Problem 53
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
Verified Answer
In a 0.100 M solution of cyanic acid (HCNO) that is 5.9% ionized, the concentrations of CNO⁻ and H⁺ are both 0.0059 M, while the concentration of non-ionized HCNO is 0.0941 M. The ionization constant, Ka, for cyanic acid is approximately \(3.70 \times 10^{-4}\).
1Step 1: Calculate the ionized concentration of HCNO
First, we need to calculate the ionized concentration of HCNO. To do this, we simply multiply the initial concentration by the percentage ionized (as a decimal).
Ionized concentration of HCNO = (Initial concentration) × (Percentage ionized)
Ionized concentration of HCNO = (0.100 M) × (5.9/100)
\[ Ionized \ concentration \ of \ HCNO = 0.0059 \ M \]
2Step 2: Determine the concentrations of CNO⁻ and H⁺
Since HCNO is a weak monoprotic acid, we can assume that every ionized HCNO molecule produced one H⁺ ion and one CNO⁻ ion. Therefore, the concentrations of CNO⁻ and H⁺ are equal to the ionized concentration of HCNO.
\[ [CNO^-] = [H^+] = 0.0059 \ M \]
3Step 3: Calculate the concentration of non-ionized HCNO
To find the concentration of non-ionized HCNO, we can subtract the ionized concentration from the initial concentration:
Non-ionized HCNO concentration = (Initial concentration) - (Ionized concentration)
\[ [HCNO] = 0.100 \ M - 0.0059 \ M = 0.0941 \ M \]
4Step 4: Calculate the ionization constant (Ka)
Now that we have the concentrations of HCNO, CNO⁻, and H⁺, we can find Ka using the ionization equilibrium formula:
\[ K_{a} = \frac{[CNO^-][H^+]}{[HCNO]} \]
Plugging in the concentrations, we get:
\[ K_{a} = \frac{(0.0059 \ M)(0.0059 \ M)}{0.0941 \ M} \]
Calculating the Ka value:
\[ K_{a} ≈ 3.70 \times 10^{-4} \]
In conclusion, the concentrations of CNO⁻, H⁺, and HCNO are 0.0059 M, 0.0059 M, and 0.0941 M, respectively. The ionization constant, Ka, for cyanic acid is approximately 3.70 x 10⁻⁴.
Key Concepts
Ionization Constant (Ka)Cyanic AcidEquilibrium Concentration Calculation
Ionization Constant (Ka)
In the context of weak acids, the ionization constant, known as Ka, plays a crucial role. It measures the strength of the acid in a solution. Specifically, Ka quantifies how much the acid dissociates into its ions in water. A higher Ka value indicates a stronger weak acid, meaning it ionizes more in water.
The formula for calculating the ionization constant is:
The formula for calculating the ionization constant is:
- \[ K_{a} = \frac{[H^+][A^-]}{[HA]} \]
- \([H^+]\) is the concentration of hydrogen ions.
- \([A^-]\) stands for the concentration of the conjugate base, in this case, the cyanic ion \[CNO^-\].
- \([HA]\) represents the concentration of the undissociated acid (HCNO).
Cyanic Acid
Cyanic acid (HCNO) is a weak acid, meaning it doesn’t completely dissociate in solution. Instead, only a small percentage of its molecules release protons (H⁺) when dissolved in water. This property distinguishes weak acids from strong acids, which ionize completely.
This particular acid contains a cyano group, which is a carbon triple-bonded to a nitrogen atom. Despite its weak nature, understanding the percentage of HCNO that ionizes is critical for calculating concentrations of its ions in solution. In the given exercise, HCNO is 5.9% ionized. This tells us that 5.9% of the HCNO molecules in a 0.100 M solution dissociate to produce H⁺ and CNO⁻ ions.
Knowing the ionization level helps in calculating other concentrations and the equilibrium constant, thereby providing a snapshot of the acid's behavior in aqueous solution.
This particular acid contains a cyano group, which is a carbon triple-bonded to a nitrogen atom. Despite its weak nature, understanding the percentage of HCNO that ionizes is critical for calculating concentrations of its ions in solution. In the given exercise, HCNO is 5.9% ionized. This tells us that 5.9% of the HCNO molecules in a 0.100 M solution dissociate to produce H⁺ and CNO⁻ ions.
Knowing the ionization level helps in calculating other concentrations and the equilibrium constant, thereby providing a snapshot of the acid's behavior in aqueous solution.
Equilibrium Concentration Calculation
After understanding the ionization percentage, the next step is to calculate the equilibrium concentrations of species in the solution. This involves a few straightforward calculations:
First, calculate the concentration of ionized HCNO using the percentage ionized:
First, calculate the concentration of ionized HCNO using the percentage ionized:
- Ionized HCNO Concentration = Initial Concentration × Percentage Ionized
- \[ = 0.100 \, \text{M} \times \frac{5.9}{100} = 0.0059 \, \text{M} \]
- \([H^+] = [CNO^-] = 0.0059 \, \text{M}\)
- \( [HCNO] = 0.100 \, \text{M} - 0.0059 \, \text{M} = 0.0941 \, \text{M} \)
Other exercises in this chapter
Problem 51
Lactic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right)\) has one acidic hydrogen. A \(0.10 \mathrm{M}\) solution of lactic acid has a
View solution Problem 52
Phenylacetic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\right)\) is one of the substances that accumulates in the blood of people
View solution Problem 54
A \(0.100 \mathrm{M}\) solution of bromoacetic acid \(\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)\) is 13.2\% ionized. Calculate \(\left[\mathrm{H}^{+}\right],
View solution Problem 55
A particular sample of vinegar has a pH of 2.20 . If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\), calculate the
View solution