Problem 42 - Chemistry Matter and Change

Question

Calculate the pH of the following solutions. \(\begin{array}{ll}{\text { a. } 1.0 \mathrm{MHI}} & {\text { c. } 1.0 \mathrm{MKOH}} \\ {\text { b. } 0.050 \mathrm{MHNO}_{3}} & {\text { d. } 2.4 \times 10^{-5} \mathrm{MMg}(\mathrm{OH})_{2}}\end{array}\)

Step-by-Step Solution

Answer

The pH values for the given solutions are: a) 1.0 M HI has pH = 0, b) 0.050 M HNO3 has pH ≈ 1.3, c) 1.0 M KOH has pH = 14, and d) 2.4 x 10^-5 M Mg(OH)2 has pH ≈ 9.7.

Step 1: Identify the chemicals and reaction equations

In order to calculate the pH, we will first need to identify the corresponding acids and bases used in the given solutions. a) 1.0 M HI - This is a strong acid that dissociates completely in water: \[HI \rightarrow H^+ + I^-\] b) 0.050 M HNO3 - This is also a strong acid that dissociates completely in water: \[HNO_3\rightarrow H^+ + NO_3^-\] c) 1.0 M KOH - This is a strong base, which dissociates completely in water: \[KOH \rightarrow K^+ + OH^-\] d) 2.4 x 10^-5 M Mg(OH)2 - This is a weak base, dissociation equation shows that 2 moles of OH- are produced per molecule of Mg(OH)2, so that needs to be considered: \[Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^-\]

Step 2: Calculate [H+] or [OH-] for each solution

Using the dissociation equations, we can now determine the concentration of H+ or OH- ions present in each solution. a) 1.0 M HI -> [H+] = 1.0 M b) 0.050 M HNO3 -> [H+] = 0.050 M c) 1.0 M KOH -> [OH-] = 1.0 M d) 2.4 x 10^-5 M Mg(OH)2 -> [OH-] = (2.4 x 10^-5 M) * 2 = 4.8 x 10^-5 M

Step 3: Calculate pH or pOH for each solution

Now that we know the concentration of H+ or OH- ions in each solution, we can calculate pH or pOH using the following formulas: pH = -log[H+], pOH = -log[OH-], and the relationship between pH and pOH (pH + pOH = 14). a) 1.0 M HI -> pH = -log(1.0) = 0 b) 0.050 M HNO3 -> pH = -log(0.050) ≈ 1.3 c) 1.0 M KOH -> pOH = -log(1.0) = 0 -> pH = 14 - pOH = 14 d) 4.8 x 10^-5 M Mg(OH)2 -> pOH = -log(4.8 x 10^-5) ≈ 4.3 -> pH = 14 - pOH ≈ 9.7

Step 4: Summarize the results

Finally, we can summarize our pH results for each solution. a) The pH of 1.0 M HI is 0. b) The pH of 0.050 M HNO3 is 1.3. c) The pH of 1.0 M KOH is 14. d) The pH of 2.4 x 10^-5 M Mg(OH)2 is 9.7.

Key Concepts

Strong Acids

Strong acids are substances that completely dissociate into their ions when dissolved in water. This means they break apart entirely, releasing hydrogen ions (H^+ ) into the solution.

Common examples of strong acids include:

• Hydrochloric acid (HCl)
• Nitric acid (HNO₃)
• Hydrobromic acid (HBr)

In the exercise, HI and HNO₃ are both strong acids.

The dissociation equations look like this:

• HI ightarrow H^+ + I^-
• HNO_3 ightarrow H^+ + NO_3^-

Because they completely dissociate, their concentration directly gives the [H^+] concentration. Knowing this makes calculating the pH straightforward: pH = - log[ H^+].

Strong Bases

Strong bases are compounds that fully dissociate in water, releasing hydroxide ions (OH^- ).

Common strong bases include:

• Sodium hydroxide (NaOH)
• Potassium hydroxide (KOH)
• Calcium hydroxide (Ca(OH)₂)

In the task above, KOH is a strong base.

The dissociation equation is:
KOH ightarrow K^+ + OH^-

This complete dissociation tells us that the concentration of KOH gives the [OH^-] concentration directly.

To find the pH, first calculate pOH: pOH = - log[ OH^- ].

Then use the relationship pH + pOH = 14 to find the pH.

Weak Bases

Weak bases only partially dissociate in water, meaning they don't fully release their ions.

This behavior makes them different from strong bases, and calculating pH can be more complex.

Some examples of weak bases are:

• Ammonia (NH₃)
• Magnesium hydroxide (Mg(OH)₂)
• Baking soda (sodium bicarbonate)

In the exercise, Mg(OH)₂ is a weak base. Its dissociation equation looks like this:
Mg(OH)_2 ightarrow Mg^{2+} + 2OH^-

Notice each Mg(OH)₂ produces two OH^- ions, so you must multiply the concentration by 2.

This small degree of ionization means the base's concentration doesn't equal [OH^-] directly, requiring careful consideration in calculations.

Dissociation Equations

Dissociation equations illustrate how acids or bases split into ions when dissolved in water. These equations help in understanding the behavior of the substance in the solution.

The general form is:
AB ightarrow A^+ + B^-

For strong acids and bases, dissociation is complete, meaning all molecules break apart. For weak bases, only some molecules dissociate.

Consider these examples from the exercise:

• Strong Acid HI: HI ightarrow H^+ + I^-
• Strong Base KOH: KOH ightarrow K^+ + OH^-
• Weak Base Mg(OH)_2: Mg(OH)_2 ightarrow Mg^{2+} + 2OH^-

Understanding these equations is crucial for predicting ion concentration and calculating pH or pOH in solutions.