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21.4: Multimedia Resources for Chapter 21

Difficulty Level: At Grade Created by: CK-12
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Copy and distribute the lesson worksheets. Ask students to complete the worksheets alone or in pairs as a review of lesson content.

The Mathematics of Salt Hydrolysis Worksheet

Part I

Reactions between acids and bases produce water and a salt.

\begin{align*}HNO_3~+~NH_4OH~ \rightarrow ~ H_2O~+~NH_4NO_3\end{align*}

In the reaction above, nitric reacts with ammonium hydroxide to produce water and ammonium nitrate. Ammonium nitrate is the salt in this reaction.

When salts are dissolved in water, the resulting solution could be neutral but it is also possible for the solution to be acidic or basic. Recall that the conjugate base of a strong acid, the anion in the acid, has virtually no affinity for hydrogen ions in water. That is why strong acids completely dissociate in aqueous solutions. Similarly, the cations of strong bases have no affinity for hydroxide ions in water.

The anions of weak acids, on the other hand, have significant affinity for hydrogen ions. Consider what happens when the salt sodium acetate is dissolved in water. The water, before the salt is added, has equal molarity of hydrogen and hydroxide ions. When the salt dissolves, the sodium ions (cation of strong base) will not attach to hydroxide ions but some of the acetate ion (anion of weak acid) will attach to hydrogen ions in the solution. When some of the hydrogen ions are thus removed from the solution, the resulting solution will have a higher molarity of hydroxide ions than hydrogen ions and the solution will be basic.

In the same way, when salts containing cations of weak bases and anions of strong acids are dissolved in water, the some of the cations of the weak base will attach to hydroxide ions. When hydroxide ions are removed from the solution, the solution will have excess hydrogen ions and the solution will be acidic.

The table below shows the strong acids and strong bases. All other acids and bases are weak.

Strong Acids and Bases
Strong Acid Formula Strong Base Formula
Hydrochloric Acid \begin{align*}HCl\end{align*} Lithium Hydroxide \begin{align*}LiOH\end{align*}
Hydrobromic Acid \begin{align*}HBr\end{align*} Sodium Hydroxide \begin{align*}NaOH\end{align*}
Hydroiodic Acid \begin{align*}HI\end{align*} Potassium Hydroxide \begin{align*}KOH\end{align*}
Nitric Acid \begin{align*}HNO_3\end{align*} Rubidium Hydroxide \begin{align*}RbOH\end{align*}
Perchloric Acid \begin{align*}HClO_4\end{align*} Cesium Hydroxide \begin{align*}CsOH\end{align*}
Sulfuric Acid \begin{align*}H_2SO_4\end{align*} Calcium Hydroxide \begin{align*}Ca(OH)_2\end{align*}
Strontium Hydroxide \begin{align*}Sr(OH)_2\end{align*}
Barium Hydroxide \begin{align*}Ba(OH)_2\end{align*}

The table below shows the acid or base properties of various salt solutions.

Acidity of Salt Solutions
Parent Acid Strength Parent Base Strength Acidity/Basicity of Salt Solution
Strong Strong Neutral
Strong Weak Acidic
Weak Strong Basic
Weak Weak Depends on \begin{align*}K_a\end{align*} and \begin{align*}K_b\end{align*}

The following table shows the \begin{align*}K_a\end{align*} and \begin{align*}K_b\end{align*} values for various weak acids and bases.

Acid-Base Dissociation Constants
Weak Acid \begin{align*}K_a\end{align*} Weak Base \begin{align*}K_b\end{align*}
\begin{align*}HC_2H_3O_2\end{align*} \begin{align*}1.8~x~10^{-5}\end{align*} \begin{align*}CH_3NH_3OH\end{align*} \begin{align*}4.4~x~10^{-4}\end{align*}
\begin{align*}HCN\end{align*} \begin{align*}6.2~x~10^{-10}\end{align*} \begin{align*}NH_4OH\end{align*} \begin{align*}1.8~x~10^{-5}\end{align*}

Consider a water solution of the salt ammonium cyanide, \begin{align*}NH_4CN\end{align*}. This is a salt of a weak acid and a weak base. The acidity-basicity of the salt solution will depend on the relative strength of the parent acid and base. The \begin{align*}K_a\end{align*} for \begin{align*}HCN\end{align*} is \begin{align*}6.2~x~10^{-10}\end{align*}. The \begin{align*}K_b\end{align*} for \begin{align*}NH_4OH\end{align*} is \begin{align*}1.8~x~10^{-5}\end{align*}. Since the \begin{align*}K_a\end{align*} is smaller than the \begin{align*}K_b\end{align*}, the cyanide ion has greater attraction for hydrogen ions than the ammonium ion has for hydroxide ions. Therefore, more hydrogen ions will be removed from the solution than hydroxide ions and the hydroxide ions will be in excess. This solution will be basic.

Part I Exercises

For the following salts, identify the parent acid and strength, the parent base and strength, and indicate whether the salt solution will be acidic or basic. Complete the table.

Part II

If we know the molarity of the salt solution and the necessary \begin{align*}K_a\end{align*} and/or \begin{align*}K_b\end{align*} values, we can calculate the pH of the salt solution.

Example 1

What is the pH of a \begin{align*}0.100~ M\end{align*} solution of ammonium nitrate, \begin{align*}NH_4NO_3\end{align*}?

Solution

\begin{align*}NH_4NO_3\end{align*} is the salt product of the reaction between a strong acid, \begin{align*}HNO_3\end{align*}, and a weak base, \begin{align*}NH_4OH\end{align*}. Therefore, this salt solution will be acidic. The actual pH can be calculated as shown below.

The ammonium ion in the solution acts as an acid by removing hydroxide ions from solution and thus produces hydrogen ions.

\begin{align*}NH_4^+~+~H_2O~ \leftrightarrows ~NH_4OH~+~H^+\end{align*}

Since the ammonium ion is acting as an acid in this reaction, the equilibrium constant for the reaction is a \begin{align*}K_a\end{align*}.

\begin{align*}K_a~=~ \frac {[NH_4OH][H^+]} {[NH_4^+]}\end{align*}

The \begin{align*}K_a\end{align*} for ammonium ion acting as an acid is calculated by dividing the \begin{align*}K_w\end{align*} by the \begin{align*}K_b\end{align*} for ammonium hydroxide.

\begin{align*}K_a~ \text{for}~NH_4^+~=~ \frac {K_w} {K_b}~=~ \frac {1.0~x~10^{-14}} {1.8~x~10^{-5}}~=~5.6~x~10^{-10}\end{align*}

Therefore,

\begin{align*}K_a~=~ \frac {[NH_4OH][H^+]} {[NH_4^+]}~=~5.6~x~10^{-10}\end{align*}

Since the ammonium nitrate dissociates 100%, the initial concentration of the of the ammonium ion is \begin{align*}0.100~M\end{align*}. As usual, we let \begin{align*}x\end{align*} represent the molarity of the ammonium hydroxide formed and therefore, \begin{align*}x\end{align*} will also represent the molarity of the hydrogen ion formed. The molarity of the ammonium ion at equilibrium will \begin{align*}0.100~M~-~x\end{align*} but the \begin{align*}x\end{align*} in this expression is much smaller than \begin{align*}0.100~M\end{align*} and can be neglected.

Therefore,

\begin{align*}5.6~x~10^{-10}~=~ \frac {(x)(x)} {(0.100}\end{align*}

And

\begin{align*}[H^+]~=~x~=~7.5~x~10^{-5}~ \text{and}~pH~=~4.1\end{align*}

You should always check to make sure that your original determination of whether the solution will be acidic or basic is in agreement with the pH you calculate. In this case, we decided that the solution would be acidic and the pH of 4.1 is in agreement with the solution being acidic.

Example 2

What is the pH of a </math>1.00~M</math> solution of sodium cyanide, \begin{align*}NaCN\end{align*}?

Solution

Sodium cyanide is the salt of a weak acid and strong base. The cyanide ion in solution will act as a base, attaching to hydrogen ions and leaving an excess of hydroxide ions in the solution. The solution will be basic.

\begin{align*}CN^-~+~H_2O~ \leftrightarrows ~HCN~+~OH^-\end{align*}

The \begin{align*}K_b\end{align*} for cyanide ion can be calculated by dividing the \begin{align*}K_w\end{align*} by the \begin{align*}K_a\end{align*} for hydrocyanic acid.

\begin{align*}K_b~=~\frac {K_w} {K_b}~=~ \frac {1.0~x~10^{-14}} {6.2~x~10^{-10}}~=~1.6~x~10^{-5}\end{align*}
\begin{align*}K_b~=~1.6~x~10^{-5}~=~ \frac {[HCN][OH^-]} {[CN^-]}~=~ \frac {(x)(x)} {1.00}\end{align*}
\begin{align*}[OH^-]~=~x~=~4.0~x~10^{-3}~M\end{align*}
\begin{align*}pOH~=~2.4 \text{and~therefore}~pH~=~11.6\end{align*}

We originally decided this solution would be basic and the calculated pH verifies that it is basic.

Exercises Part II

1. What is the pH of an \begin{align*}0.200~M\end{align*} solution of potassium acetate, \begin{align*}KC_2H_3O_2\end{align*}?

2. What is the pH of a \begin{align*}0.0100~M\end{align*} solution of ammonium bromide, \begin{align*}NH_4Br\end{align*}?

Titration Worksheet

CK12 Chemistry

Name _____________________________ Date ____________

Show ALL your work in solving these problems.

1. What is the concentration of hydroiodic acid if \begin{align*}59.69~mL\end{align*} of it is neutralized by \begin{align*}40.39~mL\end{align*} of \begin{align*}0.2968~M\end{align*} lithium hydroxide solution?

2. It requires \begin{align*}65.95~mL\end{align*} of potassium hydroxide solution to neutralize \begin{align*}89.29~mL\end{align*} of \begin{align*}0.2118~M\end{align*} nitric acid. What is the concentration of the potassium hydroxide solution?

3. What is the molarity of a \begin{align*}CuOH\end{align*} solution if \begin{align*}50.50~mL\end{align*} of the solution is titrated to the endpoint with \begin{align*}51.99~mL\end{align*} of \begin{align*}0.3574~M\end{align*} carbonic acid, \begin{align*}H_2CO_3\end{align*}?

4. What volume of \begin{align*}0.8351~M~Fe(OH)_2\end{align*} is needed to neutralize \begin{align*}98.35~mL\end{align*} of \begin{align*}0.5417~M~H_3PO_4\end{align*}

5. \begin{align*}5.000~g\end{align*} of an unknown monoprotic solid acid was titrated with \begin{align*}163.0~mL\end{align*} of \begin{align*}0.1500~M~NaOH\end{align*} solution. What is the molar mass of the acid?

6. A solution of \begin{align*}NaOH\end{align*} was standardized against oxalic acid dehydrate, \begin{align*}H_2C_2O_4 \cdot 2H_2O\end{align*} (molar mass = \begin{align*}126.066~g/mol\end{align*}). The volume of \begin{align*}NaOH\end{align*} solution required to neutralize \begin{align*}1.500~g\end{align*} of oxalic acid was \begin{align*}45.00~mL\end{align*}. Oxalic acid is a diprotic acid. The standardized \begin{align*}NaOH\end{align*} solution is then used to titrate \begin{align*}3.500~g\end{align*} of an unknown solid monoprotic acid, \begin{align*}HA\end{align*}. The titration required \begin{align*}42.00~mL\end{align*} of the \begin{align*}NaOH\end{align*} solution.

A. Calculate the molarity of the \begin{align*}NaOH\end{align*} solution.
B. Calculate the moles of \begin{align*}HA\end{align*} in the \begin{align*}3.500~g\end{align*} sample.
C. Calculate the molar mass of the unknown acid.

Buffers Worksheet

A buffer solution is one which maintains an approximately constant pH when small amounts of either a strong acid or base are added. Solutions containing either weak acids and their conjugate bases, or weak bases and their conjugate acids, can be buffering solutions. A common method of producing a buffer is to make solution of a weak acid and a salt of that weak acid.

An example of a buffer solution is a solution containing both acetic acid, \begin{align*}HC_2H_3O_2\end{align*}, and sodium acetate, \begin{align*}NaC_2H_3O_2\end{align*}. The weak acid is acetic acid and the conjugate base is the acetate ion from sodium acetate.

  • The number of moles of the conjugate pairs must be large compared to the moles of added strong acids or bases.
  • The ratio \begin{align*} \frac {[weak~acid]} {[conjugate~base]}\end{align*} or \begin{align*} \frac {[weak~base]} {[conjugate~acid]}\end{align*} should lie between 0.1 and 10, with optimal buffering of either strong acid or strong base at a 1:1 ratio.
  • The pH of a 1:1 ratio buffer is equal to the \begin{align*}pK_a\end{align*} of the weak acid or the \begin{align*}pK_b\end{align*} of the weak base. In practical buffers, the ratio of [acid] or [base] to [conjugate] is close to 1:1 and therefore the pH of the buffer is close to the \begin{align*}pK_a\end{align*} or \begin{align*}pK_b\end{align*}.

Possible Buffers

Weak acid = \begin{align*}HC_2H_3O_2\end{align*} Conjugate base = \begin{align*}C_2H_3O_2^-\end{align*} \begin{align*}K_a~=~1.8~x~10^{-5}\end{align*}

Weak base = \begin{align*}NH_4OH\end{align*} Conjugate acid = \begin{align*}NH_4^+\end{align*} \begin{align*}K_b~=~1.8~x~10^{-5}\end{align*}

Weak acid = \begin{align*}HCO_3^-\end{align*} Conjugate base = \begin{align*}CO_3^{2-}\end{align*} \begin{align*}K_a~=~5.6~x~10^{-11}\end{align*}

Weak acid = \begin{align*}H_2PO_4^-\end{align*} Conjugate base = \begin{align*}HPO_4^{2-}\end{align*} \begin{align*}K_a~=~6.2~x~10^{-8}\end{align*}

Let’s do some general calculations with a theoretical weak acid.

We will use \begin{align*}HA\end{align*} as the weak acid and write the dissolving equation and \begin{align*}K_a\end{align*} expression for this weak acid.

\begin{align*}HA~ \leftrightarrows ~H^+~~+~A^-\end{align*}
\begin{align*}K_a~=~ \frac {[H^+][A^-]} {[HA]}\end{align*}

Let’s assume the \begin{align*}K_a\end{align*} for this weak acid is \begin{align*}1.00~x~10^{-4}\end{align*} and assume we are creating this buffer solution by pouring together 100. mL of \begin{align*}0.200~M~HA\end{align*} solution and 100. mL of \begin{align*}0.200~M~NaA\end{align*} solution. One of the concepts you must keep track of is the fact that as soon as the two solutions are poured together, the concentrations of the substances are cut in half due to dilution.

The initial concentrations in the solution (before any dissolving or reaction takes place) will be \begin{align*}[HA]~=~0.100 ~M\end{align*}, \begin{align*}[H^+]~=~0~ M\end{align*}, and \begin{align*}[A^-]~=~0.100~M\end{align*}.

Some amount of the \begin{align*}[HA]\end{align*} will dissolve and produce \begin{align*}[H^+]\end{align*} and \begin{align*}[A^-]\end{align*}. Is we assign the variable \begin{align*}x\end{align*} to represent the molarity of \begin{align*}[HA]\end{align*} that dissolves, then we can represent the concentrations as follows; \begin{align*}[HA]~=~0.100~-~x\end{align*}, \begin{align*}[H^+]~=~x\end{align*}, and \begin{align*}[A^-]~=~0.100~+~x\end{align*}.

We can then substitute these quantities into the \begin{align*}K_a\end{align*} expression.

\begin{align*}K_a~=~1.00~x~10^{-4}~=~ \frac {[H^+][A^-]} {[HA]}~ =~ \frac {(x)(0.100~-~x} {0.100~+~x}\end{align*}

At this point we can make a very simplifying assumption. Weak acids do not dissociate much, perhaps one in one thousand or one in ten thousand, or even one in one hundred thousand. Therefore, the size of the value of \begin{align*}x\end{align*} is extremely small compared to \begin{align*}0.100\end{align*}. The \begin{align*}x\end{align*} when added to or subtracted from a much larger number can be neglected. Therefore, the expression above becomes (when the added and subtracted </math>x’s</math> are dropped:

\begin{align*}K_a~=~1.00~x~10^{-4}~=~ \frac {[H^+][A^-]} {[HA]}~ =~ \frac {(x)(0.100} {0.100}\end{align*}

When we solve this equation for \begin{align*}[H^+]\end{align*}, we get

\begin{align*}[H^+]~=~ \frac {(1.00~x~10^{-4})(0.100} {0.100}\end{align*}

You can easily see that the \begin{align*}[H^+]~=~ 1.00~x~10^{-4}\end{align*} and the pH = 4.

You can also see in this calculation why the mathematics is greatly simplified when the weak acid concentration and conjugate base concentration are equal.

In the equation solved for \begin{align*}[H^+]\end{align*} above, the \begin{align*}1.00~x~10^{-4}\end{align*} is the \begin{align*}K_a\end{align*}, the \begin{align*}0.100\end{align*} in the numerator is \begin{align*}[HA]\end{align*} and the \begin{align*}0.100\end{align*} in the denominator is the \begin{align*}[A^-]\end{align*}.

This provides you with an equation, called the Henderson-Hasselbalch equation, which allows you to calculate the pH of a buffer from the \begin{align*}K_a\end{align*} and the concentrations of the weak acid and conjugate base.

\begin{align*}H^+~=~(K_a)( \frac {[HA]} {[A^-]})\end{align*}

Taking the –log of both sides of the equation produces the Henderson-Hasselbalch equation.
\begin{align*}pH~=~ pK_a~+~\text{log} \frac {[HA]} {[A^-]}\end{align*}

Example Exercises

1. Use the Henderson-Hasselbalch equation to calculate the pH of a buffer produced by mixing 300. mL of \begin{align*}1.00~M~HC_2H_3O_2\end{align*} and 100. mL of \begin{align*}1.00~M~NaC_2H_3O_2\end{align*}. \begin{align*}K_a~=~1.8~x~10^{-5}\end{align*}

Solution:

Using the dilution equation to calculate the new molarities of the weak acid and conjugate base:

\begin{align*}M_N~x~V_N~=~M_O~x~V_O\end{align*}

\begin{align*}M_N~= \frac { M_O~x~V_O } { V_N }\end{align*}

\begin{align*}[ HC_2H_3O_2]_N~=~ \frac { (1.00~M)(300.~mL } { (400.~mL }~=~ 0.75~M\end{align*}

\begin{align*}[ NaC_2H_3O_2]_N~=~ \frac { (1.00~M)(100.~mL } { (400.~mL }~=~ 0.25~M\end{align*}

Plugging the \begin{align*}K_a\end{align*} and the two concentrations into the Henderson-Hasselbalch equation:

\begin{align*}pH~=~-log (1.8~x~10^{-5}~+~ \text{log} \frac {0.75} {0.25}~=~4.74~=~0.47~=~5.21\end{align*}

2. On page 1 of this worksheet, there is a list of four possible buffers. Which of the possible buffers listed above should be chosen to prepare a buffer of pH = 7?

Solution:

Since it is easiest to prepare a buffer where the pH of the buffer is close to the \begin{align*}pK_a\end{align*} or \begin{align*}pK_b\end{align*}, a buffer with a pH of 7 should be prepared from with \begin{align*} \frac {[H_2PO_4^-]} {[HPO_4^{2-}]}\end{align*}.

3. What is the pH of a buffer formed from 50.0 mL of \begin{align*}15.0~M~NH_4OH\end{align*} and 53.5 g of \begin{align*}NH_4Cl\end{align*} in enough water to make 500. mL of solution?

Solution:

Step 1: Determine the moles of the \begin{align*}NH_4^+\end{align*} ion.

\begin{align*}mols~NH_4^+~=~ \frac {g} {mm}~=~ \frac {53.5~g} {53.5~g/mol}~=~1.00~mol\end{align*}

Step 2: Determine the molarity of the \begin{align*}NH_4^+\end{align*} ion.

\begin{align*}[NH_4^+]~=~ \frac {mol} {L}~=~ \frac {1.00~mol} {0.500~L}~=~2.00~M\end{align*}

Step 2: Determine the molarity of the \begin{align*}NH_4OH\end{align*} (using dilution formula).

\begin{align*}[NH_4OH]_N~=~ \frac {(15.0~M)(0.050~L} {0.500~L}~=~1.5~M\end{align*}

Step 3: Find the \begin{align*}K_b\end{align*} for \begin{align*}[NH_4OH]\end{align*}.

\begin{align*}K_b\end{align*} for \begin{align*}[NH_4OH]~=~ 1.8~x~10^{-5}\end{align*}

Step 4: Plug the data in the Henderson-Hasselbalch equation (for base).

\begin{align*}pOH~=~pK_b~-log \frac {[NH_4OH]} {[NH_4^+]}\end{align*}
\begin{align*}pOH~=~-log (1.8~x~10^{-5})~+~-log ( \frac {1.5~M} {2.00~M})~=~4.74~+~0.12~=~4.86\end{align*}
\begin{align*}pH~=~14.00~-~4.86~=~9.14\end{align*}

Problems

1. Calculate the pH of 0.100 M \begin{align*}NH_4OH\end{align*}.

2. Calculate the pH of 40.0 mL of 0.100 M \begin{align*}NH_4OH\end{align*} that has been diluted with 20.0 mL of water.

3. Calculate the pH of 40.0 mL of 0.100 M \begin{align*}NH_4OH\end{align*} that has been mixed with 20.0 mL of 0.200 M \begin{align*}NH_4Cl\end{align*}.

4. A buffer solution contains \begin{align*}0.384 M HCO_3^-~~\text{and}~~0.239~M~CO_3^{2-}\end{align*}. What is the pH of this buffer? \begin{align*}K_a~=~5.6~x~10^{-11}\end{align*}

5. Calculate the pH of a solution that is 0.120 M in acetic acid and 0.210 M in sodium acetate. Use \begin{align*} 1.76~x~10^{-5}\end{align*} as the \begin{align*}K_a\end{align*} for acetic acid.

Answers to Worksheets

  • The worksheet answer keys are available upon request. Please send an email to teachers-requests@ck12.org to request the worksheet answer keys.

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