = 4.75 + log [(CH3COO^-)/(CH3COOH)]
Solve for the ratio.
If I didn't goof on the calculator I get approximately 0.1.
pH = pKa + log[(base)(acid)]
3.8 = 4.75 + log [(CH3COO^-)/(CH3COOH)]
Solve for the ratio.
If I didn't goof on the calculator I get approximately 0.1.