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# 20.4: Strength of Acids and Bases

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• describe the difference between a strong and weak acid.
• identify specific acids as strong or weak.
• define weak acids and bases.
• use \begin{align*}K_a\end{align*} or \begin{align*}K_b\end{align*} to find \begin{align*}[\mathrm{H}^+]\end{align*} and vice versa.
• use \begin{align*}K_a\end{align*} or \begin{align*}K_b\end{align*} to find pH.

## Vocabulary

• \begin{align*}K_a\end{align*}
• \begin{align*}K_b\end{align*}
• strong acid
• strong base
• weak acid
• weak base

## Introduction

A great number of people associate strong acids with their ability to react with skin, essentially “melting” it away from bone. On a popular crime show, this very chemistry was used as a method for a crime. The crime show used sulfuric acid. Why sulfuric acid and not acetic acid? What makes the difference? How can we tell if an acid is strong or weak? The answers to these questions will be the focus of the lesson that follows.

## Strong Acids and Bases

Strong acids and strong bases are those that completely dissociate when dissolved in water. HCl is an example of a strong acid, as seen in the equation below:

\begin{align*}\mathrm{HCl}_{(g)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightarrow \mathrm{H}_3\mathrm{O}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}\end{align*}

Notice that this equation only has a forward arrow instead of the equilibrium arrows seen in many dissociation reactions. This indicates that the reaction goes to completion, so when the reaction is over, there are essentially no HCl molecules left in solution, only \begin{align*}\mathrm{H}^+\end{align*} and \begin{align*}\mathrm{Cl}^-\end{align*} ions. Since strong acids fully dissociate, many ions are produced, making the solution a good conductor of electricity.

There are only six common strong acids. These acids are shown in Table below. Each of the acids found in this table will completely dissociate in water.

Strong Acids
Name Formula
Hydrochloric acid \begin{align*}\mathrm{HCl}\end{align*}
Hydrobromic acid \begin{align*}\mathrm{HBr}\end{align*}
Hydroiodic \begin{align*}\mathrm{HI}\end{align*}
Nitric acid \begin{align*}\mathrm{HNO}_3\end{align*}
Perchloric acid \begin{align*}\mathrm{HClO}_4\end{align*}
Sulfuric acid \begin{align*}\mathrm{H}_2\mathrm{SO}_4\end{align*}

## Weak Acids and Bases

Weak acids and weak bases do not completely dissociate when dissolved in water. The less dissociation that takes place, the weaker the acid since there will be fewer \begin{align*}\mathrm{H}^+\end{align*} ions in solution. For example, when acetic acid is placed in water, only about 5% of the acetic acid molecules separate into \begin{align*}\mathrm{H}^+\end{align*} ions and \begin{align*}\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-\end{align*} ions:

\begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{C}_2\mathrm{H}_3\mathrm{O}^-_{2(aq)}\end{align*}

Notice that we now use the equilibrium arrows in the chemical equation. At equilibrium, this solution will contain more acetic acid molecules than hydronium and acetate ions.

Let’s look at citric acid, a weak acid. Citric acid, \begin{align*}\mathrm{C}_6\mathrm{H}_8\mathrm{O}_7\end{align*}, is commonly found in everyday products like lemons, limes, and soft drinks. It is the substance responsible for making these foods taste sour. If we were to write an ionization equation for citric acid, it would appear as written below:

\begin{align*}\mathrm{C}_6\mathrm{H}_8\mathrm{O}_{7(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+_{(aq)} + \mathrm{C}_6\mathrm{H}_7\mathrm{O}^-_{7(aq)}\end{align*}

All acids that you will encounter at this level are weak acids if they are not one of the six listed in Table above are weak. Even though these weak acids are very soluble in water, they dissolve as molecules, and only a few of the molecules break into ions in the solution. Since weak acids do not produce many ions, a weak acid solution will only conduct a small electric current.

Example:

Write dissociation equations for only those acids that are weak.

1. Sulfuric acid \begin{align*}(\mathrm{H}_2\mathrm{SO}_4)\end{align*}
2. Hydrofluoric acid \begin{align*}(\mathrm{HF})\end{align*}
3. Trichloroacetic acid \begin{align*}(\mathrm{CCl}_3\mathrm{COOH})\end{align*}

Solution:

1. \begin{align*}\mathrm{H}_2\mathrm{SO}_4\end{align*} is a strong acid (one of the six).
2. \begin{align*}\mathrm{HF}\end{align*} is a weak acid (not one of the six); \begin{align*}\mathrm{HF}_{(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{F}^-_{(aq)}\end{align*}
3. \begin{align*}\mathrm{CCl}_3\mathrm{COOH}\end{align*} is a weak acid (not one of the six). \begin{align*}\mathrm{CCl}_3\mathrm{COOH}_{(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{CCl}_3\mathrm{COO}^-_{(aq)}\end{align*}

## Weak Acids and Weak Bases as Equilibrium Systems

The dissociation of a weak acid is another special type of reaction that is given its own equilibrium constant. The equilibrium constant for the dissociation of a weak acid is designated Ka. The dissociation reaction for acetic acid, one of the primary components of vinegar, is shown below.

\begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{C}_2\mathrm{H}_3\mathrm{O}^-_{2(aq)}\end{align*}

The equilibrium constant, \begin{align*}K_a\end{align*}, for acetic acid is \begin{align*}1.8 \times 10^{-5}\end{align*}. This value is small, indicating that the equilibrium position lies more to the left than to the right. In other words, there are more acetic acid molecules at equilibrium than there are acetate ions or hydronium ions. Since \begin{align*}K_a\end{align*} is less than \begin{align*}1\end{align*}, we know that \begin{align*}[\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2] > [\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-][\mathrm{H}^+]\end{align*} at equilibrium.

Similarly, the equilibrium constant for the reaction between a weak base and water is designated Kb. An example of a weak base reaction is shown below.

\begin{align*}\mathrm{NH}_{3(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}^+_{4(aq)} + \mathrm{OH}^-_{(aq)}\end{align*}

The equilibrium constant for this reaction is, coincidentally, also \begin{align*}1.8 \times 10^{-5}\end{align*}. Again, a value much less than one indicates that the equilibrium favors the reactants. In this solution, very few ammonia molecules are able to remove a proton from water to create ammonium and hydroxide ions.

Example:

Put the following acids in order of decreasing acid strength. Write an equilibrium expression for each. Remember dissociation increases with increasing acid strength (or decreasing pH).

Formic acid (\begin{align*}\mathrm{HCOOH}\end{align*}) \begin{align*}K_a = 6.3 \times 10^{-4}\end{align*}
Phosphoric acid (\begin{align*}\mathrm{H}_3\mathrm{PO}_4\end{align*}) \begin{align*}K_a = 7.2 \times 10^{-3}\end{align*}
Oxalic Acid (\begin{align*}\mathrm{HO}_2\mathrm{CCO}_2\mathrm{H}\end{align*}) \begin{align*}K_a = 5.6 \times 10^{-2}\end{align*}
Arsenic acid (\begin{align*}\mathrm{H}_3\mathrm{AsO}_4\end{align*}) \begin{align*}K_a = 6.0 \times 10^{-3}\end{align*}

Solution:

Order of decreasing acid strength: Oxalic acid > Phosphoric acid > Arsenic acid > Formic acid

Equilibrium Equations:

1. Oxalic acid: \begin{align*}\mathrm{HO}_2\mathrm{CCO}_2\mathrm{H}_{(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{HO}_2\mathrm{CCO}^-_{2(aq)}\end{align*}
2. Phosphoric acid: \begin{align*}\mathrm{H}_3\mathrm{PO}_{4(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{H}_2\mathrm{PO}^-_{4(aq)}\end{align*}
3. Arsenic acid: \begin{align*}\mathrm{H}_3\mathrm{AsO}_{4(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{H}_2\mathrm{AsO}^-_{4(aq)}\end{align*}
4. Formic acid: \begin{align*}\mathrm{HCOOH}_{(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{HCOO}^-_{(aq)}\end{align*}

## Equilibrium Constants for Acid and Base Dissociation

The pH for solutions of strong acids and strong bases can be calculated simply by knowing the original concentration of acid or base. Consider a solution that is \begin{align*}0.010 \mathrm{M} \ \mathrm{HCl}\end{align*}. \begin{align*}\mathrm{HCl}\end{align*} is a strong acid, so the acid molecules dissociate completely. At equilibrium, this solution of \begin{align*}\mathrm{HCl}\end{align*} will be \begin{align*}0.010 \ \mathrm{M}\end{align*} in \begin{align*}\mathrm{H}^+\end{align*} ions and \begin{align*}0.010 \ \mathrm{M}\end{align*} in \begin{align*}\mathrm{Cl}^-\end{align*} ions. Plugging the value of the hydrogen ion concentration into the pH formula, we can determine that this solution has a pH of 2.

Consider a \begin{align*}0.0010 \ \mathrm{M} \ \mathrm{NaOH}\end{align*} solution. \begin{align*}\mathrm{NaOH}\end{align*} is a strong base, so this solution will be \begin{align*}0.0010 \ \mathrm{M}\end{align*} in sodium ions and also in hydroxide ions. Since the solution is \begin{align*}1.0 \times 10^{-3} \mathrm{M}\end{align*} in hydroxide ions, it will be \begin{align*}1.0 \times 10^{-11} \ \mathrm{M}\end{align*} in hydrogen ions. Therefore, this solution will have a pH = 11.

A strong acid such as \begin{align*}\mathrm{H}_2\mathrm{SO}_4\end{align*} is only slightly more complicated. Suppose we wish to determine the pH of a \begin{align*}0.00010 \ \mathrm{M}\end{align*} solution of \begin{align*}\mathrm{H}_2\mathrm{SO}_4\end{align*}.

\begin{align*}\mathrm{H}_2\mathrm{SO}_{4(aq)} \rightarrow 2 \ \mathrm{H}^+ + \mathrm{SO}_4^{2-}\end{align*}

Since the original solution was \begin{align*}0.00010 \ \mathrm{M}\end{align*} and sulfuric acid is a strong acid, then complete dissociation will produce a solution that contains \begin{align*}[\mathrm{H}^+] = 0.00020 \ \mathrm{M}\end{align*}. Substituting this hydrogen ion concentration in the pH formula yields:

\begin{align*}\mathrm{pH} = -\log (2.0 \times 10^{-4}) = -(0.30 - 4) = -(-3.7) = 3.7\end{align*}

Let's now consider the process for finding the pH of weak acids and bases. In these cases, you need more information than you need for strong acids and bases. Not only do you need to know the concentration of the original acid or base solution, but you also must know the \begin{align*}K_a\end{align*} or \begin{align*}K_b\end{align*}. Suppose we wish to know the pH of a \begin{align*}1.0 \ \mathrm{M}\end{align*} solution of ascorbic acid, \begin{align*}\mathrm{H}_2\mathrm{C}_6\mathrm{H}_6\mathrm{O}_{6(aq)}\end{align*}, whose \begin{align*}K_a = 7.9 \times 10^{-5}\end{align*}.

\begin{align*}\mathrm{H}_2\mathrm{C}_6\mathrm{H}_6\mathrm{O}_{6(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{HC}_6\mathrm{H}_6\mathrm{O}^-_{6(aq)} \ \ \ \ \ \ \ \ K_a = 7.9 \times 10^{-5}\end{align*}

The \begin{align*}K_a\end{align*} expression for this reaction would be written

\begin{align*}K_a = \frac {[\mathrm{H}^+][\mathrm{HC}_6\mathrm{H}_6\mathrm{O}_6^-]} {[\mathrm{H}_2\mathrm{C}_6\mathrm{H}_6\mathrm{O}_6]}\end{align*}

To find the hydrogen ion concentration from this \begin{align*}K_a\end{align*} expression and the original concentration of the acid, we need a little algebra. Let the molarity of the acid that has already dissociated be represented by \begin{align*}x\end{align*}. In other words, the molarity of the hydrogen ions and ascorbate ions in solution will also be represented by \begin{align*}x\end{align*}. The molarity of the undissociated acid must therefore be \begin{align*}1.0 - x\end{align*}. We can now substitute these variables into the \begin{align*}K_a\end{align*} expression and set it equal to the given \begin{align*}K_a\end{align*} value.

\begin{align*}K_a = \frac {(x)(x)} {(1.0 - x)} = 7.9 \times 10^{-5}\end{align*}

When this equation is simplified, we find that it is a quadratic equation, which, of course, can be solved by the quadratic formula.

\begin{align*}x^2 + (7.9 \times 10^{-5})x - (7.9 \times 10^{-5}) = 0\end{align*}
\begin{align*}x = 8.9 \times 10^{-3} \ \mathrm{M}\end{align*}

However, there is a shortcut available to solve this problem that simplifies the math greatly. It involves significant figures and adding or subtracting extremely small numbers from large numbers. If you are working to 3 significant figures and you are required to subtract \begin{align*}0.00005\end{align*} from \begin{align*}1.00\end{align*}, when you carry out the subtraction and round to 3 significant figures, you discover that you get the original number before you subtracted.

\begin{align*}1.00 - 0.00005 = 0.99995\end{align*}, which to 3 significant figures is \begin{align*}1.00\end{align*}.

In the problem we solved above about ascorbic acid, the \begin{align*}K_a\end{align*} value is very small, \begin{align*}1.8 \times 10^{-5}\end{align*}. This indicates that the amount of ascorbic acid that dissociates, represented by \begin{align*}x\end{align*}, is tiny. When we assigned the variables in that problem, we see that the molarity of ascorbic acid remaining after dissociation is represented by \begin{align*}1.0 - x\end{align*}. Since this \begin{align*}x\end{align*} is very tiny, the result of this subtraction will still be \begin{align*}1.0 \ \mathrm{M}\end{align*}. Therefore, the \begin{align*}K_a\end{align*} expression from above,

\begin{align*}K_a = \frac {(x)(x)} {(1.0 - x)} = 7.9 \times 10^{-5}\end{align*}

can quite safely be written as

\begin{align*}K_a = \frac {(x)(x)} {(1.0)} = 7.9 \times 10^{-5}\end{align*}

We are assuming \begin{align*}x\end{align*} is so small that it will not alter the value of \begin{align*}1.0\end{align*} when we subtract and then round to proper significant figures. This is a safe assumption when the value of the \begin{align*}K_a\end{align*} is very small (less than \begin{align*}1 \times 10^{-3}\end{align*}).

When we solve this second expression,

\begin{align*} \frac {(x)(x)} {(1.0)} = 7.9 \times 10^{-5}\end{align*}
\begin{align*}x^2 = 7.9 \times 10^{-5}\end{align*}
\begin{align*}x = 8.9 \times 10^{-3} \ \mathrm{M}\end{align*}

We no longer need to use the quadratic formula, and note that the answer (when rounded to the proper number of significant figures) is exactly the same as when the expression was solved with the quadratic formula.

Let's go through another example using a hypothetical weak acid, \begin{align*}0.10 \ \mathrm{M} \ HA\end{align*}, whose \begin{align*}K_a = 4.0 \times 10^{-7}\end{align*}.

\begin{align*}HA_{(aq)} \rightleftharpoons H^+ + A^- \ \ K_a = 4.0 \times 10^{-7}\end{align*}
\begin{align*}K_a = \frac {[H^+][A^-]} {[HA]} = 4.0 \times 10^{-7}\end{align*}

Let \begin{align*}x\end{align*} represent the molarity of \begin{align*}HA\end{align*} that dissociates, so \begin{align*}[H^+] = [A^-] = x\end{align*}, and \begin{align*}[HA] = 0.10 - x\end{align*}.

\begin{align*}K_a = \frac {[x][x]} {[0.10 - x]} = 4.0 \times 10^{-7}\end{align*}

Once again, since the \begin{align*}K_a\end{align*} value is very small, the \begin{align*}x\end{align*} subtracted in the denominator can be neglected. and the equation becomes

\begin{align*}K_a = \frac {[x][x]} {[0.10]} = 4.0 \times 10^{-7}\end{align*}
\begin{align*}x^2 = 4.0 \times 10^{-8}\end{align*}
\begin{align*}x = 2.0 \times 10^{-4} \ \mathrm{M}\end{align*}

Therefore, the hydrogen ion concentration in this solution is \begin{align*}2.0 \times 10^{-4} \ \mathrm{M}\end{align*}. Substituting this value into the pH formula yields:

\begin{align*}\mathrm{pH} = -\log (2.0 \times 10^{-4}) = -(0.3 - 4) = -(-3.7) = 3.7\end{align*}

The same process is used for weak bases. There is one additional step when working with weak bases: once the hydroxide ion concentration is determined, you must then find the hydrogen ion concentration before substituting the value into the pH formula. \begin{align*}K_b\end{align*} represents the equilibrium constant for the reduction between a weak base and water. Consider dimethylamine (a weak base used in making detergents). We will calculate the pH of a 1.0 M solution of this weak base.

\begin{align*}(\mathrm{CH}_3)_2\mathrm{NH}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons (\mathrm{CH}_3)_2\mathrm{NH}^+_{2(aq)} + \mathrm{OH}^-_{(aq)} \ \ \ K_b = 5.9 \times 10^{-7}\end{align*}
\begin{align*}K_b = \frac {[(\mathrm{CH}_3)_2\mathrm{NH}^+_2][\mathrm{OH}^-} {[(\mathrm{CH}_3)_2\mathrm{NH}} = 5.9 \times 10^{-7}\end{align*}

Allowing \begin{align*}x\end{align*} to represent the molarity of \begin{align*}(\mathrm{CH}_3)_2\mathrm{NH}\end{align*} that dissociates results in \begin{align*}[(\mathrm{CH}_3)_2\mathrm{NH}_2^+] = x\end{align*} and \begin{align*}[\mathrm{OH}^-] = x\end{align*}. The molarity of undissociated \begin{align*}(\mathrm{CH}_3)_2\mathrm{NH}\end{align*} will be \begin{align*}1.0 - x\end{align*}.

Substituting the variables into the \begin{align*}K_b\end{align*} expression yields

\begin{align*}K_b = \frac {[x][x]} {[1.0 - x]} = 5.9 \times 10^{-7}\end{align*}

and neglecting the \begin{align*}x\end{align*} in the denominator because it is beyond the significant figures of the problem yields

\begin{align*}K_b = \frac {[x][x]} {[1.0]} = 5.9 \times 10^{-7}\end{align*}

Therefore,

\begin{align*}x^2 = 5.9 \times 10^{-7}\end{align*}
\begin{align*}x = 7.7 \times 10^{-4} \ \mathrm{M}\end{align*}

Now that we know the hydroxide ion concentration in the solution, we calculate the hydrogen ion concentration by dividing the \begin{align*}[\mathrm{OH}^-]\end{align*} into the \begin{align*}K_w\end{align*}. This will yield \begin{align*}[\mathrm{H}^+] = 1.3 \times 10^{-11} \ \mathrm{M}\end{align*}. The final step is to plug the hydrogen ion concentration into the pH formula.

\begin{align*}\mathrm{pH} = -\log (1.3 \times 10^{-11}) = 10.9\end{align*}

Example:

Acetic acid is mixed with water to form a \begin{align*}0.10 \ \mathrm{mol/L} \ \mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(aq)}\end{align*} solution at \begin{align*}25^\circ\mathrm{C}\end{align*}. If the equilibrium concentrations of \begin{align*}\mathrm{H}_3\mathrm{O}^+_{(aq)}\end{align*} and \begin{align*}\mathrm{C}_2\mathrm{H}_3\mathrm{O}^-_{2(aq)}\end{align*} are both \begin{align*}1.34 \times 10^{-3} \ \mathrm{mol/L}\end{align*} and the equilibrium concentration of \begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(aq)}\end{align*} is \begin{align*}0.0999 \ \mathrm{mol/L}\end{align*}, determine the \begin{align*}K_a\end{align*} and the pH of the solution at equilibrium.

Solution:

\begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(aq)} \rightleftharpoons \mathrm{H}^+_{(aq)} + \mathrm{C}_2\mathrm{H}_3\mathrm{O}^-_{2(aq)}\end{align*}
\begin{align*}K_a~= \frac {[\mathrm{H}^+][\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-]} {[\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2]} = \frac {(1.34 \times 10^{-3})(1.34 \times 10^{-3})} {0.0999} = 1.80 \times 10^{-5}\end{align*}
\begin{align*}\text{pH} = -\log [\mathrm{H}^+] = -\log (1.34 \times 10^{-3}) = 2.87\end{align*}

This video shows an example of how to plug values into the Ka for a weak acid to find [H+] and pH (5c): http://www.youtube.com/watch?v=MOv7Z16FMK0 (3:26).

## Lesson Summary

• Strong acids and bases undergo 100% dissociation in water.
• Weak acids undergo less than 100% dissociation in water.
• It is safe to assume that an acid is weak unless it is one of the six strong acids listed in the chapter.
• \begin{align*}K_a\end{align*} represents the equilibrium constant for the dissociation of a weak acid.
• \begin{align*}K_b\end{align*} represents the equilibrium constant for the dissociation of a weak base.
• Equilibrium calculations are the same for weak acids and bases as they were for all other equilibrium reactions.

## Review Questions

1. What makes weak acids and bases a special case for equilibrium reactions?
2. What do the constants \begin{align*}K_a\end{align*} and \begin{align*}K_b\end{align*} represent?
3. Oxalic acid is a weak acid. Its ionization reaction is represented as: \begin{align*}\mathrm{H}_2\mathrm{C}_2\mathrm{O}_{4(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_3\mathrm{O}{^{+}}_{(aq)} + \mathrm{HC}_2\mathrm{O}_4{^{-}}_{(aq)}\end{align*}. Which of the following best represents the acid ionization constant expression, \begin{align*}K_a\end{align*}?
1. \begin{align*}K_a = \frac {[\mathrm{H}_3\mathrm{O}^+][\mathrm{HC}_2\mathrm{O}_4^-]} {[\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4][\mathrm{H}_2\mathrm{O}]}\end{align*}
2. \begin{align*}K_a = \frac {[\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4][\mathrm{H}_2\mathrm{O}]} {[\mathrm{H}_3\mathrm{O}^+][\mathrm{HC}_2\mathrm{O}_4^-]}\end{align*}
3. \begin{align*}K_a = \frac {[\mathrm{H}_3\mathrm{O}^+][\mathrm{HC}_2\mathrm{O}_4^-]} {[\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4]}\end{align*}
4. \begin{align*}K_a = \frac {[\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4]} {[\mathrm{H}_3\mathrm{O}^+][\mathrm{HC}_2\mathrm{O}_4^-]}\end{align*}
4. Choose the weakest acid from the list below.
1. \begin{align*}\mathrm{HNO}_{2(aq)}; K_a = 5.6 \times 10^{-3} \end{align*}
2. \begin{align*}\mathrm{HF}_{(aq)}; K_a = 6.6 \times 10^{-4} \end{align*}
3. \begin{align*}\mathrm{H}_3\mathrm{PO}_{4(aq)}; K_a = 6.9 \times 10^{-3}\end{align*}
4. \begin{align*}\mathrm{HCOOH}_{(aq)}; K_a = 1.8 \times 10^{-4}\end{align*}
5. Choose the reaction that would have an equilibrium constant represented by \begin{align*}K_b\end{align*}.
1. \begin{align*}\mathrm{H}_2\mathrm{PO}_4{^{-}}_{(aq)}+ \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{HPO}_4{^{-2}}_{(aq)} + \mathrm{H}_3\mathrm{O}{^{+}}_{(aq)}\end{align*}
2. \begin{align*}\mathrm{NH}_4{^{+}}_{(aq)}+ \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}_{3(aq)} + \mathrm{H}_3\mathrm{O}{^{+}}_{(aq)}\end{align*}
3. \begin{align*}\mathrm{NH}_4{^{-}}_{(aq)}+ \mathrm{OH}{^{-}}_{(aq)}\rightleftharpoons \mathrm{NH}_{3(aq)} + \mathrm{H}_2\mathrm{O}_{(l)}\end{align*}
4. \begin{align*}\mathrm{F}{^{-}}_{(aq)}+ \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{HF}_{(aq)}+ \mathrm{OH}{^{-}}_{(aq)}\end{align*}
6. A \begin{align*}0.150 \ \mathrm{mol/L}\end{align*} solution of a weak acid having the general formula \begin{align*}HA\end{align*} is \begin{align*}15.0\%\end{align*} ionized in aqueous solution. Which expression best represents the calculation of the acid ionization constant \begin{align*}K_a\end{align*} for this acid?
1. \begin{align*}K_a = \frac {(0.150)(0.150)} {(0.150)}\end{align*}
2. \begin{align*}K_a = \frac {(0.0225)(0.0225)} {(0.128)}\end{align*}
3. Not enough information is given.
7. Put the following bases in order of increasing base strength. Write equilibrium reactions for each.
1. ethanolamine \begin{align*}(\mathrm{HOCH}_2\mathrm{CH}_2\mathrm{NH}_2), \ K_b = 3.2 \times 10^{-5}\end{align*}
2. piperidine \begin{align*}(\mathrm{C}_5\mathrm{H}_{10}\mathrm{NH}), \ K_b = 1.3 \times 10^{-3}\end{align*}
3. triethylamine \begin{align*}((\mathrm{CH}_3\mathrm{CH}_2)_3\mathrm{N}), \ K_b = 5.2 \times 10^{-4}\end{align*}
4. ethylenediamine \begin{align*}(\mathrm{H}_2\mathrm{NCH}_2\mathrm{CH}_2\mathrm{NH}_2), \ K_b = 8.5 \times 10^{-5}\end{align*}

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