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Chapter 16: Problem 77

(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 \(\mathrm{F}^{-}\).

Short Answer

Expert verified
(a) Hydrofluoric acid is the stronger acid. (b) The cyanate ion (NCO^-) is the stronger base. (c) Kb values for NCO- and F- are \(2.86 \times 10^{-11}\) and \(1.47 \times 10^{-11}\), respectively.

Step by step solution

01

Determine the stronger acid

Given the Ka values for cyanic acid (\(3.5 \times 10^{-4}\)) and hydrofluoric acid (\(6.8 \times 10^{-4}\)), we can determine which one is a stronger acid. The larger the Ka value, the stronger the acid. Therefore, we can compare the Ka values: Cyanic acid Ka: \(3.5 \times 10^{-4}\) Hydrofluoric acid Ka: \(6.8 \times 10^{-4}\) From these values, we can see that hydrofluoric acid has a larger Ka value and is, therefore, the stronger acid.
02

Determine the stronger base

The corresponding bases for cyanic acid and hydrofluoric acid are the cyanate ion (NCO^-) and the fluoride ion (F^-), respectively. To determine which one is a stronger base, we can compare their Kb values. However, we only have their Ka values at the moment. To find the Kb values, we will use the ion-product constant for water, Kw. The relationship between the Ka, Kb and Kw can be stated as: Kw = Ka × Kb Where Kw (the ion-product constant for water) is equal to \(1.0 \times 10^{-14}\) at 25°C.
03

Calculate Kb values for cyanate and fluoride ions

Using the relation Kw = Ka × Kb, we'll calculate the Kb values for both NCO- and F-. First, rearrange the equation to solve for Kb: Kb = Kw / Ka For the cyanate ion (NCO^-): Kb(NCO^-) = \(1.0 \times 10^{-14} / 3.5 \times 10^{-4}\) For the fluoride ion (F^-): Kb(F^-) = \(1.0 \times 10^{-14} / 6.8 \times 10^{-4}\)
04

Compare Kb values and determine the stronger base

Now, let's compare the Kb values and determine which ion is the stronger base. First, calculate the numerical values: Kb(NCO^-) = \(2.86 \times 10^{-11}\) Kb(F^-) = \(1.47 \times 10^{-11}\) Since the Kb value for the cyanate ion (NCO^-) is larger, it is the stronger base compared to the fluoride ion (F^-). In summary: (a) Hydrofluoric acid is the stronger acid. (b) The cyanate ion (NCO^-) is the stronger base. (c) Kb values for NCO- and F- are \(2.86 \times 10^{-11}\) and \(1.47 \times 10^{-11}\), respectively.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) Acid strength in a series of \(\mathrm{H}-\mathrm{A}\) molecules increases with increasing size of A. (b) For acids of the same general structure but differing electronegativities of the central atoms, acid strength decreases with increasing electronegativity of the central atom. (c) The strongest acid known is HF because fluorine is the most electronegative element.

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},\) (b) \(3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in 500.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 10.0 \mathrm{~mL}\) of $0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\( diluted to \)500.0 \mathrm{~mL},(\mathbf{d})\( a solution formed by mixing \)20.0 \mathrm{~mL}$ of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of $8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}$.

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} \cdot(\mathbf{a})$ What is the \(\mathrm{pH}\) 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: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{p} K_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the pH of a \(0.25 \mathrm{M}\) solution of this substance?

Calculate the molar concentration of \(\mathrm{OH}^{-}\) in a \(0.724 \mathrm{M}\) solution of hypobromite ion $\left(\mathrm{BrO}^{-} ; K_{b}=4.0 \times 10^{-6}\right) .\( What is the \)\mathrm{pH}$ of this solution?
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