Question

A beaker with 145 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.50 mL of a 0.350 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
Express your answer numerically to two decimal places. Use a minus ( - ) sign if the pH has decreased.
delta pH=
Please show all work thanks

Answer 1

Let's first determine how much acetic acid and acetate we have in the buffer:

pH = pKa + log (base/acid)>>>>

5.000 = 4.760 + log (base/acid) >>>>

0.240 = log (base/acid)>>>

10^0.240 = 10^{log (base/acid)} >>>

base/acid = 1.73780

145 mL buffer (0.100 M) = 14.5 mmols of Acid + Base        (A + B from now on)

B/A = 1.737800829 >>>>>>

B = 1.73780(A)

14.5 mmol = A + B

14.5 mmol = A + 1.737800829(A) >>>>  

14.5 mmol = 2.737800829A >>>>

A = 5.296221641 mmol = acetic acid

Amount of B: 14.5 mmol - 5.296221641 mmol = 9.203778359 mmol B = acetate

Now that we have our mmol of A and B, we can see what remains after HCl is added:

8.50 mL HCl (0.350 M) = 2.975 mmol HCl added

                    Acetate         +           HCl        --->    Acetic acid            +             H2O

Before            9.203778359            2.975                5.296221641

Change          -2.975                      -2.975                +2.975

Final              6.228778359              0                      8.271221641

We still have a buffer, so we can use pH = pKa + log (base/acid) again to find the pH:

pH = 4.760 + log (6.228778359/8.271221641) = 4.6368

ΔpH: 4.6368 - 5.000 = -0.3632 = -0.363