Question

Determine the PH for a 0.267 M NaCH3H5O3 solution at 25 degrees Celsius, the Ka for HO3H5CH3 is 1.38x10-4

Answer 1

Concentration of CH,CH,0,Na is 0.267M К, of CH,CH,Cн,0,Н %31.38х10* К, of CH,CH,CH,0,н к, 1.00x 10-14 1.38x10 = 7.25 x1011 Base hydrolysis reaction: CH CH,O, + н,о > сн,CH,о,н + он 1(м): 0.267 0 0 С (М): x -x E(M) 0.267 -x х [CH,CH,0,H OH K = [снCH.о, ] 7.25x 10-1 0.267 x Assume x is much smaller than 0.267, so (0.267 -x) = 0.267 7.25x10 0.267 x2 7.25x 101x 0.267 =19.3x 10-12 x=19.3x1012 =4.40x 10 M From the equivalence table, OH x = 4.40 x10 M - -log[OH =-log (4.40x10 - 6.00 log (4.40) РОН %3D = 5.36 рH %3D 14.00 - РОН = 14.00 5.36 8.64 Hence, the pH of the resulting solution is 8.64