A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a

Question

A beaker with 155 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 8.70 mL of a 0.380 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.

Part A A beaker with 155 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 8.70 mL of a 0.380 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. ApH- 4.636 Submit Hints Mv Answers Give Up Review Part Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

buffer milimoles = 155 x 0.1 = 15.5 = acid + salt

acid = acetic acid, salt = sodium acetate

pH = pKa + log [salt/acid]

5.0 = 4.74 + log [salt/acid]

salt/acid = 1.82

salt = 1.82 acid

1.82 acid + acid = 15.5

acid = 5.5 millimoles

salt = 10 millimoles

8.70 mL of a 0.380 M HCl added

millimoles of HCl = C = 8.70 x 0.380 = 3.306

on addition of ’ C’ millimoles of acid to acidic buffer salt moles decreases and acid moles increases

so

pH = pKa + log [Salt –C/acid + C]

pH = 4.74 + log [10 –3.306 / 5.5 + 3.306 ]

pH = 4.62

change in pH = final pH - initial pH

                      = 4.62 - 5.0

                     = - 0.38

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