Answer :
Kₐ = 5.8*10⁻⁶
C = 0.45 mol/L
[H⁺] = √(KₐC)
pH = -lg[H⁺] = -lg{√(KₐC)}
pH = -lg{√(5.8*10⁻⁶*0.45)} = 2.79
C = 0.45 mol/L
[H⁺] = √(KₐC)
pH = -lg[H⁺] = -lg{√(KₐC)}
pH = -lg{√(5.8*10⁻⁶*0.45)} = 2.79