Homework Soursepreparation of 100 mL of a 0100 M acetate buffer required so
preparation of 100 mL of a 0.100 M acetate buffer, required solution pH 4.95. pKa= 4.757
After mixing 3.907mL of 1.0M acetic acid, 6.093mL of 1.0M sodium acetate and 90.0mL of water, the pH measured was 4.69. In order to increase the pH, 2.9mL of NaOH was added to the buffer solution which make the pH = 4.91, but the total volume is now change from 100mL to 102.9mL, which mean the concentration of buffer is not 0.100 M anymore. Calculate the new concentration for the buffer assuming that both of acetic acid and acetate sodium concentration were incorrect.
I know that we have to use the ICE table and the pKa buffer equation for this but, i\'m not sure how to calculate 2 unknown variables. I could get the answer by assuming one of the concentration is correct, but i needed to do the harder way. Please help.
After mixing 3.907mL of 1.0M acetic acid, 6.093mL of 1.0M sodium acetate and 90.0mL of water, the pH measured was 4.69. In order to increase the pH, 2.9mL of NaOH was added to the buffer solution which make the pH = 4.91, but the total volume is now change from 100mL to 102.9mL, which mean the concentration of buffer is not 0.100 M anymore. Calculate the new concentration for the buffer assuming that both of acetic acid and acetate sodium concentration were incorrect.
I know that we have to use the ICE table and the pKa buffer equation for this but, i\'m not sure how to calculate 2 unknown variables. I could get the answer by assuming one of the concentration is correct, but i needed to do the harder way. Please help.
Solution
when NaOH is added [CH3COOH] = ( 3.907-2.9M)/102.9 , where M- molarity of NaoH ,
[CH3COO-] = ( 6.093+2.9M)/102.9, using henderson eq
4.95= 4.757 + log{6.093+2.9M)/(3.907-2.9M),
M= 0.000258 Molar NaoH ,
new [CH3COO} = ( 3.907-2.9(0.00258)/102.9 = 0.03796,
[CH3COO-] = (6.093+2.9(0.00258)/102.9 = 0.05922 M,
conc of buffer = [CH3COOH] +[CH3COOna]= 0.0971 M
