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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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Answer:

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

Lanuel

Based on the calculations, the pH change is equal to 5.21.

Given the following data:

  • Volume of acetic acid buffer = 160 mL.
  • pH of acetic acid buffer = 5.000.
  • Total molarity = 0.100 M.
  • Volume of HCl = 7.10 mL.
  • Molarity of HCl = 0.460 M.
  • pKa of acetic acid = 4.740.

How to determine the pH change.

First of all, we would determine the molarities of the base and the acid by applying Henderson-Hasselbalch equation:

[tex]pH =pK_a+ log_{10} \frac{A^-}{HA}\\\\5.000=4.740+log_{10} (\frac{x}{0.100-x})\\\\log_{10} (\frac{x}{0.100-x})=5.000-4.740\\\\log_{10} (\frac{x}{0.100-x})=0.26\\\\\frac{x}{0.100-x}=log^{-1}0.26\\\\\frac{x}{0.100-x}=1.8197\\\\x=0.18197-1.8197x\\\\x+1.8197x=0.18197\\\\2.8197x=0.18197\\\\x=\frac{0.18197}{2.8197}[/tex]

x = 0.0645 M (acetate concentration).

For acetic acid concentration:

[tex]0.100-x=HA\\\\HA=0.100-0.0645[/tex]

HA = 0.0355 M.

Next, we would determine the moles of both acetic acid, acetate and HCl:

For acetic acid:

[tex]Number \;of \;moles = 0.0355 \times 0.160[/tex]

Moles = 0.00568 mol.

For acetate:

[tex]Number \;of \;moles = 0.0645 \times 0.160[/tex]

Moles = ‭0.01032‬ mol.

For HCl:

[tex]Number \;of \;moles = 0.460 \times 0.00710[/tex]

Moles = ‭0.003266‬ mol.

By stoichiometry (1:1):

Acetic acid = [tex]0.00568-0.003266[/tex] = 0.002414 mol.

Acetate = [tex]0.01032-0.003266[/tex] = 0.007054 mol.

Now, we can determine the pH change:

[tex]pH =pK_a+ log_{10} \frac{A^-}{HA}\\\\pH =4.740+ log_{10} \frac{0.007054}{0.002414}\\\\pH =4.740+ log_{10} (2.9221)\\\\pH =4.740+ 0.4657[/tex]

pH = 5.2057 ≈ 5.21.

Read more on pH change here: https://brainly.com/question/491373

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