Question

Part A

A beaker with 1.00×10² 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 5.90 mL of a 0.370M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

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

100 mL buffer (0.100 M) = 10 mmols of Acid + Base        (A + B from now on)

B/A = 1.737800829 >>>>>>

B = 1.73780(A)

10 mmol = A + B

14.5 mmol = A + 1.737800829(A) >>>>  

10 mmol = 2.737800829A >>>>

A = 3.653 mmol = acetic acid

Amount of B: 10 mmol - 3.653 mmol = 6.347 mmol B = acetate

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

5.9 mL HCl (0.370 M) = 2.183 mmol HCl added

                    Acetate         +           HCl        --->    Acetic acid            +             H2O

Before            6.347            2.183                3.653

Change          -2.183                      -2.183                +2.183

Final              4.164              0                      5.836

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

pH = 4.760 + log (4.164/5.836) = 4.613

ΔpH: 4.613 - 5.000 = -0.387

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