Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.63-M HCl to 440. mL of each of the following solutions.

a.) .181 M C2H3O2 1-

ph before mixing: 9.00 (correct)

pH after mixing: ??

b.) a buffer solution that is 0.181 M in each C2H3O2 1- and HC2H3O2

pH before mixing: ??

pH after mixing: ??

Thank you!! Sincerely, a confused college student

Answer 1

10.0 mL 3.22 M HCl + 420 mL water

pH before mixing= -0.51 for HCl and 7 for water
pH after mixing= 1.13
pH change= 1.64

b) 10.0 mL 3.22 M HCl + 420 mL 0.160 M CH3COO-
reaction: HCl + CH3COO- --> CH3COOH + Cl-

we have 32.22 mmoles of HCl and 67.20 mmoles of CH3COO-, so we will obtain 67.20-32.22=34.98 mmoles of CH3COO and 32.22 mmoles of CH3COOH in a total of 430 mL. This is a buffer solution.

pH before mixing=-0.51 for HCl and x2/0.160=1.8E-5 and pOH=2.77 and pH=11.23 for CH3COO-
pH after mixing= pH=pKa-log[Cb/Ca]=4.74-log[(34.98/430)/(…
pH change=-6.52

c) we already have 10 mL 32.22 mmoles of HCl and we add 420 mL 0.160 M CH3COOH in this case.
pH before mixing= -0.51 for HCl and x2/0.160=1.8E-5 and pH=2.77 for CH3COOH
pH after mixing= we have a strong acid accompanied by a weak acid We will sum the H(+) ions then:
H(+) ions from HCl=32.22 mmoles
H(+) ions from CH3COOH= x2/0.160=1.8E-5, x=0.0017 M 420 mL=0.712 mmoles
total=32.932 mmoles
total volume=430 milliliters
concentration= 0.0766 M, pH=1.12
pH change= 1.12-2.77=-1.65

d) pH before mixing= -0.51 for HCl and what about the pH of buffer solution?
pH=pKa-log(Cb/Ca)=4.74-log(1)=4.74 for the buffer solution.

pH after mixing=?
the basic component in the buffer solution will react with HCl first:
HCl + CH3COO- --> CH3COOH + H2O
We have 32.22 mmoles of HCl and 67.2 mmoles of CH3COO (and CH3COOH, but we will not use the latter now). So, we will react these two to obtain 32.22 mmoles of CH3COOH and we will have 34.98 mmoles of CH3COO remaining. We had 67.2 mmoles of CH3COOH before, so the total amount of CH3COOH will be 67.2+34.98=102.18 mmoles in 430 mL of a solution, so the concentration will be 0.238 M of CH3COOH.

1.8E-5=x2/0.238 and x (H+ ion concentration)=0.002 M, so the pH=2.68
pH change= 2.68-4.74=-2.06