From the equilibrium concentrations given, calculate Ka for...

From the equilibrium concentrations given, calculate $K_{a}$ for each of the weak acids and $K_{b}$ for each of the weak bases. $$ \begin{array}{l}{\text { (a) } \mathrm{NH}_{3} :\left[\mathrm{OH}^{-}\right]=3.1 \times 10^{-3} M} \\ {\left[\mathrm{NH}_{4}^{+}\right]=3.1 \times 10^{-3} M} \\ {\left[\mathrm{NH}_{3}\right]=0.533 M}\end{array} $$ $$ \begin{array}{l}{\text { (b) HNO }_{2} :\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.011 \mathrm{M} ;} \\ {\left[\mathrm{NO}_{2}-\right]=0.0438 \mathrm{M} ;} \\ {\left[\mathrm{HNO}_{2}\right]=1.07 \mathrm{M}}\end{array} $$ $$ \begin{array}{l}{\text { (c) }\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} :\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\right]=0.25 M} \\ {\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\right]=4.3 \times 10^{-3} M ;} \\ {\left[\mathrm{OH}^{-}\right]=4.3 \times 10^{-3} \mathrm{M}}\end{array} $$ $$ \begin{array}{l}{\text { (d) } \mathrm{NH}_{4}^{+} :\left[\mathrm{NH}_{4}^{+}\right]=0.100 \mathrm{M}} \\ {\left[\mathrm{NH}_{3}\right]=7.5 \times 10^{-6} M} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.5 \times 10^{-6} \mathrm{M}}\end{array} $$

From the equilibrium concentrations given, calculate $K_{a}$ for each of the weak acids and $K_{b}$ for each of the weak bases.
$$
\begin{array}{l}{\text { (a) } \mathrm{NH}_{3} :\left[\mathrm{OH}^{-}\right]=3.1 \times 10^{-3} M} \\ {\left[\mathrm{NH}_{4}^{+}\right]=3.1 \times 10^{-3} M} \\ {\left[\mathrm{NH}_{3}\right]=0.533 M}\end{array}
$$
$$
\begin{array}{l}{\text { (b) HNO }_{2} :\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.011 \mathrm{M} ;} \\ {\left[\mathrm{NO}_{2}-\right]=0.0438 \mathrm{M} ;} \\ {\left[\mathrm{HNO}_{2}\right]=1.07 \mathrm{M}}\end{array}
$$
$$
\begin{array}{l}{\text { (c) }\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} :\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\right]=0.25 M} \\ {\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\right]=4.3 \times 10^{-3} M ;} \\ {\left[\mathrm{OH}^{-}\right]=4.3 \times 10^{-3} \mathrm{M}}\end{array}
$$
$$
\begin{array}{l}{\text { (d) } \mathrm{NH}_{4}^{+} :\left[\mathrm{NH}_{4}^{+}\right]=0.100 \mathrm{M}} \\ {\left[\mathrm{NH}_{3}\right]=7.5 \times 10^{-6} M} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.5 \times 10^{-6} \mathrm{M}}\end{array}
$$

Key Concepts

Equilibrium Constant Expression

This concept involves writing an expression for chemical equilibrium based on the reaction’s balanced equation, where the concentrations of products are divided by those of reactants, with each concentration raised to the power of its stoichiometric coefficient. It is fundamental for calculating any equilibrium constant value when the concentrations at equilibrium are known.

Acid Dissociation Constant (Ka)

Ka represents the equilibrium constant for the dissociation of a weak acid in water into its conjugate base and a hydrogen ion. It quantifies the strength of an acid; a smaller Ka indicates a weaker acid with less dissociation. Calculating Ka using equilibrium concentrations involves the ratio of the product concentrations (typically hydrogen ion and conjugate base) to the concentration of the weak acid.

Base Dissociation Constant (Kb)

Kb is the equilibrium constant for the reaction of a weak base with water to form its conjugate acid and hydroxide ion. It measures how strongly a base reacts with water; a smaller Kb reflects a weaker base with less dissociation. Determining Kb involves the ratio of the concentrations of the hydroxide ion and the conjugate acid to the concentration of the initial base.

Transcript

00:01 Problem 57 from chapter 14 is asking us to calculate the ka or kb of several molecular compounds.

00:10 So to answer this question, we must recall that the ionization constant of either an acid or a base, i'll just leave it as indiscriminate at the moment, is equal to the product, the concentration of the product, divided by the concentration of the reactant.

00:34 That's a square bracket.

00:44 And you can have multiple terms on either the nominator or the nominator, but i'll just leave it as one term for now.

00:53 So product, concentration of product divided by the concentration of the reactant is going to give us either the ionization acid ionization constant or the base ionization constant.

01:03 So for part a of this question, we are asked to calculate the ionization constant of ammonia.

01:12 And the product of this reaction is ammonium and the hydroxide ion.

01:18 So we can infer that we are going to be calculating the kb of ammonia because ammonia is accepting a proton to become ammonium.

01:34 Of nh3.

01:36 So let's go ahead and equal to the concentration of ammonium ions multiply by the concentration of hydroxide ions.

02:01 And this is going to be divided by the concentration of ammonia provided in the question.

02:19 Once we go ahead and plug the values in that we, we were given in the question.

02:25 We are, we generate a value of the kb of ammonia as 1 .8 times 10 to the minus 5.

02:42 All right, so that's a kb of ammonia.

02:45 Next, we are asked to calculate either the kb or kb for nitrous acid.

02:56 So let's go ahead and take a look at the different constituents of this question.

03:06 So we have nitrous acid and nitrogen dioxide.

03:13 And as we can see, the nitrous acid loses a proton.

03:21 Therefore, we are going to be calculating the k -a of nitrous acid because nitrous acid is losing a proton and there are hydronium ions in solution, suggesting that the reactant is an acid.

03:45 So we are going to plug in the values for nitrogen dioxide, the concentration of nitrogen dioxide, multiplied by hydrogen ion concentration.

03:59 And then that is divided by the concentration of our reactant, which is nitrous acid.

04:12 Whoops.

04:19 H .n.

04:25 When we plug the concentrations that we are given in the question, we are left with 4 .5 times 10 to the minus 4 as the kb, or excuse me, the ka.

04:47 4 .5 times 10 to the minus 4...

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