Datos : m = 0.
0560 g CH₃⁻COOH solución = 50.
0 ml sol Ka = 1.
8 * 10⁻⁵ Ecuación química CH₃ - COOH→ H⁺ + CH₃ - COO⁻ Se calcula la molaridad : Peso molecular ácido acético C = 12 * 1 = 12 H = 1 * 4 = 4 O = 16 * 2 = 32 ___ Pm = 60 g / mol 50.
0 ml * 1 litro / 1000 ml = 0.
05 litros sol Formula de molaridad M = gramos de soluto / ( Pm * V sol (l) ) M = 0.
0560 g / ( 60 g / mol * 0.
05 Lts ) M = 0.
0186 mol / Lts Se calcula la concentración de H⁺ partiendo de la ecuación de Ka : [ H⁺ ] * [ CH₃₋COO⁻ ] Ka = _____________________ [ CH₃₋COOH ] [ H⁺ ] = [ CH₃₋COO⁻ ] = M * α [ CH₃₋COOH ] = M ( M * α ) * ( M * α ) Ka = _________________ M Ka = ( M * α )² / M ( M * α )² = Ka * M ( M * α ) = √ ( Ka * M ) ( M * α ) = √ ( 1.
8 * 10⁻⁵ * 0.
0186 ) ( M * α ) = √ 3.
348 * 10⁻⁷ ( M * α ) = 5.
78 * 10⁻⁴ [ H⁺ ] = 5.
78 * 10⁻⁴ [ CH₃₋COO⁻ ] = 5.
78 * 10⁻⁴ [ CH₃₋ COOH ] = 0.
0186.