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Chapter 17: Problem 20

Explain why a mixture formed by mixing \(100 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(50 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) will act as a buffer.

Short Answer

Expert verified
The mixture of 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M NaOH will act as a buffer because after the neutralization reaction, it contains significant concentrations of both CH3COOH (0.0333 M) and its conjugate base CH3COONa (0.0333 M). This allows the mixture to absorb small amounts of added acid or base without significantly changing the solution's pH, as the CH3COO- anions can combine with added protons, and the remaining acetic acid can neutralize any added base.

Step by step solution

01

Identify the reactants and products

The given reactants are CH3COOH (acetic acid) and NaOH (sodium hydroxide). Acetic acid is a weak acid, and sodium hydroxide is a strong base. When these two react, they will undergo an acid-base neutralization reaction, producing water and the sodium salt of the conjugate base of acetic acid (sodium acetate): \[CH_{3}COOH(aq) + NaOH(aq) \rightarrow CH_{3}COONa(aq) + H_{2}O(l)\]
02

Calculate the moles of reactants

Next, we need to calculate the amount of moles of the reactants being mixed together. We can do this using their molarities and volumes: Moles of acetic acid: \[ moles = Molarity 脳 Volume = 0.100\ M 脳 100\ mL 脳 \frac{1\ L}{1000\ mL} = 0.01\ mol\] Moles of sodium hydroxide: \[ moles = Molarity 脳 Volume = 0.100\ M 脳 50\ mL 脳 \frac{1\ L}{1000\ mL} = 0.005\ mol\]
03

Determine the moles of products formed

In this case, NaOH is the limiting reagent because there are fewer moles (0.005 mol) than CH3COOH (0.01 mol). The reaction will completely consume all of the NaOH moles and convert them into CH3COONa: Moles of CH3COONa produced: \(0.005\ mol\) This means that 0.005 mol of CH3COOH will react with 0.005 mol of NaOH, leaving 0.005 mol of CH3COOH unreacted.
04

Determine the final concentrations

To determine the final concentrations of CH3COOH and CH3COONa, we must take into account the total volume of the mixture. The final volume is the sum of the volumes of the initial solutions: \(100\ mL + 50\ mL = 150\ mL\). The concentrations of the reactants after the reaction are: Final concentration of CH3COOH: \[Molarity = \frac{moles}{volume} = \frac{0.005\ mol}{150\ mL} 脳 \frac{1000\ mL}{1\ L} = 0.0333\ M\] Final concentration of CH3COONa: \[Molarity = \frac{moles}{volume} = \frac{0.005\ mol}{150\ mL} 脳 \frac{1000\ mL}{1\ L} = 0.0333\ M\]
05

Explain how the mixture acts as a buffer

After the reaction, the mixture contains significant concentrations of both CH3COOH (0.0333 M) and its conjugate base CH3COONa (0.0333 M). This mixture can act as a buffer because it can absorb small amounts of added acid or base without significantly changing the solution's pH. Should any additional acid be added, the CH3COO- anions from sodium acetate can combine with the protons and remove them from the solution, while the addition of an extra base would be neutralized by the remaining acetic acid in the mixture. Therefore, a mixture formed by mixing 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M NaOH will act as a buffer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Neutralization
In an acid-base neutralization reaction, an acid and a base react to form water and a salt. For the exercise we're discussing, acetic acid (CH鈧僀OOH), which is a weak acid, reacts with sodium hydroxide (NaOH), a strong base. When they mix, the NaOH donates hydroxide ions ( ext{OH}鈦�) to react with the hydrogen ions ( ext{H}鈦�) from CH鈧僀OOH. This reaction produces water (H鈧侽) and sodium acetate (CH鈧僀OONa), which is the salt of the reaction. The chemical equation is: \[ ext{CH}_{3} ext{COOH}_{(aq)} + ext{NaOH}_{(aq)} ightarrow ext{CH}_{3} ext{COONa}_{(aq)} + ext{H}_2 ext{O}_{(l)} \] Neutralization is important because it results in the formation of a conjugate base in the presence of excess weak acid, which is essential for creating a buffer solution.
Weak Acid
A weak acid, such as acetic acid (CH鈧僀OOH), only partially dissociates in solution. This means it does not completely release all of its hydrogen ions into the solution. Rather, it establishes an equilibrium between its ionized and non-ionized forms.In water, a weak acid like acetic acid ionizes as follows: \[ ext{CH}_{3} ext{COOH} ightleftharpoons ext{CH}_{3} ext{COO}^{-} + ext{H}^{+} \] The degree of dissociation is characterized by the acid dissociation constant, ext{K}_a. Weak acids like acetic acid are crucial in buffer solutions because they can react with strong bases to form their conjugate bases, helping to maintain ext{pH} levels. Therefore, understanding the behavior of weak acids helps us predict how a buffer solution will respond to the addition of acids or bases.
Conjugate Base
The conjugate base of an acid is what remains after the acid donates a proton (H鈦�). For acetic acid (CH鈧僀OOH), its conjugate base is acetate ion (CH鈧僀OO鈦�). This is formed during the neutralization reaction when CH鈧僀OOH reacts with NaOH and donates a proton to the hydroxide ion ( ext{OH}鈦�), forming water. The creation of a buffer involves having both a weak acid and its conjugate base in solution. Having a conjugate base like CH鈧僀OO鈦� allows the solution to neutralize added acids. When extra protons are introduced, they can be picked up by the CH鈧僀OO鈦�, minimizing ext{pH} changes. Thus, in our exercise, having CH鈧僀OONa means there's a supply of CH鈧僀OO鈦�, the presence of acetate contributes significantly to the buffering capacity of the solution.
pH Stability
A buffer solution can maintain ext{pH} stability, meaning it can resist drastic changes in ext{pH} when small amounts of acid or base are added. This stability is achieved due to the presence of both a weak acid and its conjugate base, which work together to absorb added hydrogen or hydroxide ions. In our case, with a comparable concentration of CH鈧僀OOH and its conjugate base CH鈧僀OONa in solution, when an acid is added: - The conjugate base ( ext{CH}_3 ext{COO}鈦�) reacts with the added ext{H}鈦�, removing them from the solution. Similarly, when a base is added: - The weak acid ( ext{CH}_3 ext{COOH}) can donate protons to react with the ext{OH}鈦� ions. This dual action ensures the ext{pH} does not fluctuate significantly, maintaining ext{pH} stability essential in many biological and chemical processes.

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Most popular questions from this chapter

(a) What is the common-ion effect? (b) Give an example of a salt that can decrease the ionization of \(\mathrm{HNO}_{2}\) in solution.

Suppose you want to do a physiological experiment that calls for a \(\mathrm{pH} 6.5\) buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 M\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH} 6.50 ?\) (lgnore any volume change.)

(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89\). What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid. Do both the molecular weight and \(K_{a}\) value agree with your choice?

(a) The molar solubility of \(\mathrm{PbBr}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{s p^{2}}\) (b) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dissolves per liter of solution, calculate the solubilityproduct constant. (c) Using the appropriate \(K_{s p}\) value from Appendix D, calculate the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in grams per liter of solution.

Suggest how the cations in each of the following solution mixtures can be separated: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Cd}^{2+}\), (b) \(\mathrm{Cu}^{2+}\) and \(\mathrm{Mg}^{2+}\), (c) \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+}\), (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Hg}^{2+}\).
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