A beaker with 110 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop.?

A beaker with 110 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1 M. A student adds 9.00 mL of a 0.420 M HCLsolution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.A beaker with 110 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop.?
That pH = 5.00 means [H+] = 10^-5.00 M.

That the pKa of acetic acid is 4.76, means:

10^-4.76 = [H+]*[AC-]/[HAC], or:

[AC-]/[HAC] = 10^0.24 = 1.738

adding 1 to both side, we have:

0.1/[HAC] = ([HAC]+[AC-])/[HAC] = 2.738

Thus [HAC] = 0.1/2.738 = 0.0365 (M)

and [AC-] = 0.1000 - 0.0365 = 0.0635 (M)

Right after increasing the volume to 119ml, the concentrations are:

[HAC] = 0.0365M *(110/119) = (4.015/119)M

and [AC-] = 0.0635M *(110/119) = (6.985/119)M



Making 9.00mL of a 0.420 M HCL solution to a final 119ml would dilute the HCl solution to:

0.420M *(9.00/119) = (3.78/119)M

Mixing HCl and AC- will cause the reaction to form Cl- and HAC. Thus the new concentrations are:

[HAC] = (4.015/119)M + (3.78/119)M = (7.795/119)M

and [AC-] = (6.985/119)M - (3.78/119)M = (3.205/119)M

Hence:

Ka = 10^-4.76 = [H+]*[AC-]/[HAC] = [H+]*3.205/7.795

pH = -log([H+]) = -log(10^-4.76*7.795/3.205) = 4.37