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Calculate the change in pH that results from adding (a) \(0.100 \mathrm{mol} \mathrm{NaNO}_{2}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{2}(\mathrm{aq})\) (b) \(0.100 \mathrm{mol} \mathrm{NaNO}_{3}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}(\mathrm{aq})\) Why are the changes not the same? Explain.

Short Answer

Expert verified
The changes in pH are not the same because the addition of NaNO_2 to HNO_2 introduces a common ion that affects the equilibrium of the weak acid, thereby lowering the concentration of H+ ions and increasing pH. However, the addition of NaNO_3 to HNO_3 does not affect the pH as NO鈧- ions do not react with water.

Step by step solution

01

Identify the Chemical Reactions

For the addition of NaNO_2 to HNO_2, the weak acid dissociation reaction can be written as: HNO_2(aq) 鈬 H+(aq) + NO_2-(aq). For the addition of NaNO_3 to HNO_3, it is important to remember that HNO_3 is a strong acid and completely dissociates into ions as HNO_3 鈫 H+ + NO鈧-. Adding NaNO鈧 will result in the release of NO鈧- ions, which do not affect the pH.
02

Calculate for Part (a)

By using the formula for the acid dissociation constant (Ka) which is [H+][NO_2鈦籡/[HNO_2], and the concentrations of each species from the reaction with NaNO_2, the ionization of HNO_2 can be calculated. Adding NaNO_2 to HNO_2 introduces common ion NO_2-, and causes a shift in equilibrium left because of Le Chatelier's principle. This in turn decreases [H+] and as a result the pH increases.
03

Calculate for Part (b)

NO鈧- ions do not react with water. Therefore, the addition of NaNO鈧 does not cause any changes to the pH level of the HNO_3 solution.
04

Compare the Results

The pH change from adding NaNO_2 to HNO_2 will be different to that of adding NaNO鈧 to HNO_3 because of the difference in the reaction of the added species with water. The former adds a common ion which suppress the ionization of HNO_2, thereby lowering [H+] and increasing the pH value. The latter, however, does not affect the pH as NO鈧- ions do not react with water.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that helps us understand how a chemical equilibrium responds to changes in concentration, pressure, or temperature. According to this principle, if a system at equilibrium is subjected to a change, it will respond by shifting in a direction that counteracts the effect of the change.

In the context of pH and acid-base equilibria, Le Chatelier's Principle can help us understand how the addition of ions can affect the ion concentrations and thus the pH of a solution. For example, when NaNO鈧 is added to a solution containing HNO鈧, the common ion NO鈧傗伝 is introduced. Le Chatelier's principle predicts that the equilibrium will shift to the left, reducing the ionization of HNO鈧傗攁 weak acid.

To maintain equilibrium, the system decreases the concentration of hydrogen ions, [H鈦篯, which leads to an increase in pH. This is an example of how added species can systematically influence a chemical equilibrium, leading to noticeable changes in the pH of a solution.
Acid-Base Equilibria
Understanding acid-base equilibria is crucial for grasping how pH changes occur in solution. Equilibrium in an acid-base reaction is established when the rate of the forward reaction (acid dissociation into ions) equals the rate of the reverse reaction (recombination into the acid).

In our exercise scenario, HNO鈧傗攁 weak acid鈥攑artially dissociates in water according to the reaction: HNO鈧(aq) 鈬 H鈦(aq) + NO鈧傗伝(aq). This equilibrium situation allows us to apply the expression for the acid dissociation constant, Ka. The equilibrium constant (Ka) provides a measure of the strength of the acid in solution, informing us of the equilibrium concentrations of the ions and the undissociated acid.

When comparing this with HNO鈧, a strong acid, it dissociates completely: HNO鈧 鈫 H鈦 + NO鈧冣伝. Because HNO鈧 dissociates fully, the equilibrium concept seems less relevant, as there is almost no undissociated HNO鈧 left in solution. This stark difference is why adding NaNO鈧 doesn't change the pH of the solution鈥攊t doesn't affect this complete ionization process.
Common Ion Effect
The "common ion effect" is an important concept in acid-base chemistry. It refers to the decrease in the ionization of a weak acid or base when a salt containing an ion already present in the equilibrium is added. This effect is directly related to Le Chatelier鈥檚 principle and acid-base equilibria.

For example, when NaNO鈧 is added to the HNO鈧 solution, it provides a source of NO鈧傗伝 ions. Since NO鈧傗伝 is already a product of the dissociation of HNO鈧, its addition shifts the equilibrium position to the left according to Le Chatelier's principle. This suppresses further ionization of the HNO鈧, reducing the concentration of H鈦 ions and increasing the pH.

In contrast, adding NaNO鈧 to a solution of HNO鈧 does not result in a common ion effect. This is because HNO鈧 is a strong acid that is fully ionized in water, so additions of NO鈧冣伝 have no effect on the equilibrium of HNO鈧冣攖here's no further ionization to suppress. Therefore, the pH remains unchanged.

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Most popular questions from this chapter

Amino acids contain both an acidic carboxylic acid group \((-\mathrm{COOH})\) and a basic amino group \(\left(-\mathrm{NH}_{2}\right)\) The amino group can be protonated (that is, it has an extra proton attached) in a strongly acidic solution. This produces a diprotic acid of the form \(\mathrm{H}_{2} \mathrm{A}^{+}\), as exemplified by the protonated amino acid alanine. The protonated amino acid has two ionizable protons that can be titrated with \(\mathrm{OH}^{-}\) For the \(-\mathrm{COOH}\) group, \(\mathrm{pK}_{\mathrm{a}_{1}}=2.34 ;\) for the \(-\mathrm{NH}_{3}^{+}\) group, \(\mathrm{p} K_{\mathrm{a}_{2}}=9.69 .\) Consider the titration of a 0.500 M solution of alanine hydrochloride with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution. What is the \(\mathrm{pH}\) of \((\mathrm{a})\) the \(0.500 \mathrm{M}\) alanine hydrochloride; (b) the solution at the first half- neutralization point; (c) the solution at the first equivalence point? The dominant form of alanine present at the first equivalence point is electrically neutral despite the positive charge and negative charge it possesses. The point at which the neutral form is produced is called the isoelectric point. Confirm that the \(\mathrm{pH}\) at the isoelectric point is \(\mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}_{1}}+\mathrm{p} \mathrm{K}_{\mathrm{a}_{2}}\right)\) What is the \(\mathrm{pH}\) of the solution (d) halfway between the first and second equivalence points? (e) at the second equivalence point? (f) Calculate the pH values of the solutions when the following volumes of the \(0.500 \mathrm{M} \mathrm{NaOH}\) have been added to \(50 \mathrm{mL}\) of the \(0.500 \mathrm{M}\) alanine hydrochloride solution: \(10.0 \mathrm{mL}, 20.0 \mathrm{mL}, 30.0 \mathrm{mL}, 40.0 \mathrm{mL}, 50.0 \mathrm{mL}\) \(60.0 \mathrm{mL}, 70.0 \mathrm{mL}, 80.0 \mathrm{mL}, 90.0 \mathrm{mL}, 100.0 \mathrm{mL},\) and \(110.0 \mathrm{mL}\) (g) Sketch the titration curve for the 0.500 M solution of alanine hydrochloride, and label significant points on the curve.

Explain why the volume of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to reach the equivalence point in the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HA is the same regardless of whether HA is a strong or a weak acid, yet the \(\mathrm{pH}\) at the equivalence point is not the same.

During the titration of equal concentrations of a weak base and a strong acid, at what point would the \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} ?(\mathrm{a})\) the initial \(\mathrm{pH} ;\) (b) halfway to the equivalence point; (c) at the equivalence point; (d) past the equivalence point.

\(\begin{array}{lll}\text { Given } & 1.00 & \mathrm{L}\end{array}\) of a solution that is \(0.100 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) and \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) (a) Over what pH range will this solution be an effective buffer? (b) What is the buffer capacity of the solution? That is, how many millimoles of strong acid or strong base can be added to the solution before any significant change in pH occurs?

Carbonic acid is a weak diprotic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) with \(K_{a_{1}}=4.43 \times 10^{-7}\) and \(K_{\mathrm{a}_{2}}=4.73 \times 10^{-11} .\) The equiv- alence points for the titration come at approximately pH 4 and 9. Suitable indicators for use in titrating carbonic acid or carbonate solutions are methyl orange and phenolphthalein. (a) Sketch the titration curve that would be obtained in titrating a sample of \(\mathrm{NaHCO}_{3}(\mathrm{aq})\) with \(1.00 \mathrm{M} \mathrm{HCl}\) (b) Sketch the titration curve for \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) with 1.00 M HCl. (c) What volume of \(0.100 \mathrm{M} \mathrm{HCl}\) is required for the complete neutralization of \(1.00 \mathrm{g} \mathrm{NaHCO}_{3}(\mathrm{s}) ?\) (d) What volume of \(0.100 \mathrm{M} \mathrm{HCl}\) is required for the complete neutralization of \(1.00 \mathrm{g} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s}) ?\) (e) A sample of NaOH contains a small amount of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) For titration to the phenolphthalein end point, \(0.1000 \mathrm{g}\) of this sample requires \(23.98 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{HCl} .\) An additional \(0.78 \mathrm{mL}\) is required to reach the methyl orange end point. What is the percent \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) by mass, in the sample?

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