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Chapter 15: Problem 131

A buffer solution is prepared by mixing 75.0 \(\mathrm{mL}\) of 0.275 \(\mathrm{M}\) fluorobenzoic acid $\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{F}\right)\( with 55.0 \)\mathrm{mL}$ of 0.472 \(\mathrm{M}\) sodium fluorobenzoate. The \(\mathrm{pK}_{\mathrm{a}}\) of this weak acid is \(2.90 .\) What is the pH of the buffer solution?

Short Answer

Expert verified
The pH of the buffer solution is approximately 3.12. This is calculated using the Henderson-Hasselbalch equation, by first finding the moles and concentrations of the weak acid and conjugate base in the solution, and then plugging these values into the equation: pH ≈ 2.90 + 0.2172 = 3.1172.

Step by step solution

01

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

We will use the given volumes and molarities to find the moles of HA and A- present in the solution. Moles of HA = volume (HA) × molarity (HA) = 75.0 mL × 0.275 M = 20.625 mmol Moles of A- = volume (A-) × molarity (A-) = 55.0 mL × 0.472 M = 25.96 mmol
02

Calculate the final volume of the buffer solution

Next, we need to find the total volume of the buffer solution by adding the volumes of both components: Total volume = volume (HA) + volume (A-) = 75.0 mL + 55.0 mL = 130.0 mL
03

Find the concentrations of HA and A- in the buffer solution

Divide the moles of HA and A- by the total volume of the buffer solution to find their respective concentrations: \[Concentration \ of \ HA = \frac{20.625 \ mmol}{130.0 \ mL} = 0.1587 \ M\] \[Concentration \ of \ A^{-} = \frac{25.96 \ mmol}{130.0 \ mL} = 0.1997 \ M\]
04

Calculate the pH using the Henderson-Hasselbalch equation

We'll now use the Henderson-Hasselbalch equation with the pKa, and the concentrations of HA and A- that we found in the previous steps. pH = pKa + log10([A-]/[HA]) = 2.90 + log10(0.1997/0.1587) pH ≈ 2.90 + 0.2172 = 3.1172 So, the pH of the buffer solution is approximately 3.12.

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Most popular questions from this chapter

Th pH of blood is steady at a value of approximately 7.4 as a result of the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ The actual buffer system in blood is made up of $\mathrm{H}_{2} \mathrm{CO}_{3}\( and \)\mathrm{HCO}_{3}$ - One way the body keeps the pH of blood at 7.4 is by regulating breathing. Under what blood ph conditions will the body increase breathing and under what blood pH conditions will the body decrease breathing? Explain.

The common ion effect for weak acids is to significantly decrease the dissociation of the acid in water. Explain the common ion effect

The active ingredient in aspirin is acetylsalicylic acid. A 2.51 -g sample of acetylsalicylic acid required 27.36 \(\mathrm{mL}\) of 0.5106 $\mathrm{M} \mathrm{daOH}\( for complete reaction. Addition of 13.68 \)\mathrm{mL}$ of 0.5106\(M \mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH \(=3.48 .\) Determine the molar mass of acetylsalicylic acid and its \(K_{2}\) value. State any assumptions you must make to reach your answer.

One method for determining the purity of aspirin $\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)$ is to hydrolyze it with NaOH solution and then to titrate the remaining NaOH. The reaction of aspirin with NaOH is as follows: \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\) $$ \mathrm{C}_{7} \mathrm{H}_{3} \mathrm{O}_{3}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ A sample of aspirin with a mass of 1.427 g was boiled in 50.00 \(\mathrm{mL}\) of 0.500\(M \mathrm{NaOH}\) . After the solution was cooled, it took 31.92 \(\mathrm{mL}\) of 0.289 \(\mathrm{M} \mathrm{HCl}\) to titrate the excess NaOH. Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

What volumes of 0.50 \(\mathrm{M} \mathrm{HNO}_{2}\) and 0.50 \(\mathrm{M}\) NaNO, must be mixed to prepare 1.00 \(\mathrm{L}\) of a solution buffered at \(\mathrm{pH}=3.55 ?\)
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