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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.

Short Answer

Expert verified
[CNO鈦籡, [H鈦篯 = 0.0059 M; [HCNO] = 0.0941 M; K鈧 = 3.70 脳 10鈦烩伌.

Step by step solution

01

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: \[\text{Ionized Concentration} = 0.059 \times 0.100 \, \text{M} = 0.0059 \, \text{M}\]
02

Determine Ion Concentrations

The ionized concentration accounts for the amount of H鈦 and CNO鈦 ions produced. Therefore, \[[\text{CNO}^-] = [\text{H}^+] = 0.0059 \, \text{M}\]This is because for every HCNO that ionizes, one H鈦 and one CNO鈦 ion are produced.
03

Calculate Remaining HCNO Concentration

To find the concentration of HCNO that remains in solution, subtract the ionized concentration from the initial concentration:\[[\text{HCNO}] = 0.100 \, \text{M} - 0.0059 \, \text{M} = 0.0941 \, \text{M}\]
04

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:\[K_a = \frac{(0.0059)(0.0059)}{0.0941} = 3.70 \times 10^{-4}\]

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Key Concepts

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

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鈦) and cyanate ions (CNO鈦). 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鈦 and one CNO鈦 when it dissociates, the concentrations of H鈦 and CNO鈦 in the solution will be equal. Ionization of cyanic acid can thus be represented by the equation:

HCNO 鈬 H鈦 + CNO鈦

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鈦 concentration: 0.0059 M
- Ionized CNO鈦 concentration: 0.0059 M

These calculations are based on the fact that for every one molecule of HCNO that ionizes, one H鈦 and one CNO鈦 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.

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Most popular questions from this chapter

(a) Give the conjugate base of the following Br酶nstedLowry acids: (i) \(\mathrm{H}_{2} \mathrm{SO}_{3},\) (ii) \(\mathrm{HSO}_{3}^{-}(\mathbf{b})\) Give the conjugate acid of the following Br酶nsted-Lowry bases: (i) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\), (ii) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\).

(a) Given that \(K_{a}\) for cyanic acid is \(3.5 \times 10^{-4}\) and that for hydrofluoric acid is \(6.8 \times 10^{-4}\), which is the stronger acid? (b) Which is the stronger base, the cyanate ion or the fluoride ion? (c) Calculate \(K_{b}\) values for \(\mathrm{NCO}^{-}\) and \(\mathrm{F}^{-}\).

At \(50^{\circ} \mathrm{C}\), the ion-product constant for \(\mathrm{H}_{2} \mathrm{O}\) has the value \(K_{w}=5.48 \times 10^{-14}\). (a) What is the pH of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

(a) The hydrogen carbonate ion \(\left(\mathrm{HCO}_{3}^{-}\right)\) is amphiprotic. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water.

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{p} K_{b}\) of the butyrate ion is 9.16. (a) Calculate the \(K_{a}\) for butyric acid. (b) Calculate the pH of a \(0.075 \mathrm{M}\) solution of butyric acid. (c) Calculate the pH of a \(0.075 \mathrm{M}\) solution of sodium butyrate.

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