Question

Acetic acid has a Ka of 1.8*10^-5. Three acetic acid/ acetate buffer solutions, A,B, and C, wer made using varying concentrations: 1. [acetic acid] ten times greater than [acetate] 2. [acetate] ten times greater than [acetic acid] 3. [acetate] = [acetic acid] Match each buffer to the expected pH pH = 3.74 pH= 4.74 pH = 5.74

Answer 1

Explanation:

It is given that $K_{a}$ for acetic acid is $1.8\tiemes 10^{-5}$. And, its $pK_{a}$ value will be calculated as follows.

            $pK_{a} = -log_{10} K_{a}$

                        = $-log_{10} (1.8 \times 10^{-5}$

                        = 4.74

And, according to the Henderson-Hasselbalch equation,

             pH = $pK_{a} + log \frac{[Salt]}{[Acid]}$

(1).   When [Acetic acid] is ten times greater than [acetate]

This means that $\frac{[\text{Acetate}]}{[\text{Acetic acid}]} = \frac{1}{10}$

So,   pH = $pK_{a} + log \frac{[Acetate]}{[\text{Acetic Acid}]}$

             = $4.74 + log \frac{1}{10}$

             = 3.74

(2). When [Acetate] ten times greater than [Acetic acid]

This means that $\frac{[Acetate]}{\text{Acetic acid}}$ = $\frac{10}{1}$

             pH = $4.74 + log_{10} 10$

                   = 5.74

(3).  When [acetate] = [acetic acid]

This means that $\frac{\text{Acetate}}{\text{Acetic acid}}$ = 1

            pH = $4.74 + log_{10} 1$

                  = 4.74