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Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{\text {; }}\) (b) carbonate ion, \(\mathrm{CO}_{3}^{2-} ;\) (c) formate ion, \(\mathrm{CHO}_{2}^{-}\).

Short Answer

Expert verified
(a) Dimethylamine reaction with water: \((CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-\), \(K_b = \frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}\). (b) Carbonate ion reaction with water: \(CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-\), \(K_b = \frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}\). (c) Formate ion reaction with water: \(CHO_2^- + H_2O \rightleftharpoons HCHO_2 + OH^-\), \(K_b = \frac{[HCHO_2][OH^-]}{[CHO_2^-]}\).

Step by step solution

01

(a) Dimethylamine reaction and \(K_b\) expression

Dimethylamine (\((CH_3)_2NH\)) reacts with water (\(H_2O\)) in the following chemical equation: \( (CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-\) The \(K_b\) expression can be written as: \(K_b = \frac{[\left(CH_3\right)_2NH_2^+][OH^-]}{[\left(CH_3\right)_2NH]}\)
02

(b) Carbonate ion reaction and \(K_b\) expression

The carbonate ion (\(CO_3^{2-}\)) reacts with water (\(H_2O\)) in the following chemical equation: \(CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-\) The \(K_b\) expression can be written as: \(K_b = \frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}\)
03

(c) Formate ion reaction and \(K_b\) expression

The formate ion (\(CHO_2^{-}\)) reacts with water (\(H_2O\)) in the following chemical equation: \(CHO_2^- + H_2O \rightleftharpoons HCHO_2 + OH^-\) The \(K_b\) expression can be written as: \(K_b = \frac{[HCHO_2][OH^-]}{[CHO_2^-]}\)

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Key Concepts

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

Base Reaction with Water
When a base is dissolved in water, it undergoes a process known as hydrolysis, which is a specific type of chemical reaction where water interacts with the base. This reaction involves the transfer of a proton from water to the base, generating a hydroxide ion \(OH^-\) and the corresponding conjugate acid of the base. For instance, dimethylamine \(\left(CH_3\right)_2NH\) in water reacts to form its conjugate acid \(\left(CH_3\right)_2NH_2^+\) and hydroxide ions. Similarly, other bases like carbonate \(CO_3^{2-}\) and formate ions \(CHO_2^{-}\) also react with water to produce bicarbonate \(HCO_3^-\), formic acid \(HCHO_2\), and more hydroxide ions.

It's important to know that these reactions are reversible and reach a state of chemical equilibrium, where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. This balance is crucial as it maintains the concentrations of ions in the solution, which is essential for predicting the behavior of the base in various chemical environments.
Equilibrium Constant Expression
The equilibrium constant expression is a mathematical way of expressing the balance between the products and reactants in a reversible chemical reaction at equilibrium. For base hydrolysis reactions, the equilibrium constant is referred to as \(K_b\), the base dissociation constant. It numerically describes the strength of the base in water. The expression for \(K_b\) involves the molar concentrations of the products over the reactants, each raised to the power of their stoichiometric coefficients.

An important point to remember is that water, being the solvent and present in large excess, typically does not appear in the \(K_b\) expression. As seen in the solutions provided for dimethylamine, carbonate ion, and formate ion, their respective \(K_b\) expressions are ratios that compare the concentration of the hydroxide ion and the conjugate acid to the concentration of the base itself. By calculating \(K_b\), you can infer how likely a base is to accept a proton in the presence of water and thus establish its basic potency.
Weak Base Kb
The value of the base dissociation constant, \(K_b\), gives insight into the base's strength. A weak base has a relatively small \(K_b\) value and does not dissociate completely in aqueous solution. This implies that in the reversible base reaction with water, the concentration of undissociated base is significant in comparison with its conjugate acid and hydroxide ions.

For example, when observing weak bases such as dimethylamine, carbonate ion, and formate ion, their \(K_b\) values would be less than one. This suggests that at equilibrium, the majority of the base remains in its undissociated form. As a result, in solutions of weak bases, the concentration of hydroxide ions is not as high as it would be with strong bases, which means the pH of the solution won't be significantly high. Understanding \(K_b\) is critical for predicting how a base will behave in different chemical processes, and it is vital for scientific tasks ranging from designing buffer solutions to pharmaceutical drug formulation.

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Most popular questions from this chapter

Hemoglobin plays a part in a series of equilibria involving protonation- deprotonation and oxygenation-deoxygenation. The overall reaction is approximately as follows: $$ \mathrm{HbH}^{+}(a q)+\mathrm{O}_{2}(a q) \rightleftharpoons \mathrm{HbO}_{2}(a q)+\mathrm{H}^{+}(a q) $$ where \(\mathrm{Hb}\) stands for hemoglobin and \(\mathrm{HbO}_{2}\) for oxyhemoglobin. (a) The concentration of \(\mathrm{O}_{2}\) is higher in the lungs and lower in the tissues. What effect does high \(\left[\mathrm{O}_{2}\right]\) have on the position of this equilibrium? (b) The normal \(\mathrm{pH}\) of blood is 7.4. Is the blood acidic, basic, or neutral? (c) If the blood \(\mathrm{pH}\) is lowered by the presence of large amounts of acidic metabolism products, a condition known as acidosis results. What effect does lowering blood pH have on the ability of hemoglobin to transport \(\mathrm{O}_{2}\) ?

Calculate the pH of each of the following strong acid solutions: (a) \(8.5 \times 10^{-3} \mathrm{M} \mathrm{HBr}\), (b) \(1.52 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) in \(575 \mathrm{~mL}\) of solution, (c) \(5.00 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HClO}_{4}\) diluted to \(50.0 \mathrm{~mL}\), (d) a solution formed by mixing \(10.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HBr}\) with \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\).

Consider the base hydroxylamine, \(\mathrm{NH}_{2} \mathrm{OH}\). (a) What is the conjugate acid of hydroxylamine? (b) When it acts as a base, which atom in hydroxylamine accepts a proton? (c) There are two atoms in hydroxylamine that have nonbonding electron pairs that could act as proton acceptors. Use Lewis structures and formal charges ers (Section 8.5) to rationalize why one of these two atoms is a much better proton acceptor than the other.

The amino acid glycine \(\left(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)\) can participate in the following equilibria in water: \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{\mathrm{a}}=4.3 \times 10^{-3}\) \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \({ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH} \quad K_{\mathrm{b}}=6.0 \times 10^{-5}\) (a) Use the values of \(K_{a}\) and \(K_{b}\) to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-} $$

If a solution of \(\mathrm{HF}\left(K_{a}=6.8 \times 10^{-4}\right)\) has a pH of \(3.65\), calculate the concentration of hydrofluoric acid.

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