Problem 53 - Chemistry The Central Science In Si Units

Question

A \(0.100 \mathrm{M}\) solution of cyanic acid (HCNO) is \(5.9 \%\) ionized. Using this information, calculate [CNO^ \(\left.^{-}\right],\left[\mathrm{H}^{+}\right],[\mathrm{HCNO}],\) and \(K_{a}\) for cyanic acid.

Step-by-Step Solution

Answer

[CNO⁻], [H⁺] = 0.0059 M; [HCNO] = 0.0941 M; Kₐ = 3.70 × 10⁻⁴.

Step 1: Calculate Ionized Concentration

The question states that the cyanic acid solution is 5.9% ionized. This means that 5.9% of the initial concentration of HCNO is converted into ions. Calculate the ionized concentration by multiplying the percentage (as a decimal) with the initial concentration: \[\text{Ionized Concentration} = 0.059 \times 0.100 \, \text{M} = 0.0059 \, \text{M}\]

Step 2: Determine Ion Concentrations

The ionized concentration accounts for the amount of H⁺ and CNO⁻ ions produced. Therefore, \[[\text{CNO}^-] = [\text{H}^+] = 0.0059 \, \text{M}\]This is because for every HCNO that ionizes, one H⁺ and one CNO⁻ ion are produced.

Step 3: Calculate Remaining HCNO Concentration

To find the concentration of HCNO that remains in solution, subtract the ionized concentration from the initial concentration:\[[\text{HCNO}] = 0.100 \, \text{M} - 0.0059 \, \text{M} = 0.0941 \, \text{M}\]

Step 4: Calculate Acid Dissociation Constant \(K_a\)

The acid dissociation constant \(K_a\) can be calculated using the formula:\[K_a = \frac{[\text{H}^+][\text{CNO}^-]}{[\text{HCNO}]}\]Substitute the known values:\[K_a = \frac{(0.0059)(0.0059)}{0.0941} = 3.70 \times 10^{-4}\]

Key Concepts

Cyanic Acid Ionization

Understanding cyanic acid ionization is essential when studying acid-base reactions. Cyanic acid, symbolized as HCNO, ionizes in water to produce hydrogen ions (H⁺) and cyanate ions (CNO⁻). This process is called ionization. When cyanic acid is dissolved in water, a percentage of the molecules dissociates (splits) into ions. The extent of this ionization depends on various factors including the nature of the acid and the solution concentration. In our case, a 5.9% ionization means that only 5.9% of the initial HCNO molecules split into ions.

Because each HCNO molecule produces one H⁺ and one CNO⁻ when it dissociates, the concentrations of H⁺ and CNO⁻ in the solution will be equal. Ionization of cyanic acid can thus be represented by the equation:

HCNO ⇌ H⁺ + CNO⁻

The ionization degree gives us an idea of how many molecules are ionized compared to those that remain intact.

Concentration Calculation

Calculating concentrations accurately is crucial in chemistry to determine how much of each species is present in a solution. Given that 5.9% of HCNO is ionized, we can calculate the concentrations of the ions and the un-ionized acid. Starting with an initial concentration of 0.100 M HCNO, to find how much of it ionizes, we multiply the initial concentration by the ionization percentage expressed as a decimal (0.059). This results in an ionized concentration of 0.0059 M for each ion formed:

- Ionized H⁺ concentration: 0.0059 M
- Ionized CNO⁻ concentration: 0.0059 M

These calculations are based on the fact that for every one molecule of HCNO that ionizes, one H⁺ and one CNO⁻ ion are formed. Thus, the ion concentrations are equal, assuming full disassociation into these ions when it does ionize.

Initial and Ionized Concentration

Before cyanic acid starts to ionize, it has an initial concentration of 0.100 M. Ionization results in the formation of ions, leaving behind a portion of cyanic acid that remains non-ionized. After determining the ionized concentration (0.0059 M), we subtract it from the initial 0.100 M concentration to find how much HCNO remains un-ionized. This remaining concentration is given by:

HCNO remaining: 0.100 M - 0.0059 M = 0.0941 M

Knowing the initial and ionized concentrations is critical for calculating the acid dissociation constant (\(K_a\)), which is a measure of the strength of an acid in solution. The expression to calculate \(K_a\) using the concentrations of the ions (\([H^+]\) and \([CNO^-]\)) and the remaining HCNO (\([HCNO]\)) is:

\[ K_a = \frac{[H^+][CNO^-]}{[HCNO]} \]

Substituting in our known values gives the \(K_a\), a representation of the acid's capability to donate protons.