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Chapter 15: Problem 2

A friend asks the following: “Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like NaOH is added, the HA reacts with the OH2 to form A2. Thus the amount of acid (HA) is decreased, and the amount of base (A2) is increased. Analogously, adding HCl to the buffered solution forms more of the acid (HA) by reacting with the base (A2). Thus how can we claim that a buffered solution resists changes in the pH of the solution?” How would you explain buffering to this friend?

Short Answer

Expert verified
In short, a buffered solution resists changes in pH when small amounts of acid or base are added because of the equilibrium between the weak acid (HA) and its conjugate base (A-). When a strong base or acid is added, the weak acid and conjugate base consume the added base or acid, respectively, leading to minimal change in the concentrations of HA and A-. This equilibrium maintains the pH within a relatively narrow range, as per the Henderson-Hasselbalch equation: \[pH = pK_a + \log \frac{[A^-]}{[HA]}\]

Step by step solution

01

1. Defining Buffered Solutions

A buffered solution is a solution that resists significant changes in pH when a small amount of acid or base is added. This resistance is achieved by the presence of a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, the buffered solution consists of the weak acid, HA, and its salt, NaA.
02

2. Buffer Components: Weak Acid and its Conjugate Base

Buffered solutions take advantage of the equilibrium between a weak acid and its conjugate base. The weak acid, HA, can donate a proton (H+) and form the conjugate base A-. The balance between HA and A- determines the pH of the buffered solution. The equilibrium expression for the dissociation of a weak acid (HA) is: \[HA \rightleftharpoons H^+ + A^-\] The pH of the solution can be determined using the Henderson-Hasselbalch equation: \[pH = pK_a + \log \frac{[A^-]}{[HA]}\] Here, \(pK_a\) is the negative log of the acid dissociation constant and represents the ability of the weak acid to donate a proton.
03

3. Addition of a Strong Base

When a strong base like NaOH is added to the buffered solution, the hydroxide ions (OH-) will react with the weak acid (HA), forming water (H2O) and the conjugate base (A-): \[HA + OH^- \rightarrow A^- + H_2O\] As a result, the amount of HA decreases while the amount of A- increases. This may lead to a slight shift in the pH of the buffered solution. However, due to the presence of both HA and A-, pH will not change dramatically. The buffered solution will consume the added base without affecting the pH significantly.
04

4. Addition of a Strong Acid

Similarly, when a strong acid like HCl is added to the buffered solution, the chloride ions (Cl-) will not participate in the buffer reaction, but excess H+ ions will react with the conjugate base (A-) to reform the weak acid (HA): \[A^- + H^+ \rightarrow HA\] The amount of A- decreases while the amount of HA increases, leading to a slight change in the pH of the solution. However, once again, due to the presence of both HA and A- in the solution, the pH will not change dramatically. The buffered solution neutralizes the added acid with minimal pH change.
05

5. Explaining Buffering

In conclusion, a buffered solution resists changes in pH when a small amount of acid or base is added because the weak acid (HA) and its conjugate base (A-) can consume the added acid or base without significantly altering their concentrations. Due to the equilibrium between HA and A-, adjustments in their amounts allow for the maintenance of the solution's pH within a relatively narrow range.

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Most popular questions from this chapter

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

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{1}}=5.28 $$ In two separate experiments the pH was measured during the titration of 5.00 \(\mathrm{mmol}\) of each acid with 0.200 \(\mathrm{M} \mathrm{NaOH}\) . Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at 25.00 \(\mathrm{mL}\) added NaOH, and in the other experiment the stoichiometric point was at 50.00 \(\mathrm{mL}\) NaOH. Explain these results.

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 $\mathrm{M} \mathrm{HONH}_{2}$ by 0.100 $\mathrm{M} \mathrm{HCl} .\left(K_{\mathrm{b}} \text { for } \mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)$ a. Calculate the \(\mathrm{pH}\) after 0.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. b. Calculate the \(\mathrm{pH}\) after 25.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. c. Calculate the \(\mathrm{pH}\) after 70.0 \(\mathrm{mL}\) of HCl has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 300.0 \(\mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the pH = 6.04?

Two drops of indicator HIn \(\left(K_{2}=1.0 \times 10^{-9}\right),\) where HIn is yellow and In - is blue, are placed in 100.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{MCl}\) . a. What color is the solution initially? b. The solution is titrated with 0.10\(M\) NaOH. At what pH will the color change (yellow to greenish yellow) occur? c. What color will the solution be after 200.0 \(\mathrm{mL}\) NaOH has been added?

Calculate the ph of each of the following solutions. $$ \begin{array}{l}{\text { a. } 0.100 M \text { propanoic acid }\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)} \\ {\text { b. } 0.100 M \text { sodium propanoate }\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)} \\ {\text { c. pure } \mathrm{H}_{2} \mathrm{O}}\end{array} $$ $$ \begin{array}{l}{\text { d. a mixture containing } 0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} \text { and } 0.100 \mathrm{M}} \\\ {\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}}\end{array} $$
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