What is [ H ⁺ ] in mol / L of a solution that is 0.20 M in CH ₃ COONa and 0.10 M in CH ₃ COOH ? ( K _ a . for CH ₃ COOH =1.8 10⁻⁵ )

Step-by-step solution

Key Idea CH ₃ COOH (weak acid) and CH ₃ COONa (conjugated salt) form acidic buffer and for acidic buffer, pH = pK _ a + [ salt ] [ acid ] and aligned [ H ⁺ ] & =- antilog pH pH & =- K _ a + [ salt ] [ acid ] & . pK _ a =- K _ a ] & =- (1.8 10⁻⁵ )+ (0.20) (0.10) & =4.74+ 2 & =4.74+0.3010=5.041 aligned Now, gathered [ H ⁺ ]= antilog (-5.045) =9.0 10⁻⁶ ~mol / L gathered