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Chapter 6: Problem 7

Use the systematic approach to equilibrium problems to calculate the \(\mathrm{pH}\) of the following solutions. Be sure to state and justify any assumptions you make in solving the problems. a. \(0.050 \mathrm{M} \mathrm{HClO}_{4}\) b. \(1.00 \times 10^{-7} \mathrm{M} \mathrm{HCl}\) c. \(0.025 \mathrm{M} \mathrm{HClO}\) d. \(0.010 \mathrm{M} \mathrm{HCOOH}\) e. \(0.050 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) f. \(0.010 \mathrm{M} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\)

Short Answer

Expert verified
The pH values are: a) 1.30, b) 6.70, c) 4.07, d) 2.87, e) 13.00, f) 8.62.

Step by step solution

01

Understand the Nature of Each Solution

To solve these problems, recognize the nature of each solution: - **a.** Perchloric acid (\(\text{HClO}_4\)) is a strong acid.- **b.** Hydrochloric acid (\(\text{HCl}\)) is a strong acid at low concentrations, but near neutral\( \text{H}_2\text{O} \), hydrolysis must be considered.- **c.** Hypochlorous acid (\(\text{HClO}\)) is a weak acid.- **d.** Formic acid (\(\text{HCOOH}\)) is a weak acid.- **e.** Barium hydroxide (\(\text{Ba(OH)}_{2}\)) is a strong base.- **f.** Pyridine (\(\text{C}_5\text{H}_5\text{N}\)) is a weak base.
02

Calculate pH for Strong Acid (0.050 M HClO4)

\(\text{HClO}_4\) ionizes completely:\[\text{HClO}_4 \rightarrow \text{H}^+ + \text{ClO}_4^-\]Since it is fully ionized, \([\text{H}^+] = 0.050\text{ M}\).Use \( \text{pH} = -\log [\text{H}^+]\):\[\text{pH} = -\log(0.050) \approx 1.30\]
03

Special Case for Dilute Strong Acid (1.00 x 10^-7 M HCl)

For this highly diluted HCl solution, the contribution of \([\text{H}^+ ]\) from water is comparable:\(\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-\). Thus:\([ ext{H}^+]_{\text{total}} = 1.00 \times 10^{-7}\text{ M} + 1.00 \times 10^{-7}\text{ M} = 2.00 \times 10^{-7}\text{ M}\)Calculate the \(\text{pH} \):\[\text{pH} = -\log(2.00 \times 10^{-7}) \approx 6.70\]
04

Calculate pH for Weak Acid (0.025 M HClO)

Use the equilibrium expression for weak acid dissociation:\[\text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^-\]Given \(K_a = 2.9 \times 10^{-8}\):\[K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]_{\text{initial}} - [\text{H}^+]}\]Assume \([\text{H}^+] \approx \sqrt{K_a \times [\text{HClO}]_{\text{initial}}}\):\[[\text{H}^+] \approx \sqrt{2.9 \times 10^{-8} \times 0.025} \approx 8.54 \times 10^{-5}\]Then calculate \(\text{pH} \):\[\text{pH} = -\log(8.54 \times 10^{-5}) \approx 4.07\]
05

Calculate pH for Weak Acid (0.010 M HCOOH)

For formic acid with \(K_a = 1.8 \times 10^{-4}\):\[\text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^-\]Apply the formula similar as previous:\[[\text{H}^+] \approx \sqrt{K_a \times [\text{HCOOH}]_{\text{initial}} }\]\[[\text{H}^+] \approx \sqrt{1.8 \times 10^{-4} \times 0.010} \approx 1.34 \times 10^{-3}\]Calculate \(\text{pH} \):\[\text{pH} = -\log(1.34 \times 10^{-3}) \approx 2.87\]
06

Calculate pH for Strong Base (0.050 M Ba(OH)2)

\(\text{Ba(OH)}_2\) is a strong base offering \([\text{OH}^-] = 2 \times 0.050 = 0.100\text{ M}\). Calculate the \([\text{H}^+]\) via:\[\text{pOH} = -\log(0.100) = 1.00\]Then convert \( ext{pOH}\) to \( ext{pH}\):\[\text{pH} = 14 - \text{pOH} = 14 - 1.00 = 13.00\]
07

Calculate pH for Weak Base (0.010 M C5H5N)

Use the equilibrium:\[\text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} \rightleftharpoons \text{C}_5\text{H}_5\text{NH}^+ + \text{OH}^-\]With \(K_b = 1.7 \times 10^{-9}\):\[K_b = \frac{[\text{C}_5\text{H}_5\text{NH}^+][\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}]_{\text{initial}} - [\text{OH}^-]}\]Assume:\([ ext{OH}^-] \approx \sqrt{K_b \times [\text{C}_5\text{H}_5\text{N}]_{\text{initial}}}\):\[[\text{OH}^-] \approx \sqrt{1.7 \times 10^{-9} \times 0.010} \approx 4.12 \times 10^{-6}\]Compute \( ext{pOH}\) and corresponding \( ext{pH}\):\[\text{pOH} = -\log(4.12 \times 10^{-6}) \approx 5.38\]\[\text{pH} = 14 - 5.38 = 8.62\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Problems
Equilibrium problems often involve determining the concentration of ions in a solution at equilibrium. These problems require you to understand the balance reached in a solution when weak acids or bases partially dissociate. To tackle equilibrium problems:
  • Start by writing the balanced chemical equation for the dissociation.
  • Set up an expression for the equilibrium constant (\( K_a \) or \( K_b \)).
  • Use the initial concentrations and equilibrium expressions to solve for the unknowns, typically the \([\text{H}^+]\) or \([\text{OH}^-])\) concentration.
  • Approximate the concentrations using small x approximations when appropriate.
For accurate results, verify assumptions (like neglecting x in the denominator) and use approximations wisely.
Strong Acids and Bases
Strong acids and bases fully dissociate in water. This means for any concentration of a strong acid like HClO4 or a strong base like Ba(OH)2:
  • The concentration of hydronium ions \([\text{H}^+])\) directly matches the concentration of the acid.
  • The concentration of hydroxide ions \([\text{OH}^-])\) directly correlates to the base's dissociation.
This complete ionization simplifies \( \text{pH} \) calculations:
  • For strong acids, use \( \text{pH} = -\log[\text{H}^+] \).
  • For strong bases, first find \( \text{pOH} = -\log[\text{OH}^-]) \), then calculate \( \text{pH} \) as \( \text{pH} = 14 - \text{pOH} \).
Remember, this only applies when the acid or base is truly strong, meaning it dissociates completely in the solution.
Weak Acids and Bases
Weak acids and bases partially dissociate in water, making their \( \text{pH} \) calculation a bit more complex. Examples include HClO or HCOOH for weak acids, and C5H5N for weak bases. To handle these cases:
  • Formulate their equilibrium reactions and set up the equilibrium constant expressions (\( K_a \) for acids, \( K_b \) for bases).
  • Weak acids dissociate into \([\text{H}^+] \) and their conjugate base, while weak bases acquire \([\text{H}^+] \) to form their conjugate acids.
To find the \([\text{H}^+] \) or \([\text{OH}^-]) \):
  • Use the approximation \( x \approx \sqrt{K_a \times [HA]_{initial}} \) or \( x \approx \sqrt{K_b \times [B]_{initial}} \) when \( x \) is negligible compared to initial concentrations.
These steps will yield the degree of dissociation necessary to calculate the \( \text{pH} \) or \( \text{pOH} \).
Acid-Base Reactions
Acid-base reactions involve the transfer of protons between species:
  • An acid donates \([\text{H}^+] \) ions.
  • A base accepts \([\text{H}^+] \) ions.
In aqueous solutions, water acts as a reference substance:
  • Acids increase the \([\text{H}^+] \), reducing the \( \text{pH} \).
  • Bases increase \([\text{OH}^-]) \), thus raising the \( \text{pH} \).
In the context of calculations, ionic products, equilibrium constants, and the relationship \( \text{pH} + \text{pOH} = 14 \) are critical. Efficient problem-solving often relies on understanding these fundamental interactions and leveraging them to find unknown values in a solution setup.

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Most popular questions from this chapter

One analytical method for determining the concentration of sulfur is to oxidize it to \(\mathrm{SO}_{4}^{2-}\) and then precipitate it as \(\mathrm{BaSO}_{4}\) by adding \(\mathrm{BaCl}_{2}\). The mass of the resulting precipitate is proportional to the amount of sulfur in the original sample. The accuracy of this method depends on the solubility of \(\mathrm{BaSO}_{4}\), the reaction for which is shown here. $$ \mathrm{BaSO}_{4}(s)=\mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) $$ For each of the following, predict the affect on the solubility of \(\mathrm{Ba} \mathrm{SO}_{4}\) : (a) decreasing the solution's \(\mathrm{pH}\); (b) adding more \(\mathrm{BaCl}_{2}\); and (c) increasing the solution's volume by adding \(\mathrm{H}_{2} \mathrm{O}\).

Consider the following hypothetical complexation reaction between a metal, \(\mathrm{M},\) and a ligand, \(\mathrm{L}\) $$ \mathrm{M}(a q)+\mathrm{L}(a q)=\mathrm{ML}(a q) $$ for which the formation constant is \(1.5 \times 10^{8}\). (a) Derive an equation similar to the Henderson-Hasselbalch equation that relates \(\mathrm{pM}\) to the concentrations of \(\mathrm{L}\) and \(\mathrm{ML}\). (b) What is the \(\mathrm{pM}\) for a solution that contains \(0.010 \mathrm{~mol}\) of \(\mathrm{M}\) and \(0.020 \mathrm{~mol}\) of \(\mathrm{L}\) ? (c) What is pM if you add \(0.002 \mathrm{~mol}\) of \(\mathrm{M}\) to this solution? Be sure to state and justify any assumptions you make in solving the problem.

A redox buffer contains an oxidizing agent and its conjugate reducing agent. Calculate the potential of a solution that contains \(0.010 \mathrm{~mol}\) of \(\mathrm{Fe}^{3+}\) and \(0.015 \mathrm{~mol}\) of \(\mathrm{Fe}^{2+}\). What is the potential if you add sufficient oxidizing agent to convert \(0.002 \mathrm{~mol}\) of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+} ?\) Be sure to state and justify any assumptions you make in solving the problem.

Calculate the ionic strength of the following solutions a. \(0.050 \mathrm{M} \mathrm{NaCl}\) b. \(0.025 \mathrm{M} \mathrm{CuCl}_{2}\) c. \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\)

Calculate the standard state potential and the equilibrium constant for each of the following redox reactions. Assume that \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) is \(1.0 \mathrm{M}\) for an acidic solution and that \(\left[\mathrm{OH}^{-}\right]\) is \(1.0 \mathrm{M}\) for a basic solution. Note that these reactions are not balanced. Reactions (a) and (b) are in acidic solution; reaction (c) is in a basic solution. a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{SO}_{3}(a q)=\mathrm{Mn}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) b. \(\quad \mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q)=\mathrm{I}_{2}(a q)\) c. \(\mathrm{ClO}^{-}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\)
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