A beaker with 200 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 8.00 mL of a 0.280 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.



Please show your work. Thank you!A beaker with 200 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop...?
The ionisation of a weak acid given by



HA ==%26gt; H+ + A- with Ka = [H+] [A-] / [HA]. Rearranging Ka and taking logs it is easy to show pH = pKa + log ([A-] / [HA]) Eqn 1.



The corresponding charge balance is



[Na+] + [H+] ===%26gt; [OH-] + [A-]



Given that [Na+] %26gt;%26gt;%26gt; [H+] and [A-] %26gt;%26gt;%26gt; [OH-] the charge balance approximates to [Na+] = [A-]



Substituting pH=5.0, pKa = 4.76 into Eqn 1 leads to log ([A-] / [HA]) = 0.24 or [A-] = 1.738 [HA]. Eqn 2



Given is [A-] + [HA] = 0.1 Eqn 3



Solving Eqn 2 %26amp; 3 simultaneously leads to [A-] = 0.0635 M. Hence [Na+] = 0.0635 M and [HA] = 0.0365 M



Now add HCl



The dilution factor becomes 200 / (200 + 8) = 0.962



Thus

1. [Na+] = 0.0635 x 0.962 = 0.0611 M



2. [A-] + [HA] = 0.1 x 0.962 = 0.0962 M



3. [Cl-] = 8 x 0.28 /(200 + 8) = 0.0108 M



The new charge balance becomes



[Na+] + [H+] ===%26gt; [OH-] + [A-] + [Cl-] which simplifies (for the same reasons as before) to



[Na+] ===%26gt; [A-] + [Cl-] . Thus [A-] = [Na+] - [Cl-] = 0.0611 -0.0108 = 0.0503 M



Hence [HA] = 0.0962 - 0.0503 = 0.0459M



From Eqn 1 and substituting values



pH = 4.76 + log (0.0503/0.0459) = 4.80A beaker with 200 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop...?
The ';a'; and ';b'; start at 0.02 moles each.

Now work out how many moles of HCl are added: 8 x 0.28/1000.



';a'; goes up by this amount, and ';b'; goes down by this amount.



Now work out the new number of moles of ';a'; and ';b';.



Now apply [H+] = Ka x a/b.



The new volume, 208, cancels out.