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Chapter 14: Problem 82

a. Calculate the pH of a buffered solution that is 0.100 \(M\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 M\) 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 $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q)$$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the 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 \(b\) and c agree? Explain.

Short Answer

Expert verified
The pH of the buffered solution is 4.19. After converting 20% of benzoic acid to benzoate anion, the pH is 4.37 using the dissociation equilibrium. Using the given equilibrium involving benzoate anion and water, the pH is 4.97. The disagreement between the results is due to the assumption that the concentration of \(H_3O^+\) and \(OH^-\) has negligible effects on the pH in the Henderson-Hasselbalch equation.

Step by step solution

01

Use the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \[pH = pK_a + log\frac{[A^-]}{[HA]}\] where, \(pH\) - pH of the solution \(pK_a\) - \(-log(K_a)\) \([A^-]\) - molar concentration of benzoate anion (\(C_6H_5CO_2^-\)) \([HA]\) - molar concentration of benzoic acid (\(C_6H_5CO_2H\))
02

Calculate \(pK_a\)

Given \(K_a = 6.4 \times 10^{-5}\), we can find \(pK_a\) using the formula \(pK_a = -log(K_a)\): \(pK_a = -log(6.4 \times 10^{-5}) = 4.19\)
03

Calculate the pH

Since the concentration of benzoic acid and benzoate ion are both 0.100 \(M\), we can use the Henderson-Hasselbalch equation to find the pH. \(pH = 4.19+ log\frac{0.100}{0.100} = 4.19+0 = 4.19\) So, the pH of the buffered solution is 4.19. b. Calculate the pH after 20% of benzoic acid is converted to benzoate anion
04

Update the concentrations

When 20% of benzoic acid is converted to benzoate anion, the concentrations change: \([C_6H_5CO_2H] = 0.100 \times 0.8 = 0.080\, M\) \([C_6H_5CO_2^-] = 0.100 + 0.020 = 0.120\, M\)
05

Calculate the new pH

Now we will use the new concentrations of benzoic acid and benzoate anion to find the new pH using the Henderson-Hasselbalch equation. \(pH = 4.19 + log\frac{0.120}{0.080} = 4.19 + 0.18 = 4.37\) So, the pH after converting 20% of benzoic acid to benzoate anion is 4.37. c. Calculate the pH using the given equilibrium involving benzoate anion and water
06

Write the equilibrium expression

The equilibrium expression for the given reaction is: \(K_b = \frac{[C_6H_5CO_2H][OH^-]}{[C_6H_5CO_2^-]}\)
07

Calculate the \(K_b\)

Using the relationship \(K_a \times K_b = K_w\), we can find the \(K_b\) for benzoate anion: \(K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-5}} = 1.56 \times 10^{-10}\)
08

Calculate \([OH^-]\) using the equilibrium expression

Assuming the initial concentration of hydroxide is 0, \(K_b = \frac{(0.020)(x)}{(0.120)}\) Solving for x, we get: \[x = [OH^-] = 9.36 \times 10^{-10}\]
09

Calculate new pH

Finally, we can calculate the new pH using the relationship \(pH = 14 - pOH\): \(pOH = -log[OH^-] = -log(9.36 \times 10^{-10}) = 9.03\) \(pH = 14 - pOH = 14 - 9.03 = 4.97\) So, the pH calculated using this equilibrium is 4.97. d. Comparing the results The pH calculated using the dissociation equilibrium (part b, pH = 4.37) and the pH calculated using the equilibrium involving benzoate anion and water (part c, pH = 4.97) are not equal. This disagreement is because the Henderson-Hasselbalch equation assumes that the concentration of \(H_3O^+\) and \(OH^-\) has negligible effects on the pH. The presence of \(OH^-\) affects the concentrations of benzoic acid and benzoate anion and should not be neglected in a more accurate calculation.

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Most popular questions from this chapter

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

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between 7.1 and 7.2 a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15 ?\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125 \(M\) NaOH, requiring \(16.00 \mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding \(2.00 \mathrm{mL}\) NaOH was \(6.912 .\) Calculate \(K_{\mathrm{a}}\) for the weak acid.

Calculate the \(\mathrm{pH}\) of a solution that is \(0.40 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) and \(0.80 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3} .\) In order for this buffer to have \(\mathrm{pH}=\) \(\mathrm{p} K_{\mathrm{a}},\) would you add HCl or NaOH? What quantity (moles) of which reagent would you add to \(1.0 \mathrm{L}\) of the original buffer so that the resulting solution has \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} ?\)

A best buffer has about equal quantities of weak acid and conjugate base present as well as having a large concentration of each species present. Explain.
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