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Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?

Short Answer

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Mixing acetic acid (CH3COOH) and sodium hydroxide (NaOH) forms a buffered solution containing acetic acid and its conjugate base, acetate ions (CH3COO-). A buffer solution resists changes in pH when small amounts of acid or base are added. The effectiveness of a buffer solution depends on the concentrations of the weak acid and its conjugate base. The Henderson-Hasselbalch equation illustrates this relationship: \( pH = pK_a + log_{10} \frac{[CH_3COO^-]}{[CH_3COOH]} \). By adjusting the amount of acetic acid and sodium hydroxide added, we can control the pH of the resulting buffered solution. If the concentrations of acetic acid and acetate ions are nearly equal, the buffer solution will have a maximum buffering capacity, making it more effective at resisting changes in pH.

Step by step solution

01

Understand the concept of a buffer solution

A buffer solution is a solution that resists changes in pH when small amounts of an acid or a base are added to it. It consists of a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid.
02

Understand the reaction between acetic acid and sodium hydroxide

When acetic acid (CH3COOH), a weak acid, is mixed with sodium hydroxide (NaOH), a strong base, a neutralization reaction occurs, producing sodium acetate (CH3COONa) and water (H2O). The balanced equation for this reaction is: \( CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \) Sodium acetate, formed during this reaction, dissociates in water to form acetate ions (CH3COO-) and sodium ions (Na+). The acetate ion is the conjugate base of acetic acid. The acidic component of the buffer system is acetic acid, and the basic component is the acetate ion.
03

Formation of the buffer solution

When acetic acid and sodium hydroxide are mixed, the resulting solution contains both acetic acid and its conjugate base, acetate ions: \( CH_3COOH \leftrightarrow H^+ + CH_3COO^- \) Since both acidic and basic components are present in the solution, it forms a buffer system that resists changes in pH when small amounts of acid or base are added.
04

Effect of the amount of each solution on buffer effectiveness

The effectiveness of a buffer solution depends on the concentrations of the weak acid and its conjugate base. The Henderson-Hasselbalch equation describes the relationship between the pH of a buffer solution and the concentrations of the acidic and basic components: \( pH = pK_a + log_{10} \frac{[CH_3COO^-]}{[CH_3COOH]} \) According to this equation, the pH of the buffer solution depends on the ratio of the concentrations of acetate ions (CH3COO-) and acetic acid (CH3COOH). By adjusting the amount of acetic acid and sodium hydroxide added to the solution, we can control the ratio of these two components and, consequently, the pH of the resulting buffered solution. If the concentrations of acetic acid and acetate ions are nearly equal, the buffer solution will be most effective at resisting changes in pH, as it will have a maximum buffering capacity. If the concentration of one component is much higher than the other, the buffering capacity will be decreased, making the buffer solution less effective at resisting changes in pH.

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Most popular questions from this chapter

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.

A buffered solution is made by adding \(50.0 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}\) to \(1.00 \mathrm{~L}\) of a \(0.75 M\) solution of \(\mathrm{NH}_{3}\). Calculate the \(\mathrm{pH}\) of the final solution. (Assume no volume change.)

You have \(75.0 \mathrm{~mL}\) of \(0.10 M\) HA. After adding \(30.0 \mathrm{~mL}\) of \(0.10 M\) \(\mathrm{NaOH}\), the \(\mathrm{pH}\) is \(5.50\). What is the \(K_{\mathrm{u}}\) value of \(\mathrm{HA}\) ?

An aqueous solution contains dissolved \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\). The concentration of \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{2}\) is \(0.50 \mathrm{M}\) and \(\mathrm{pH}\) is \(4.20\). a. Calculate the concentration of \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{3}^{+}\) in this buffer solution. b. Calculate the \(\mathrm{pH}\) after \(4.0 \mathrm{~g} \mathrm{NaOH}(s)\) is added to \(1.0 \mathrm{~L}\) of this solution. (Neglect any volume change.)

Consider a solution formed by mixing \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HOCl}, 25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{NaOH}\). \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\), and \(10.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of this solution.

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