Calculate the ionization constant for each of the following...

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid: $$\begin{array}{l}{\text { (a) HTe - (as a base) }} \\ {\text { (b) }\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}} \\ {\text { (c) HAsO_ }_{4}^{3-} \text { (as a base) }}\end{array}$$ $$\begin{array}{l}{\text { (d) } \mathrm{HO}_{2}^{-} \text { (as a base) }} \\ {\text { (e) } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}+} \\ {\text { (f) } \mathrm{HSO}_{3}-\text { (as a base) }}\end{array}$$

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
$$\begin{array}{l}{\text { (a) HTe - (as a base) }} \\ {\text { (b) }\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}} \\ {\text { (c) HAsO_ }_{4}^{3-} \text { (as a base) }}\end{array}$$
$$\begin{array}{l}{\text { (d) } \mathrm{HO}_{2}^{-} \text { (as a base) }} \\ {\text { (e) } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}+} \\ {\text { (f) } \mathrm{HSO}_{3}-\text { (as a base) }}\end{array}$$

Key Concepts

Acid-Base Equilibrium

Acid-base equilibrium refers to the dynamic balance between the forward and reverse reactions of proton transfer in aqueous solutions. This equilibrium is established when the rates of proton donation and acceptance are equal, and it is characterized by the relevant ionization constants (Ka or Kb). Understanding this equilibrium is key to predicting the behavior of acids and bases in various chemical reactions.

Acid Ionization Constant (Ka)

The acid ionization constant, Ka, quantifies the extent to which a weak acid donates a proton in an aqueous solution. It is defined by the equilibrium concentrations of the products and reactants when the acid dissociates, and it provides a measure of the acid’s strength—the larger the Ka, the stronger the acid.

Base Ionization Constant (Kb)

The base ionization constant, Kb, describes the extent to which a weak base accepts a proton when dissolved in water. It is defined by the equilibrium between the base, water, and the conjugate acid formed. A higher Kb indicates a stronger base, reflecting a greater tendency to accept a proton.

Relationship between Ka and Kb

For any conjugate acid-base pair, the product of the acid ionization constant (Ka) and the base ionization constant (Kb) is equal to the ionization constant of water (Kw). This relationship, expressed as Ka × Kb = Kw, allows one to calculate the ionization constant of an acid if the constant for its conjugate base is known, and vice versa.

Conjugate Acid-Base Pairs

A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton. When an acid donates a proton, it forms its conjugate base, and when a base accepts a proton, it forms its conjugate acid. The strength of an acid or base is inversely related to the strength of its conjugate partner.

Water Ionization Constant (Kw)

The water ionization constant, Kw, is the equilibrium constant for the self-ionization of water into hydronium (or hydrogen) and hydroxide ions. At 25°C, Kw has a value of 1.0 × 10?¹? mol² L?². This constant is fundamental in relating acid and base strengths through the relationship Ka × Kb = Kw in aqueous solutions.

Transcript

00:01 Problem 61 from chapter 14 is asking us to calculate the ionization constant, meaning the k -a or kb, for each of the following compounds, based on the ionization constant of its conjugate base or conjugate acid.

00:18 So to answer this question, we must remember that the kw is equal to k -a, acid ionization constant, multiplied by the kb, or the base ionization constant, where k -w is equal to 1 times 10 to the minus 14.

00:52 K -w equals 1 times 10 to the minus 14.

01:06 So for each one of these compounds, what we're going to do is we're going to refer.

01:11 To their ionization constants of either their conjugate acid or conjugate base depending on the compound and that information can be found in the appendix of the textbook and then we're going to solve for either the k a or kb so let's go ahead and get started with hydrogen telluride so it's as a base so we're going to be looking at at the k -a of its conjugate acid.

01:49 H -2, sorry, let's see, h -t -e minus, we want it as a base, so we are going to be looking for its conjugate acid.

02:04 And the k -a of hydrogen -tolerides, of hydrogen -toleride is equal to 2 .3 times 10 to the minus 3.

02:23 All we have to do is just rearrange this equation to give us the kb.

02:31 So ab is going to be equal to kw, which is 1 times 10 to the minus 14, divided by this value for the ka, 2 .3 times 10.

02:52 And what that is going to give us is 4 .3.

03:00 So the kb is going to equal 4 .3 times 10 to the minus 12.

03:22 Great.

03:22 So that's the first compound.

03:24 Next, let's move to the second compound.

03:38 The tri -methylamine cation.

03:43 So that's going to be.

03:53 3 and h4 plus.