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A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

Short Answer

Expert verified
#tag_title#Reaction of buffer components with H鈦 and OH鈦 ions#tag_content# 3. Bicarbonate (HCO鈧冣伝) neutralizing added H鈦 ions: \[ \mathrm{HCO_{3}^{-}(aq) + H^{+}(aq) \rightarrow H_{2}CO_{3}(aq)} \] 4. Carbonate (CO鈧兟测伝) neutralizing added H鈦 ions: \[ \mathrm{CO_{3}^{2-}(aq) + 2H^{+}(aq) \rightarrow H_{2}CO_{3}(aq)} \] 5. Bicarbonate (HCO鈧冣伝) neutralizing added OH鈦 ions: \[ \mathrm{HCO_{3}^{-}(aq) + OH^{-}(aq) \rightarrow CO_{3}^{2-}(aq) + H_{2}O(l)} \] In summary, the buffer made from NaHCO鈧 and Na鈧侰O鈧 neutralizes added H鈦 and OH鈦 ions through the reactions between HCO鈧冣伝 and CO鈧兟测伝 ions with H鈦 and OH鈦 ions, forming H鈧侰O鈧, CO鈧兟测伝, and H鈧侽 as products.

Step by step solution

01

Recognize the dissociation of the compounds in water interaction with H鈦 and OH鈦 ions

First, we need to show how NaHCO鈧 and Na鈧侰O鈧 dissociate in water to form bicarbonate (HCO鈧冣伝) and carbonate (CO鈧兟测伝) ions. This will be helpful in showing how they interact with the added H鈦 and OH鈦 ions. 1. NaHCO鈧 dissociation in water: \[ \mathrm{NaHCO_{3}(s) \rightarrow Na^{+}(aq) + HCO_{3}^{-}(aq)} \] 2. Na鈧侰O鈧 dissociation in water: \[ \mathrm{Na_{2}CO_{3}(s) \rightarrow 2Na^{+}(aq) + CO_{3}^{2-}(aq)} \] Now we can show how bicarbonate and carbonate ions interact with the added H鈦 and OH鈦 ions to neutralize them.

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Key Concepts

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

Dissociation of NaHCO3
To understand how a buffer works, we need to start by looking at the dissociation of the buffering agents in water. Sodium bicarbonate (\texttt{NaHCO}_3) is a common buffering agent that, when dissolved in water, separates into sodium ions (\texttt{Na}^+) and bicarbonate ions (\texttt{HCO}_3^-).

This dissolution is crucial for the buffering action as the bicarbonate ion plays a significant role in neutralizing added acids (H鈦 ions) or bases (OH鈦 ions) in a solution. The dissociation can be represented by the chemical equation:
\[\texttt{NaHCO}_3(s) \rightarrow \texttt{Na}^{+}(aq) + \texttt{HCO}_3^{-}(aq)\]

Dissociation allows \texttt{NaHCO}_3 not only to release bicarbonate ions but also to maintain a certain level of pH by moderating changes caused by the addition of strong acids or bases.
Dissociation of Na2CO3
Moving on from sodium bicarbonate, we turn our attention to sodium carbonate (\texttt{Na}_2\texttt{CO}_3). When it dissociates in water, it produces two sodium ions (\texttt{Na}^+) for each formula unit and carbonate ions (\texttt{CO}_3^{2-}).

The carbonate ion serves a similar purpose as bicarbonate鈥攈elping to buffer the solution. However, it has the ability to absorb two protons (H鈦), making it doubly effective against acids. The decomposition in water can be written as:
\[\texttt{Na}_2\texttt{CO}_3(s) \rightarrow 2\texttt{Na}^{+}(aq) + \texttt{CO}_3^{2-}(aq)\]

The presence of both \texttt{NaHCO}_3 and \texttt{Na}_2\texttt{CO}_3 in the buffer solution offers a two-tiered defense mechanism against pH changes, with both bicarbonate and carbonate ions working in tandem to neutralize different amounts of acid or base.
Bicarbonate and Carbonate Ions
Bicarbonate (\texttt{HCO}_3^-) and carbonate (\texttt{CO}_3^{2-}) ions are the cornerstone of the buffer system in the discussed solution.

When an acid is added to the buffer, the bicarbonate ion can react with the hydrogen ions (H鈦) to form carbonic acid (\texttt{H}_2\texttt{CO}_3), which is relatively weak and does not significantly lower the pH of the solution. On the other hand, carbonate ions can react with hydrogen ions to form bicarbonate, which can then further react with more hydrogen ions in the manner just described.

Neutralizing Acids

For bicarbonate, the reaction with added acid can be shown as:
\[\texttt{HCO}_3^{-} + \texttt{H}^{+} \rightarrow \texttt{H}_2\texttt{CO}_3\]

For carbonate, the reaction can accommodate two protons:
\[\texttt{CO}_3^{2-} + \texttt{H}^{+} \rightarrow \texttt{HCO}_3^{-}\]
\[\texttt{HCO}_3^{-} + \texttt{H}^{+} \rightarrow \texttt{H}_2\texttt{CO}_3\]

Neutralizing Bases

Conversely, when a base is added, the carbonic acid formed can dissociate to yield hydrogen ions, thereby neutralizing the hydroxide ions (OH鈦):
\[\texttt{H}_2\texttt{CO}_3 \rightarrow \texttt{HCO}_3^{-} + \texttt{H}^{+}\]\[\texttt{H}^{+} + \texttt{OH}^{-} \rightarrow \texttt{H}_2\texttt{O}\]

The ability of bicarbonate and carbonate ions to engage in these reversible reactions is key to maintaining a stable pH in the buffer solution, protecting it against drastic changes when acids or bases are introduced.

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Most popular questions from this chapter

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

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?

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by \(0.100 M \mathrm{KOH}\). Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. a. \(0.0 \mathrm{~mL}\) b. \(50.0 \mathrm{~mL}\) c. \(100.0 \mathrm{~mL}\) d. \(150.0 \mathrm{~mL}\) e. \(200.0 \mathrm{~mL}\) f. \(250.0 \mathrm{~mL}\)

Carbonate buffers are important in regulating the \(\mathrm{pH}\) of blood at \(7.40\). If the carbonic acid concentration in a sample of blood is \(0.0012 M\), determine the bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\). \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7}\)

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \begin{array}{l} \mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} \\ \text { Adipic acid } \quad \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{2}}=5.28 \end{array} $$ In two separate experiments the pH was measured during the titration of \(5.00\) mmol of each acid with \(0.200 M \mathrm{NaOH}\). 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 \(\mathrm{NaOH}\), and in the other experiment the stoichiometric point was at \(50.00 \mathrm{~mL} \mathrm{NaOH}\). Explain these results. (See Exercise \(113 .\) )

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