Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathbf{a})\) a solution formed by adding \(30.0 \mathrm{~g}\) of furoic acid and \(25.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.300 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) $\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\( and \)30.0 \mathrm{~mL}\( of \)0.250 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$ and diluting the total volume to \(125 \mathrm{~mL},(\mathbf{c})\) a solution prepared by adding $25.0 \mathrm{~mL}\( of \)1.00 \mathrm{M} \mathrm{NaOH}\( solution to \)100.0 \mathrm{~mL}\( of \)0.100 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$

Step by step solution

01

(a) Calculate the concentrations of furoic acid and sodium furoate

First, we need to find the moles of furoic acid (HC₅H₃O₃) and sodium furoate (NaC₅H₃O₃). The molar masses of furoic acid and sodium furoate are 112.11 g/mol and 128.11 g/mol, respectively. For furoic acid: Moles = mass / molar mass = 30.0 g / 112.11 g/mol = \(0.267\) moles For sodium furoate: Moles = mass / molar mass = 25.0 g / 128.11 g/mol = \(0.195\) moles To determine the concentrations of furoic acid and sodium furoate, divide the moles by the volume of the solution (0.300 L). \[[\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]=\frac{0.267 \, \text{moles}}{0.300 \,\text{L}}=0.890 \, \text{M} \] \[[\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]=\frac{0.195 \, \text{moles}}{0.300 \,\text{L}}=0.650 \, \text{M} \]

02

(a) Calculate the pH using the Henderson-Hasselbalch equation

Next, we calculate the pKₐ: $$ \mathrm{p}K_{a} = -\log K_{a} = -\log\left(6.76\times10^{-4}\right) = 3.17 $$ Using the Henderson-Hasselbalch equation, \[\mathrm{pH} = \mathrm{p}K_{a} + \log \frac{[\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]}{[\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]} = 3.17 + \log \frac{0.650}{0.890} = 3.17 - 0.29 = 2.88\] So, the pH of the solution for part (a) is 2.88.

03

(b) Calculate the moles and concentrations of furoic acid and sodium furoate

For furoic acid (HC₅H₃O₃, 0.200 M solution): Moles = concentration × volume = 0.200 M × 0.020 L = \(0.004\) moles For sodium furoate (NaC₅H₃O₃, 0.250 M solution): Moles = concentration × volume = 0.250 M × 0.030 L = \(0.0075\) moles Diluting to a final volume of 0.125 L: \[[\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]=\frac{0.004 \, \text{moles}}{0.125 \, \text{L}}=0.032 \, \text{M} \] \[[\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]=\frac{0.0075 \, \text{moles}}{0.125 \, \text{L}}=0.060 \, \text{M} \]

04

(b) Calculate the pH using the Henderson-Hasselbalch equation

Using the Henderson-Hasselbalch equation, \[\mathrm{pH} = \mathrm{p}K_{a} + \log \frac{[\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]}{[\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]}=3.17+\log \frac{0.060}{0.032}=3.17+0.57 = 3.74\] So, the pH of the solution for part (b) is 3.74.

05

(c) Calculate the moles of furoic acid and sodium hydroxide

For furoic acid (HC₅H₃O₃, 0.100 M solution): Moles = concentration × volume = 0.100 M × 0.100 L = \(0.010\) moles Sodium Hydroxide (NaOH, 1.00 M solution): Moles = concentration × volume = 1.00 M × 0.025 L = \(0.025\) moles

06

(c) Determine the moles of furoic acid and sodium furoate after the reaction

Since furoic acid (HC₅H₃O₃) and sodium hydroxide (NaOH) react in a 1:1 ratio, we can subtract the moles of NaOH from the moles of HC₅H₃O₃ to find the remaining moles of furoic acid and obtain the moles of sodium furoate formed: New moles of furoic acid: \(0.010 - 0.025 = -0.015\). Since we cannot have negative moles, we say that all furoic acid reacted, and there are 0 moles left. Moles of sodium furoate formed: \(0.025 - 0.010 = 0.015\) moles

07

(c) Calculate the concentrations of furoic acid and sodium furoate

Diluting to a final volume of 0.125 L: \[[\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]=\frac{0 \, \text{moles}}{0.125 \, \text{L}}=0 \, \text{M} \] \[[\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}]=\frac{0.015 \, \text{moles}}{0.125 \, \text{L}}=0.120 \, \text{M} \]

08

(c) Calculate the pH using the Henderson-Hasselbalch equation

Since there is no furoic acid left in the solution, we cannot use the Henderson-Hasselbalch equation directly, but we can approximate the pH using the simplified version of the equation: \[\mathrm{pH} \approx \mathrm{p}K_{a} + \log [\mathrm{A}^{-}]=3.17 + \log 0.120 = 3.17 - 0.92 = 2.25\] So, the pH of the solution for part (c) is 2.25.

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