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Chapter 15: Problem 33

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.

Short Answer

Expert verified
The pH of the buffer solution prepared by dissolving 21.5g benzoic acid and 37.7g sodium benzoate in 200.0 mL of solution is approximately 4.51.

Step by step solution

01

Calculate moles of benzoic acid and sodium benzoate

To determine the moles of benzoic acid and sodium benzoate in the solution, first find their molar masses and then divide the given masses of each substance by their respective molar masses. The molar mass of benzoic acid, HC7H3O2, is: \(122.12\mathrm{g/mol}\) The molar mass of sodium benzoate, NaC7H5O2, is: \(144.11\mathrm{g/mol}\) Moles of benzoic acid: \(\frac{21.5\mathrm{g}}{122.12\mathrm{g/mol}} \approx 0.176 \mathrm{mol}\) Moles of sodium benzoate: \(\frac{37.7\mathrm{g}}{144.11\mathrm{g/mol}} \approx 0.261 \mathrm{mol}\)
02

Calculate molar concentrations of benzoic acid and sodium benzoate

To find the molar concentrations of benzoic acid and sodium benzoate, divide their respective moles by the volume of the solution in liters. The volume of the solution is 200.0 mL, which is equivalent to 0.200 L. Molar concentration of benzoic acid: \(\frac{0.176 \mathrm{mol}}{0.200 \mathrm{L}} = 0.880 \mathrm{M}\) Molar concentration of sodium benzoate: \(\frac{0.261 \mathrm{mol}}{0.200 \mathrm{L}} = 1.305 \mathrm{M}\)
03

Determine the pKa of benzoic acid

In order to use the Henderson-Hasselbalch equation, you will need to know the pKa of benzoic acid. The pKa of benzoic acid is 4.19.
04

Apply the Henderson-Hasselbalch equation

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation. This equation relates the pH, pKa, and the concentrations of the acid (HA) and its conjugate base (A-) in a buffer solution as follows: pH = pKa + log\(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\) In this problem: pKa = 4.19 [HA] = 0.880 M (concentration of benzoic acid) [A-] = 1.305 M (concentration of sodium benzoate)
05

Calculate the pH of the buffer solution

Substitute the values from steps 3 and 2 into the Henderson-Hasselbalch equation and solve for pH. pH = 4.19 + log\(\frac{1.305\mathrm{M}}{0.880\mathrm{M}}\) pH ≈ 4.19 + 0.32 pH ≈ 4.51 The pH of the buffer solution is approximately 4.51.

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Most popular questions from this chapter

Sketch a \(\mathrm{pH}\) curve for the titration of a weak acid (HA) with a strong base \((\mathrm{NaOH})\). List the major species and explain how you would go about calculating the \(\mathrm{pH}\) of the solution at various points, including the halfway point and the equivalence point.

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

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the \(\mathrm{pH}\) after \(20.0 \%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the \(\mathrm{pH}\). c. Do the same as in part \(\mathrm{b}\), but use the following equilibrium to calculate the \(\mathrm{pH}\) : $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(\mathrm{b}\) and \(\mathrm{c}\) agree? Explain.

Repeat the procedure in Exercise 55 , but for the titration of \(25.0\) \(\mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with \(0.100 \mathrm{M} \mathrm{HCl}\).

Calculate the \(\mathrm{pH}\) of a solution that is \(0.60 \mathrm{M} \mathrm{HF}\) and \(1.00 \mathrm{M} \mathrm{KF}\).
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