Question

A beaker with 1.90×10² 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 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

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

190 mL buffer (0.100 M) = 19 mmols of Acid + Base        (A + B from now on)

B/A = 1.737800829 >>>>>>

B = 1.73780(A)

19 mmol = A + B

19 mmol = A + 1.737800829(A) >>>>  

19 mmol = 2.737800829A >>>>

A = 6.939878 mmol = acetic acid

Amount of B: 19 mmol - 6.939878 mmol = 12.06012 mmol B = acetate

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

5.90 mL HCl (0.410 M) = 2.419 mmol HCl added

                    Acetate         +           HCl        --->    Acetic acid            +             H2O

Before 12.06012 2.419 6.939878

Change          -2.419 -2.419 +2.419

Final 9.64112 0 9.358878

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

pH = 4.760 + log (9.64112/9.358878) = 4.7729

ΔpH: 4.7729 - 5.000 = -0.227 = -0.227

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