What is pH of weak dibasic acid, that is 2 % dissociated in its M / 100 solution at 298 K ?

Step-by-step solution

To determine the pH of a weak dibasic acid solution at concentration C = 0.01 M with 2% dissociation, the following approach applies. As each dibasic acid molecule H₂A contributes two H^+ ions upon dissociation, the hydrogen ion concentration is [H^+] = 2 C 0.02 = 2 0.01 0.02 = 0.0004 M. Thus, pH = - ₁₀(0.0004) = - ₁₀(4 10⁻⁴) = -( ₁₀4 + ₁₀10⁻⁴) . Since ₁₀4 0.6020 and ₁₀10⁻⁴ = -4 , we have pH = -(0.6020 - 4) = -(-3.3980) = 3.3980 . This value closely matches option C .