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Chapter 15: Problem 31

Calculate the pH of each of the following buffered solutions. a. \(0.10 M\) acetic acid \(/ 0.25 M\) sodium acetate b. \(0.25 M\) acetic acid \(/ 0.10 M\) sodium acetate c. \(0.080 M\) acetic acid \(/ 0.20 M\) sodium acetate d. \(0.20 M\) acetic acid \(0.080 M\) sodium acetate

Short Answer

Expert verified
The pH values for the given buffered solutions are: a. \(pH \approx 5.15\) b. \(pH \approx 4.35\) c. \(pH \approx 5.15\) d. \(pH \approx 4.35\)

Step by step solution

01

The equation is given as follows: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] where \(pH\) is the pH value, \(pKa\) is the acidic dissociation constant, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the weak acid. #Step 2: Determine the pKa value of acetic acid#

The pKa value of acetic acid is \(4.75\). Now, let us calculate the pH of each buffered solution. #a. 0.10 M acetic acid / 0.25 M sodium acetate# #Step 3a: Plug in the concentration values in the equation#
02

We have \([HA] = 0.10 M\) and \([A^-] = 0.25 M\). By using the Henderson-Hasselbalch equation, we have \[pH = 4.75 + \log \frac{0.25}{0.10}\] #Step 4a: Solve for pH#

Calculate the value in the \(\log\) term: \[pH = 4.75 + \log(2.5)\] \[pH \approx 4.75 + 0.40 = 5.15\] The pH of solution a is approximately 5.15. #b. 0.25 M acetic acid / 0.10 M sodium acetate# #Step 3b: Plug in the concentration values in the equation#
03

We have \([HA] = 0.25 M\) and \([A^-] = 0.10 M\). By using the Henderson-Hasselbalch equation, we have \[pH = 4.75 + \log \frac{0.10}{0.25}\] #Step 4b: Solve for pH#

Calculate the value in the \(\log\) term: \[pH = 4.75 + \log(0.4)\] \[pH \approx 4.75 - 0.40 = 4.35\] The pH of solution b is approximately 4.35. #c. 0.080 M acetic acid / 0.20 M sodium acetate# #Step 3c: Plug in the concentration values in the equation#
04

We have \([HA] = 0.080 M\) and \([A^-] = 0.20 M\). By using the Henderson-Hasselbalch equation, we have \[pH = 4.75 + \log \frac{0.20}{0.080}\] #Step 4c: Solve for pH#

Calculate the value in the \(\log\) term: \[pH = 4.75 + \log(2.5)\] \[pH \approx 4.75 + 0.40 = 5.15\] The pH of solution c is approximately 5.15. #d. 0.20 M acetic acid 0.080 M sodium acetate# #Step 3d: Plug in the concentration values in the equation#
05

We have \([HA] = 0.20 M\) and \([A^-] = 0.080 M\). By using the Henderson-Hasselbalch equation, we have \[pH = 4.75 + \log \frac{0.080}{0.20}\] #Step 4d: Solve for pH#

Calculate the value in the \(\log\) term: \[pH = 4.75 + \log(0.4)\] \[pH \approx 4.75 - 0.40 = 4.35\] The pH of solution d is approximately 4.35.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

Calculate the mass of sodium acetate that must be added to \(500.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) acetic acid to form a \(\mathrm{pH}=5.00\) buffer solution.

Calculate the volume of \(1.50 \times 10^{-2} \mathrm{M} \mathrm{NaOH}\) that must be added to \(500.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15\)

Carbonate buffers are important in regulating the \(\mathrm{pH}\) of blood at 7.40. If the carbonic acid concentration in a sample of blood is \(0.0012 M\), determine the bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\). \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7}\)

A certain acetic acid solution has \(\mathrm{pH}=2.68\). Calculate the volume of \(0.0975 M \mathrm{KOH}\) required to reach the equivalence point in the titration of \(25.0 \mathrm{~mL}\) of the acetic acid solution.
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