Question

0.1M of Acetic acid and 0.25 M sodium acetate. Ka of acetic acid = 1.76X10^-5

Answer 1

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 :

$P_H= P_{Ka} + log (\dfrac{base}{acid})$

$P_H= 4.76 + log(\dfrac{0.1}{0.25})$

$P_H=4.362$