A beaker with 1.60*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 7.10 mL of a 0.460 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. View Available Hint (s) ΔpH = Previous Answers Incorrect; Try Again; 16 attempts remaining Provide Feedback.

Question

Chemistry

Colt Byrd

19 October, 16:54

A beaker with 1.60*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 7.10 mL of a 0.460 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. View Available Hint (s) ΔpH = Previous Answers Incorrect; Try Again; 16 attempts remaining Provide Feedback.

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Answers (1)

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Isabela Bruce · Answer 1

The pH will change 0.16 (from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 = 1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x = 0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 = 0.0645 M = concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M = 0.00568 moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A - + H + = HA

moles conjugate base = 0.01032 - 0.003266 = 0.007054 moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946 / 0.1671 = 0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054 / 0.1671 = 0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535 / 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16