Martin P. answered • 06/18/22
graduate chemistry work, Doctorate degree, Former College Professor
This is a Buffer question.
The buffer solution is made up of a weak acid CH3COOH and its conjugate base CH3COO-
to be added to a strong acid HCL.
Step 1) The first decision you need to make is what part of the buffer will add to the strong acid.
You can't add an acid to an acid to obtain a reaction, but added the conjugate base CH3COO- can accept the H^+1 from the acid. Therefore the base in moles CH3COO- will go down and the acid in moles will go up.
equation:
CH3COO- + HCL -----------> CH3COOH
The next step: convert each substance to moles.
mol CH3COOH = 0.1 L x 0.1 m/L. = 0.01 mol
mol CH3COO- = 0.1 L. x. 0.05 m/L = 5.0 x 10^-3 mol
mol HCL = 0.003 L x 0.01 m/L = 3.0 x 10^-5 mol
Step 3) Since you are buffering a strong acid this reaction goes to completion. Set up a BCA table
B = before any reaction took place
C = the change after the reaction
A = after the reaction went to completion
you determine the limiting reactant by the smaller compound with the smaller # of moles and since that limiting reactant will always in a reaction that goes to completion go to zero it needs to be subtracted on the reactant side of the equation and added to the product side of the equation. In a properly set up buffer solution the number of moles, in this case the HCL will always be the limiting reactant.
CH3COO- + HCL -----------> CH3COOH
B 0.005 3.0 x 10^-5 0.01
C - 3.0 x 10^-5 - 3.0 x 10^-5 + 3.0 x 10^-5
A 4.97 x10^-3 0.00 0.01003
step 4) Use the Henderson Hasselbalch equation:
pH = PKa ( -log of the Ka) + log (base)/(acid)
the Ka for CH3COOH (acetic acid) = 1.8 x 10^-5
PKa = -log 1.8 10^-5 = 4.45
pH = 4.45 + log 4.97 x10^-3/0.01003
pH = 4.45 + log 0.497
pH = 4.45 - 0.3
pH = 4.45
