Question

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

Short 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 $3.70\times {10}^{-4}$.

Step-by-step solution

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) $$IonizedconcentrationofHCNO=0.0059M$$

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. $$[CN{O}^{-}]=[H^{+}]=0.0059M$$

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) $$[HCNO]=0.100M-0.0059M=0.0941M$$

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: $$K_a=\frac{[CN{O}^{-}][H^{+}]}{[HCNO]}$$ Plugging in the concentrations, we get: $$K_a=\frac{(0.0059M)(0.0059M)}{0.0941M}$$ Calculating the Ka value: $$K_a\approx 3.70\times {10}^{-4}$$ 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)

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:

• $$K_a=\frac{[H^{+}][A^{-}]}{[HA]}$$

Here,

• $[H^{+}]$ is the concentration of hydrogen ions.
• $[A^{-}]$ stands for the concentration of the conjugate base, in this case, the cyanic ion $$CN{O}^{-}$$.
• $[HA]$ represents the concentration of the undissociated acid (HCNO).

Plugging the values into this formula helps us understand how much of the acid dissociates at equilibrium. This understanding is foundational for predicting the behavior of weak acids in various chemical environments.

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.

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:

• Ionized HCNO Concentration = Initial Concentration × Percentage Ionized

Using the provided values,

• $$=0.100\text{M}\times \frac{5.9}{100}=0.0059\text{M}$$

Next, since each ionized molecule of HCNO gives rise to one H⁺ ion and one CNO⁻ ion, the concentrations of both ions will be equal to the ionized HCNO concentration:

• $[H^{+}]=[CN{O}^{-}]=0.0059\text{M}$

To find the concentration of non-ionized HCNO, subtract the ionized concentration from the initial concentration:

• $[HCNO]=0.100\text{M}-0.0059\text{M}=0.0941\text{M}$

Finally, these concentrations can be used to determine the acid's ionization constant (Ka), yielding valuable insights into its dissociation behavior under equilibrium conditions.