Question

What is the equilibrium pH of an initially 0.350M for a monoprotic acid with a Ka = 1.066E-4 Submit Answer Tries 0/98

Answer 1

let monopoolic acid be -> HA. ICE table- concentration cm) HA a) -> H+ + A 0.350 -8 to ulo ta 0. 350-8 - a e 0.350_1 - K. = CHACA - Goca L = 1.0663164 - CHI : a = 0.350-(1.066x104] x = 3.731xcoF-1.066x16 4x4 a +1.066x1ö91 – 3.731x105 =0 at + bx+ c to = x=-67562_4ac 2a :.-1.066 X164+31.068x10m) un E-3.721st 2CO

cix=-1.066 xcö't 0.0122 2 .:x= +6.055 Xlo Xlo (03) - 6- 15x103 hot occur. ACH) = x= 6.055X0² - take tue solution since ve concentrations ..pH = -log CHI)=-log (6.055X10²) ani pH = [2-22)