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Chapter 15: Problem 109

A buffer solution is prepared by mixing \(75.0 \mathrm{~mL}\) of \(0.275 \mathrm{M}\) fluorobenzoic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{3} \mathrm{O}_{2} \mathrm{~F}\right)\) with \(55.0 \mathrm{~mL}\) of \(0.472 \mathrm{M}\) sodium fluorobenzoate. The \(\mathrm{p} K_{\mathrm{a}}\) of this weak acid is \(2.90\). What is the \(\mathrm{pH}\) of the buffer solution?

Short Answer

Expert verified
After performing the calculations, the pH of the buffer solution can be determined as follows: \[ \mathrm{pH} = 2.90 + \log{\frac{\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}}{\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}}} \] By solving this equation, we find the pH of the buffer solution to be approximately 2.99.

Step by step solution

01

Calculate the moles of weak acid (HA) and conjugate base (A-)

To find the moles of fluorobenzoic acid (HA) and sodium fluorobenzoate (A-), we can multiply the initial volume by the initial concentration: Moles of HA = Volume of HA × Concentration of HA Moles of HA = \(75.0 \mathrm{~mL} \times 0.275 \mathrm{M}\) Moles of A- = Volume of A- × Concentration of A- Moles of A- = \(55.0 \mathrm{~mL} \times 0.472 \mathrm{M}\)
02

Calculate the total volume of the solution

Total Volume = Volume of HA + Volume of A- Total Volume = \(75.0 \mathrm{~mL} + 55.0 \mathrm{~mL}\)
03

Calculate the final concentrations of HA and A- in the buffer solution

To find the final concentrations of HA and A-, we will divide the moles of each by the total volume of the solution (in liters): Final concentration of HA = Moles of HA / Total Volume Final concentration of HA = \(\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}\) Final concentration of A- = Moles of A- / Total Volume Final concentration of A- = \(\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}\)
04

Apply the Henderson-Hasselbalch equation

Now we can apply the Henderson-Hasselbalch equation to find the pH of the buffer solution: \[\mathrm{pH} = \mathrm{p} K_{a} + \log{\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}}\] Where, - \(\mathrm{p} K_{a}=2.90\) - \([\mathrm{HA}]=\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}\) - \([\mathrm{A}^{-}]=\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}\) \[ \mathrm{pH} = 2.90 + \log{\frac{\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}}{\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}}} \]
05

Calculate the pH

Now, we can plug in the values we determined for the concentrations of A- and HA into the Henderson-Hasselbalch equation and solve for the pH. \[ \mathrm{pH} = 2.90 + \log{\frac{\frac{55.0 \mathrm{~mL} \times 0.472 \mathrm{M}}{130.0 \mathrm{~mL}}}{\frac{75.0 \mathrm{~mL} \times 0.275 \mathrm{M}}{130.0 \mathrm{~mL}}}} \] After performing these calculations, we will find the pH of the buffer solution.

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Most popular questions from this chapter

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 \(\mathrm{pH}=7.40\). \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7}\)

Calculate the pH of a buffer solution prepared by dissolving 21.5 \(\mathrm{g}\) benzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) and \(37.7 \mathrm{~g}\) sodium benzoate in \(200.0 \mathrm{~mL}\) of solution.

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 \(0.25 \mathrm{M} \mathrm{NaOH}\)

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A \(25.0-\mathrm{mL}\) sample of \(0.100 M\) lactic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}, \mathrm{p} K_{\mathrm{a}}=\right.\) \(3.86\) ) is titrated with \(0.100 M\) NaOH solution. Calculate the \(\mathrm{pH}\) after the addition of \(0.0 \mathrm{~mL}, 4.0 \mathrm{~mL}, 8.0 \mathrm{~mL}, 12.5 \mathrm{~mL}, 20.0 \mathrm{~mL}\), \(24.0 \mathrm{~mL}, 24.5 \mathrm{~mL}, 24.9 \mathrm{~mL}, 25.0 \mathrm{~mL}, 25.1 \mathrm{~mL}, 26.0 \mathrm{~mL}, 28.0 \mathrm{~mL}\) and \(30.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}\). Plot the results of your calculations as pH versus milliliters of \(\mathrm{NaOH}\) added.

A \(10.00-g\) sample of the ionic compound \(\mathrm{NaA}\), where \(\mathrm{A}^{-}\) is the anion of a weak acid, was dissolved in enough water to make \(100.0 \mathrm{~mL}\) of solution and was then titrated with \(0.100 \mathrm{M} \mathrm{HCl}\). After \(500.0 \mathrm{~mL}\) HCl was added, the \(\mathrm{pH}\) was \(5.00\). The experimenter found that \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration.
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