Problem 63 - Chemistry and Chemical Reactivity

Question

Sodium cyanide is the salt of the weak acid HCN. Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{OH}^{-}\) HCN, and \(\mathrm{Na}^{+}\) in a solution prepared by dissolving 10.8 g of NaCN in enough water to make \(5.00 \times 10^{2} \mathrm{mL}\) of solution at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

\([\text{H}_3\text{O}^+] = 1.06 \times 10^{-11} \text{ M}\), \([\text{OH}^-] = 9.47 \times 10^{-4} \text{ M}\), \([\text{HCN}]=9.47 \times 10^{-4} \text{ M}\), \([\text{Na}^+] = 0.44 \text{ M}\).

Step 1: Determine Moles of NaCN

First, calculate the number of moles of NaCN. With a given mass of NaCN as 10.8 g and its molar mass approximately 49 g/mol, use the formula: \(\text{Moles of NaCN} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{10.8\text{ g}}{49\text{ g/mol}}\). This gives \(0.22\) moles of NaCN.

Step 2: Calculate Initial Concentration of NaCN

Now, find the concentration of NaCN in mol/L. The volume of the solution is 0.500 L (since \(500 \text{ mL} = 0.500 \text{ L}\)). Using \([\text{NaCN}] = \frac{\text{Moles of NaCN}}{\text{Volume of solution in L}} = \frac{0.22 \text{ moles}}{0.500 \text{ L}}\), we get \(0.44 \text{ M NaCN}\).

Step 3: Use Hydrolysis to Find [HCN]

NaCN completely dissociates into \(\text{Na}^+\) and \(\text{CN}^-\) ions. The \(\text{CN}^-\) ion undergoes hydrolysis: \(\text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^-\). Use \(K_w = 1.0 \times 10^{-14}\) and \(K_a = 6.17 \times 10^{-10}\) (for HCN) to calculate the base constant \(K_b\): \(K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.17 \times 10^{-10}}\approx 1.62 \times 10^{-5}\).

Step 4: Equilibrium Concentrations using ICE Table

Set up an ICE table to find \([\text{OH}^-]\):- Initial for \(\text{CN}^-\) is 0.44 M, others 0.- Change: let \(x\) be the change for \(\text{OH}^-\) and \text{HCN}, so decrease in \(\text{CN}^-\) is \(-x\).- Equilibrium: \([\text{CN}^-]=0.44-x\), \([\text{HCN}] = x \), and \([\text{OH}^-]=x\).Apply \(K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]}\), yielding \(1.62 \times 10^{-5} = \frac{x^2}{0.44}\). Solving for \(x\) gives \([\text{OH}^-]=[\text{HCN}]=9.47 \times 10^{-4} \text{ M}\).

Step 5: Determine [H3O+]

For an aqueous solution, the relationship \([\text{H}_3\text{O}^+][\text{OH}^-]=K_w\) holds. Using \([\text{OH}^-] = 9.47 \times 10^{-4} \text{ M}\), solve \([\text{H}_3\text{O}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{9.47 \times 10^{-4}}\), thus \([\text{H}_3\text{O}^+] = 1.06 \times 10^{-11} \text{ M}\).

Step 6: Solving for [Na+]

Since NaCN dissociates completely into \(\text{Na}^+\) and \(\text{CN}^-\), \([\text{Na}^+]\) is equal to the initial concentration of NaCN, which is \([\text{Na}^+] = 0.44 \text{ M}\).

Key Concepts

Dissociation

Dissociation is the process where ionic compounds separate into their individual ions in a solution. When sodium cyanide (NaCN), a common compound, dissolves in water, it dissociates into sodium ions \(\text{Na}^+\) and cyanide ions \(\text{CN}^-\). This is because NaCN is an ionic compound composed of positively charged sodium ions and negatively charged cyanide ions.

• Dissociation results in a solution where the ions are free to move and interact with the solvent molecules.
• This complete dissociation of NaCN is crucial for further chemical reactions, such as hydrolysis.

Understanding dissociation helps in grasping how certain compounds behave in water and how the individual ions affect the solution's properties.

Hydrolysis

Hydrolysis involves the reaction of an ion with water, leading to the formation of new compounds. In the case of sodium cyanide, the cyanide ion \(\text{CN}^-\) undergoes hydrolysis. When \(\text{CN}^-\) interacts with water, it tends to form hydrogen cyanide \(\text{HCN}\) and hydroxide ions \(\text{OH}^-\). This is an equilibrium reaction, represented as:
\(\text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^-\).

• Hydrolysis affects the pH of the solution due to the generation of \(\text{OH}^-\) ions, making it more basic.
• This reaction's extent is determined by the base constant \(K_b\), which is calculated using the water dissociation constant \(K_w\) and the acid dissociation constant \(K_a\) of HCN.

Comprehending hydrolysis is key to predicting the behavior of solutions containing salts of weak acids and bases.

Equilibrium Calculations

Equilibrium calculations allow us to determine the concentrations of different species in a solution at equilibrium. With sodium cyanide, we are interested in the concentrations of \(\text{HCN}\) and hydroxide ions \(\text{OH}^-\) after hydrolysis. To approach this, we use an ICE (Initial, Change, Equilibrium) table.

• The ICE table starts by outlining the initial concentration of each species involved in the reaction.
• Then, we denote the change as the reaction moves towards equilibrium, which is often represented by the variable \(x\).
• Finally, we use the equilibrium constant \(K_b\) to solve for \(x\) and find the equilibrium concentrations.

Equilibrium calculations are crucial for understanding how a system reaches stability and predicting the concentration of ions in solutions.

Molarity

Molarity, often symbolized as M, is used to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. For sodium cyanide, once we have calculated the moles from its mass using its molar mass, we can determine the molarity.

• Molarity is calculated using the formula: \[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\].
• It provides a straightforward way to understand the concentration of ions that result from dissociation, such as \(\text{Na}^+\) and \(\text{CN}^-\) ions.

Grasping molarity helps in performing accurate calculations related to solution concentrations, crucial for any chemical analysis.

pH

The pH of a solution is a measure of its acidity or basicity. It is calculated based on the concentration of hydrogen ions \(\text{H}_3\text{O}^+\) in the solution. In our sodium cyanide solution, the pH is influenced by the hydroxide ions \(\text{OH}^-\) produced through hydrolysis. Since \(\text{OH}^-\) increases, the solution becomes more basic.

• The relationship between hydrogen ions and hydroxide ions in water is expressed as: \(\text{[H}_3\text{O}^+] \cdot [\text{OH}^-] = K_w\).
• From this, we can calculate \(\text{[H}_3\text{O}^+]\), and subsequently, the pH can be found using the formula \[\text{pH} = -\log_{10}(\text{[H}_3\text{O}^+] )\].

Understanding pH is essential for predicting how solutions will respond in different chemical environments.