Question

An acetic acid solution with a pH of 4.06 has a molar analytical concentration of 0.105. Calculate the amount of the acetate ion that is present in this solution

Answer 1

Answer 2

Given:

• pH = 4.06

• **Molar concentration of acetic acid (${\text{CH}}_3\text{COOH}$), $C_a=0.105\text{M}$

• $p{K}_a$ of acetic acid = 4.76 (standard value)

Step 1: Find $[H^{+}]$ from pH

$$[H^{+}]=10^{-\text{pH}}=10^{-4.06}=8.71\times 10^{-5}\text{M}$$

Step 2: Use $K_a$ Expression for Acetic Acid

$$K_a=10^{-p{K}_a}=10^{-4.76}=1.74\times 10^{-5}$$$$K_a=\frac{[H^{+}][{\text{CH}}_3{\text{COO}}^{-}]}{[{\text{CH}}_3\text{COOH}]}$$

Assume $[H^{+}]\approx [{\text{CH}}_3{\text{COO}}^{-}]$ (since dissociation is small):

$$1.74\times 10^{-5}=\frac{(8.71\times 10^{-5})[{\text{CH}}_3{\text{COO}}^{-}]}{0.105-[{\text{CH}}_3{\text{COO}}^{-}]}$$

Step 3: Solve for $[{\text{CH}}_3{\text{COO}}^{-}]$

Approximate $0.105-[{\text{CH}}_3{\text{COO}}^{-}]\approx 0.105$