0.1M of Acetic acid and 0.25 M sodium acetate. Ka of acetic acid = 1.76X10^-5
At very first assume that the solution is of "1l"
Now:
Comparing the "K_a" of acetic acid, "1.76X10^-5" , these solutions aren’t very large comparative to .25M and .1M, it’s a buffer equation because there is a acid and conjugate base, so you’re able to use the Henderson Hasselbalch equation.
Now :
"P_H= P_{Ka} + log (\\dfrac{base}{acid})"
"P_H= 4.76 + log(\\dfrac{0.1}{0.25})"
"P_H=4.362"
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