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Chapter 16: Problem 1

Write an equation for the ionization of hydrogen cyanide, \(\mathrm{HCN}\), in aqueous solution. What is the equilibrium expression \(K_{a}\) for this acid ionization?

Short Answer

Expert verified
The equation for HCN ionization is \( \text{HCN} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CN}^- \). Equilibrium expression is \( K_a = \frac{[\text{H}_3\text{O}^+][\text{CN}^-]}{[\text{HCN}]} \).

Step by step solution

01

Identify the Reactants and Products

Hydrogen cyanide (HCN) ionizes in water to form hydronium ions (\( \text{H}^+ \)) and cyanide ions (\( \text{CN}^- \)). This can be written in the equation format where \( \text{HCN} \) is the reactant, and \( \text{H}^+ \) and \( \text{CN}^- \) are the products. Because it's in aqueous solution, water \( \text{H}_2\text{O} \) is also a reactant to help dissolve \( \text{HCN} \).
02

Write the Ionization Equation

The ionization of \( \text{HCN} \) in water is represented by the equation: \( \text{HCN}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{H}_3\text{O}^+_{(aq)} + \text{CN}^-_{(aq)} \). This equation shows the dissociation of HCN into a proton \( \text{H}^+ \) and \( \text{CN}^- \) ion, facilitated by water to form hydronium ion \( \text{H}_3\text{O}^+ \).
03

Define the Equilibrium Constant Expression

The equilibrium expression, \( K_a \), for the acid ionization is defined as the ratio of the concentrations of the products over the reactants, excluding pure water: \( K_a = \frac{[\text{H}_3\text{O}^+][\text{CN}^-]}{[\text{HCN}]} \). This formula represents how the acid ionization constant measures the strength of HCN in donating protons in solution.

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Key Concepts

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

Ionization Equation
In chemistry, an ionization equation is vital because it shows how a substance dissociates in a solution. For hydrogen cyanide (HCN) ionizing in water, this process creates ions. Here's how it works:

The equation representing this ionization is written as:
\[ \text{HCN}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{H}_3\text{O}^+_{(aq)} + \text{CN}^-_{(aq)}\]
  • HCN is the reactant that ionizes in water.
  • The water, \(\text{H}_2\text{O}\), acts as a solvent aiding the process.
  • Hydronium ion \(\text{H}_3\text{O}^+\) and cyanide ion \(\text{CN}^-\) are the products.
This equation reflects the reversible reaction occurring when HCN dissolves, producing hydronium and cyanide ions. Understanding ionization is key to grasping how acids behave in aqueous solutions. The ions formed can influence various chemical reactions and the pH of the solution.
Equilibrium Constant
The acid ionization constant, symbolized as \(K_a\), indicates the degree to which an acid ionizes in water, revealing its strength. Here's how we define the equilibrium constant for HCN:

The expression for \(K_a\) is:
\[ K_a = \frac{[\text{H}_3\text{O}^+][\text{CN}^-]}{[\text{HCN}]} \]
  • The square brackets denote concentration at equilibrium.
  • [\(\text{H}_3\text{O}^+\)] is the concentration of hydronium ions.
  • [\(\text{CN}^-\)] is the concentration of cyanide ions.
  • [\(\text{HCN}\)] is the concentration of undissociated hydrogen cyanide.
The equilibrium constant helps in predicting how far the ionization reaction will proceed. A large \(K_a\) indicates a strong acid, meaning more dissociation and more ions in solution. For weak acids like HCN, \(K_a\) tends to be small, signifying limited ionization.
Hydronium Ion
When acids ionize in water, a crucial ion that forms is the hydronium ion \(\text{H}_3\text{O}^+\). In the case of hydrogen cyanide, water accepts a proton \(\text{H}^+\) resulting in hydronium ions:

This transformation occurs according to the ionization equation:
\[ \text{HCN} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CN}^- \]
  • Formation of hydronium ions is a hallmark of acid solutions.
  • They contribute to the acidity of the solution.
  • The more \(\text{H}_3\text{O}^+\) ions present, the lower the pH.
Hydronium ions essentially determine the acidic nature of a solution, relating directly to pH. Their presence is a clear indicator of hydrogen ions from the acid interacting with the water molecules.

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Most popular questions from this chapter

Ionization of the first proton from \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is complete \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right.\) is a strong acid) \(;\) the acid-ionization constant for the second proton is \(1.1 \times 10^{-2}\). (a) What would be the approximate hydronium-ion concentration in \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) if ionization of the second proton were ignored? (b) The ionization of the second proton must be considered for a more exact answer, however. Calculate the hydronium-ion concentration in \(0.100 M\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\), accounting for the ionization of both protons.

A 25.00-mL sample contains \(0.562 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}\). This sample is used to standardize an NaOH solution. At the equivalence point, \(42.36 \mathrm{~mL}\) of \(\mathrm{NaOH}\) has been added. a. What was the concentration of the \(\mathrm{NaOH}\) ? b. What is the \(\mathrm{pH}\) at the equivalence point? C. Which indicator, bromthymol blue, methyl violet, or alizarin yellow R, should be used in the titration? Why?

A buffer is prepared by mixing \(525 \mathrm{~mL}\) of \(0.50 M\) formic acid, \(\mathrm{HCHO}_{2}\), and \(475 \mathrm{~mL}\) of \(0.50 \mathrm{M}\) sodium formate, \(\mathrm{NaCHO}_{2} .\) Calculate the \(\mathrm{pH}\). What would be the \(\mathrm{pH}\) of \(85 \mathrm{~mL}\) of the buffer to which \(8.6 \mathrm{~mL}\) of \(0.15 M\) hydrochloric acid had been added?

The \(\mathrm{pH}\) of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) (methylamine) is \(11.8\). When the chloride salt of methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\), is added to this solution, does the pH increase or decrease? Explain, using Le Ch芒telier's principle and the common-ion effect.

Find the \(\mathrm{pH}\) of the solution obtained when \(25 \mathrm{~mL}\) of \(0.065 M\) benzylamine, \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{NH}_{2}\), is titrated to the equivalence point with \(0.050 M\) hydrochloric acid. \(K_{b}\) for benzylamine is \(4.7 \times 10^{-10}\)
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