CH3COOH +H2O ----->CH3COO- + H3O+
0.1Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0.1Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â initial
0.1-x                           Â
0.1+x      +x  Â
equlibirium
Ka= [CH3COO-][H3O+]/[CH3COOH]
1.8x10^-5= [0.1+x ][x] /[0.1-x] x=too small so ignored
x= 1.8x10^-5 x0.1 / 0.1
pH= -log 1.8x10^-5= 5-0.255 = 4.745
after adding NaOH
OH- + CH3COOH --->H2O + CH3COO-
0.001mol Â
0.01mol              Â
0.01mol initial
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0.01-.001Â Â Â Â Â Â Â Â
0.01+0.001 at eq
Ka= [CH3COO-][H] /[CH3COOH]
1.8x10^-5 = [x][ 0.011] /[0.009] =1.222x
x=1.8x10^-5/ 1.222=1.47x10^-5
pH=-log 1.47x10^-5=5-0.167=4.883
change in pH =4.883- 4.745 = 0.088
