Chapter 14 Source Answers and Examples

Reviewed solution to Problem 2

No. K_h specifies an equilibrium ratio; kinetic rate constants describe the time dependence. The original label “rate constant pK_h” conflates the two. Full coupled Henry/hydration calculations are retained enrichment.

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Reviewed solution to Problem 4

x=10⁻³·²²=0.00060256 M. Kₐ=x²/(0.100−x)=3.65279×10⁻⁶ and pKₐ=5.43738≈5.44. Supplying analytical concentration closes the original missing-data gap.

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Reviewed solution to Problem 8

Solve Kₐ=x²/(C−x): x=(−Kₐ+√(Kₐ²+4KₐC))/2=2.05226×10⁻⁵ M and pH≈4.688. The original CO₂ pressure and coupled gas/hydration model are not assumed.

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Reviewed solution to Problem 13

[OH⁻]=0.0100 M, so pOH=2.00 and pH=14.00−2.00=12.00. This complete concentration problem replaces the original missing initial amount and unvalidated slow-dissociation model.

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Reviewed solution to Problem 16

n(HCl)=(10⁻²·⁰⁰)(5.00)=0.050 mol. n(total)=PV/RT=0.817485 mol. The unrounded mole/ideal volume fraction is 0.0611635; report about 6.1% at the two significant figures set by pH 2.00. The revised final volume, temperature and pressure close the original data gaps.

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Reviewed solution to Problem 19

HCl: pH=−log₁₀4=−0.602060, pOH=14.602060. NaOH: pOH=−log₁₀5=−0.698970, pH=14.698970. The NaOH pH is higher by 0.096910. Real concentrated solutions require activity treatment.

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Worked examples and reviewed revisions

Example Problem 6.1.1

Original example and steps

Example Problem 6.1.1
The Equilibrium constant for the deprotonation of water is $K = 10^{-14}$ . Use this information to determine the pH of pure water. The equation for the deprotonation of water is:
$H_2O\leftrightarrow H^++OH^-$

Step 1: Write the Equilibrium Equation.

$K = [H^+][OH^-]$

Step 2: Set up the ICE table.

We are going to start with pure water, without any $H^+\text{ or } OH^-$ in it.

$\begin{matrix} \bold{Condition}&\bold{[H^+]}&\bold{[OH^-]}\\ Initial&0&0\\ Change&+x&+x\\ Final&x&x \end{matrix}$

Step 3: Plug in the values to the Equilibrium Equation.

$K = [H^+][OH^-]$
$K = x\cdot x$
$K = x^2$

Step 4: Determine x, and the final concentration.

$K =x^2$
$10^{-14} = x^2$
$x = \sqrt{10^{-14}}$
$x = 10^{-7}$

$[H^+] =10^{-7}M$

Step 5: Calculate pH.

$[H^+] = 10^{-7}M$
$pH = -log\{10^{-7}\}$
$pH = 7$

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Sample Problem 6.3.1

Original example and steps

Sample Problem 6.3.1
What is the pH of a 3M solution of Acetic Acid?
$CH_3COOH\leftrightarrow CH_3COO^-+H^+$ $K=10^{-4.76}$

Step 1: Draw up the ICE table.

$\begin{matrix} \bold{Condition}&\bold{CH_3COOH}&\bold{CH_3COO^-}&\bold{H^+}\\ Initial&3&0&10^{-7}\\ Change&-x&+x&+x\\ Final&3-x&x&10^{-7}+x \end{matrix}$

We have a starting concentration for $H^+$ because we are adding our acetic acid into water.

Step 2: Determine Equilibrium Equation.

$K = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

$K = \frac{x\cdot (10^{-7}+x)}{3-x}$

Step 3: Solve for x.

$10^{-4.76}=\frac {x\cdot(10^{-7}+x)}{3-x}$

$x=0.00721$

Step 4: Determine $[H^+]$

$[H^+]=x$

$[H^+]=0.00721$

Step 5: Determine pH

$pH = -log(0.00721)$

$pH = 2.14$

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Sample Problem 6.3.2

Original example and steps

Sample Problem 6.3.2
A solution of Hydrofluoric Acid ( $pK_a=3.17$ ) has a pH of 2. What is the concentration of the acid?

Step 1: Write the dissociation equation.

$HF\leftrightarrow H^++F^-$

Step 2: Plug in known values to the Henderson-Hasselbalch Equation.

$pH = pK_a+\log\frac{[F^-]}{[HF]}$

$2 = 3.17+\log\frac{[F^-]}{[HF]}$

Step 3: Write the ICE table.

We know that the HF dissociates according to the previous equation. This lets us write our ICE table:

$\begin{matrix} \bold{Condition}&\bold{[HF]}&\bold{[H^+]}&\bold{[F^-]}\\ Initial&X&10^{-7}&0\\ Change&-x&+x&+x\\ Final&X-x&10^{-7}+x&x \end{matrix}$

This looks unsolvable until we realize that we have the pH. This gives us the final value of $[H^+]$ .

Step 4: Solve for x.

$pH = 2$

$-\log[H^+]=2$

$[H^+]=0.01$

$0.01=10^{-7}+x$

$x\approx0.01$

Due to the huge difference between $0.01$ and $10^{-7}$ , we can ignore the $10^{-7}$ .

Step 5: Re-write Henderson-Hasselbalch Equation.

$2=3.17+\log\frac{[F^-]}{[HF]}$

$2=3.17+\log\frac{0.01}{X-0.01}$

Step 6: Solve for X

$\log\frac{0.01}{X-0.01}=2-3.17$

$\frac{0.01}{X-0.01}=10^{-1.17}$

$X=0.158M$

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Example Problem 6.3.3

Original example and steps

Example Problem 6.3.3
Is Ammonia or Ammonium more predominant in a solution if the pH is 10? $pK_a=9.25$ for Ammonium.

Step 1: Determine the dissociation equation

Ammonia is $NH_3$ , and ammonium is $NH_4^+$ . Therefore the dissociation equation is

$NH_4^+\leftrightarrow NH_3+H^+$

Step 2: Write our the Henderson-Hasselbalch equation.

$pH=pK_a+\log\frac{[NH_3]}{[NH_4^+]}$

$10 = 9.25 + \log\frac{[NH_3]}{[NH_4^+]}$

Now we could go through the lengthy calculations to determine the exact concentrations, or we could just look at the ratio of $[NH_3]\text{ to }[NH_4^+]$ .

Step 3: Determine the dominant species.

$10 = 9.25 + \log\frac{[NH_3]}{[NH_4^+]}$
$\log\frac{[NH_3]}{[NH_4^+]}=0.75$
$\frac{[NH_3]}{[NH_4^+]}=5.62$

As the ratio is greater than one, ammonia is the dominant molecule in the solution.

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Example Problem 6.3.4

Worked example

Example Problem 6.3.4 — AP edition reviewed replacement.

Reviewed ideal 25 °C model: a phosphoric-acid-only solution has pH=1.20. Use pKₐ₁=2.12, pKₐ₂=7.21 and pKₐ₃=12.32, K_w=10⁻¹⁴ and no additional ions. Find total phosphate-family concentration C_T. Activities matter in a real concentrated solution.

Let h=[H⁺]=10⁻¹·²=0.06309573445 M and c₀,c₁,c₂,c₃ denote H₃PO₄,H₂PO₄⁻,HPO₄²⁻,PO₄³⁻. The successive equilibria give c₁=K₁c₀/h, c₂=K₁K₂c₀/h² and c₃=K₁K₂K₃c₀/h³.

Mass balance: C_T=c₀+c₁+c₂+c₃. Charge balance: h−K_w/h=c₁+2c₂+3c₃. These equations avoid treating later proton releases as independent extra acid amounts.

Substitute the equilibrium ratios into charge balance. c₀=(h−K_w/h)/(K₁/h+2K₁K₂/h²+3K₁K₂K₃/h³)=0.524806435 M.

The species concentrations are c₁=0.0630956111 M, c₂=6.16593797×10⁻⁸ M and c₃=4.67734227×10⁻¹⁹ M. The very small c₂ and c₃ reflect the correct second and third dissociation constants at this low pH.

Adding the four species gives C_T=0.587902107 M≈0.588 M. Check: c₁+2c₂+3c₃=0.063095734448 M=h−K_w/h. The original 0.5257 M answer and its intermediate substitutions are superseded.

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Example Problem 6.3.5

Worked example

Example Problem 6.3.5 — AP edition reviewed replacement.

Reviewed leveling-effect task: why do 0.0100 M HCl and 0.0100 M HBr have approximately the same pH in the ideal dilute aqueous model at 25 °C?

Each strong acid transfers its proton essentially completely to water. HCl+H₂O→H₃O⁺+Cl⁻; HBr+H₂O→H₃O⁺+Br⁻.

For either specified solution, [H₃O⁺]≈0.0100 M and pH≈−log₁₀(0.0100)=2.00. H⁺(aq) denotes the hydrated proton.

No independent “pure hydronium dissociation percentage” or universal 98.2% strong-acid threshold is calculated by this model.

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Example 6.4.1

Original example and steps

Example 6.4.1
What is the pH of a 3M solution of NaOH?

Step 1: Determine concentration of $[OH^-]$ .

We have a 3M solution of NaOH. Sodium is an Alkali metal, therefore we can treat this as a strong base, and assume complete dissociation. Therefore, $[OH^-]=3M$ .

Step 2: Determine $pH$ .

There are two ways of doing this. We can start by finding $pOH$ and using the equation $pH + pOH=14$ . We can also use $K_w=[H^+][OH^-]=10^{-14}$ . These are practically the same approach, but it might be easier to use the $pOH$ equation, since we are more interested in $pH$ than the direct concentration of $[H^+]$ .

$pOH=-\log [OH^-]$

$pOH = -\log3$

$pOH=-0.477$

$pH + pOH = 14$

$pH = 14+0.477$

$pH = 14.477$

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Example 6.4.2

Original example and steps

Example 6.4.2
What is the pH of a 5M solution of Ammonia ( $pK_b=4.75$ ).

Step 1: Write out the base equation.

This problem involves the base reaction of ammonia. Therefore, we look to Ammonia reacting with water:

$NH_{3(aq)}+H_2O_{(l)}\leftrightarrow NH_{4(aq)}^++OH^-_{(aq)}$

Step 2: Create an ICE table.

$\begin{matrix} \bold{Condition}&\bold{[NH_3]}&\bold{NH_4^+]}&\bold{[OH^-]}\\ Initial&5&0&0\\ Change&-x&+x&+x\\ Equilibrium&5-x&x&x \end{matrix}$

Step 3: Set up the equilibrium equation.

$K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}$

$10^{-4.75}=\frac{x\cdot x}{5-x}$

Step 4: Solve for x.

We can assume that there is little dissociation, which would bean that $5-x\approx5$ .

This simplifies it to:

$10^{-4.75}=\frac{x^2}5$

$x=\sqrt{5\cdot10^{-4.75}}$

$x=0.00943$

It looks like our assumption was good. Let’s see just how close we were. To do so, plug in $x$ to see what you get for $pK_b$ .

$K_b=\frac {0.00943^2}{5-0.00943}$

$pK_b=-\log\frac{0.00943^2}{5-0.00943}$

$pK_b=4.749$

Looks like we were pretty darn close. Let’s stick with this value for $x$ , then.

Step 5: Fill in the ICE table again.

$\begin{matrix} \bold{Condition}&\bold{[NH_3]}&\bold{NH_4^+]}&\bold{[OH^-]}\\ Initial&5&0&0\\ Change&-x&+x&+x\\ Equilibrium&5-x&x&x \end{matrix}$

We know that $x=0.00943$ , so:

$\begin{matrix} \bold{Condition}&\bold{[NH_3]}&\bold{NH_4^+]}&\bold{[OH^-]}\\ Initial&5&0&0\\ Change&-0.00943&+0.00943&+0.00943\\ Equilibrium&5&0.00943&0.00943 \end{matrix}$

Notice we stuck with the approximation of $5$ for $[NH_3]$ at equilibrium.

Step 6: Using $[OH^-]$ , solve for $pH$ .

$pOH = -\log[OH^-]$

$pOH = -\log 0.00943$

$pOH=2.02$

$pH+pOH=14$

$pH = 14-2.02$

$pH = 11.98$

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Reviewed solution to Problem 2

Reviewed solution to Problem 4

Reviewed solution to Problem 8

Reviewed solution to Problem 13

Reviewed solution to Problem 16

Reviewed solution to Problem 19

Worked examples and reviewed revisions

Example Problem 6.1.1

Sample Problem 6.3.1

Sample Problem 6.3.2

Example Problem 6.3.3

Example Problem 6.3.4

Example Problem 6.3.5

Example 6.4.1

Example 6.4.2

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