Question

What change in pH should be observed if 10.0 mL of 0.100 M NaOH is added to 100 mL of a buffer that is 0.100 M in CH₃COOH and 0.100 M in NaCH₃CO₂? K_a for acetic acid is 1.8 x 10^-5.

If you can explain step by step that would be awesome!

Answer 1

CH3COOH +H2O ----->CH3COO- + H3O+

0.1              0                0.1            0    initial

0.1-x                             0.1+x       +x    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   0.01mol                0.01mol initial

0                 0.01-.001         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