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Chapter 16: Problem 46

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},(\mathbf{b}) 3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in 500.0 \(\mathrm{mL}\) of solution, (c) \(10.0 \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{~mL},\) (d) a solution formed by mixing \(20.0 \mathrm{~mL}\) of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\).

Short Answer

Expert verified
a) [OH-] = 0.182 M, pH = 13.260; b) [OH-] = 0.1128 M, pH = 13.052; c) [OH-] = 0.00042 M, pH = 10.62; d) [OH-] = 0.01547 M, pH = 12.19.

Step by step solution

01

Calculate [OH-] and pH for 0.182 M KOH

Since KOH is a strong base, it dissociates completely in the solution. Thus, the concentration of OH- is the same as the concentration of KOH: \[ [ ext{OH}^-] = 0.182 ext{ M} \]To find the pH, we first calculate the pOH:\[ ext{pOH} = - ext{log}_{10}[ ext{OH}^-] = - ext{log}_{10}(0.182) \approx 0.740 \]Now, using the relationship \( ext{pH} + ext{pOH} = 14 \), we find:\[ ext{pH} = 14 - 0.740 = 13.260 \]
02

Calculate [OH-] and pH for 3.165 g of KOH in 500.0 mL

First, convert grams of KOH to moles. The molar mass of KOH is approximately 56.11 g/mol:\[ ext{moles of KOH} = \frac{3.165}{56.11} \approx 0.0564 ext{ moles} \]Next, find the molarity of the solution by dividing moles by volume in liters:\[ ext{Molarity} = \frac{0.0564}{0.5} = 0.1128 ext{ M} \]Since KOH is a strong base, \([\text{OH}^-] = 0.1128 \text{ M}\).Calculate the pOH:\[ ext{pOH} = - ext{log}_{10}(0.1128) \approx 0.948 \]Thus, the pH is:\[ ext{pH} = 14 - 0.948 = 13.052 \]
03

Calculate [OH-] and pH for diluted Ca(OH)2 solution

Start by calculating moles of Ca(OH)鈧� in 10 mL of a 0.0105 M solution:\[ ext{moles Ca(OH)}_2 = 0.0105 ext{ M} \times 0.01 ext{ L} = 0.000105 ext{ moles} \]Each molecule of Ca(OH)鈧� produces two OH鈦� ions, so the moles of OH鈦� is twice the moles of Ca(OH)鈧�:\[ ext{moles OH}^- = 2 imes 0.000105 = 0.00021 ext{ moles OH}^- \]The final volume after dilution is 500 mL (0.5 L):\[ [ ext{OH}^-] = \frac{0.00021}{0.5} = 0.00042 ext{ M} \]Calculate the pOH:\[ ext{pOH} = - ext{log}_{10}(0.00042) \approx 3.38 \]Finally, find the pH:\[ ext{pH} = 14 - 3.38 = 10.62 \]
04

Calculate [OH-] and pH for mixed solution of Ba(OH)2 and NaOH

First, calculate the moles of OH鈦� contributed by each base. For Ba(OH)鈧�:\[ ext{moles Ba(OH)}_2 = 0.015 ext{ M} \times 0.02 ext{ L} = 0.0003 ext{ moles} \]Since Ba(OH)鈧� dissociates into two OH鈦� ions, \[ ext{moles OH}^- = 2 imes 0.0003 = 0.0006 ext{ moles OH}^- \]For NaOH:\[ ext{moles NaOH} = 8.2 imes 10^{-3} ext{ M} \times 0.04 ext{ L} = 0.000328 ext{ moles OH}^- \]Total moles of OH鈦�:\[ ext{total moles OH}^- = 0.0006 + 0.000328 = 0.000928 ext{ moles OH}^- \]The final volume of the mixed solution is 60 mL (0.06 L):\[ [ ext{OH}^-] = \frac{0.000928}{0.06} = 0.01547 ext{ M} \]Calculate the pOH:\[ ext{pOH} = - ext{log}_{10}(0.01547) \approx 1.81 \]Finally, find the pH:\[ ext{pH} = 14 - 1.81 = 12.19 \]

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Key Concepts

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

pH Calculation
The pH scale is a measure of how acidic or basic a solution is. It ranges from 0 to 14, with 7 being neutral.Understanding how to calculate the pH of a strong base solution is key for many chemistry problems. To determine the pH of a base, we need the concentration of hydroxide ions (\[\text{OH}^{-}\]). Once we have \[\text{OH}^{-}\], we can find the pOH using the formula:\[\text{pOH} = -\log_{10}[\text{OH}^{-}]\]. Then, using the relationship:
  • \(\text{pH} + \text{pOH} = 14\)
we can solve for the pH. This is because water's pH at room temperature is about 7, hence pH and pOH always add up to 14. For instance, for a solution of 0.182 M KOH, we first determine that:
  • [\text{OH}^-] = 0.182 \, \text{M}
  • \(\text{pOH} \approx 0.740\)
  • \(\text{pH} \approx 13.260\)
Hydroxide Ion Concentration
Hydroxide ion concentration is crucial when dealing with bases. In strong bases like Potassium Hydroxide (KOH) or Calcium Hydroxide (Ca(OH)\(_2\)), the dissociation is 100%, meaning each molecule breaks down completely in solution.For a basic solution, the hydroxide ion concentration \([\text{OH}^-]\) is often the same as the base's molarity. If you know a solution's hydroxide concentration, you can directly determine many essential factors about the solution including its pOH and pH.For example:
  • KOH, as a monobasic base (providing one OH鈦� per molecule), results in \([\text{OH}^-] = [\text{KOH}]\)
  • Ca(OH)\(_2\), as a dibasic base (providing two OH鈦� ions per molecule), doubles the \([\text{OH}^-]\) compared to its own molarity.
Dilution and Molarity
Dilution involves adding solvent to a solution to decrease the concentration of solutes. Molarity (M) is the number of moles of solute per liter of solution and is essential when calculating concentrations after dilution.For example, if you have a 10 mL solution of 0.0105 M Ca(OH)\(_2\) and you dilute it to 500 mL, the concentration changes.To find the new concentration after dilution, use the equation:\[C_1V_1 = C_2V_2\]where:
  • \(C_1\) and \(V_1\) are the original concentration and volume.
  • \(C_2\) and \(V_2\) are the final concentration and volume.
In our example, the volume increased (from 10 mL to 500 mL), causing the concentration to decrease. The moles of solute remain unchanged, but the volume has increased, spreading out the solute.
Mixing Solutions
Mixing solutions involves combining two or more solutions. This may result in a new concentration of solutes, specifically ions like \(\text{OH}^{-}\) in base solutions.The fundamental aspect to consider during mixing is the total moles of a solute, not just the concentrations.For instance, if you mix:
  • 20 mL of 0.015 M Ba(OH)\(_2\) with 40 mL of 8.2 脳 10鈦宦� M NaOH, calculate the moles of OH鈦� from each before mixing.
  • For Ba(OH)\(_2\), it provides two OH鈦� per molecule. Calculate total moles of OH鈦� for each solution, then combine these totals.
  • Determine the new concentration by dividing total moles of OH鈦� by the new total volume (in liters).
Finally, calculate the new pOH and consequently the pH.
KOH and Ca(OH)2
Potassium Hydroxide (KOH) and Calcium Hydroxide (Ca(OH)\(_2\)) are strong bases used often in chemistry.These compounds fully dissociate in water, making them perfect for studying the effects of basic solutions. KOH is a monobasic base, releasing one OH鈦� ion per molecule. Conversely, Ca(OH)\(_2\) is a dibasic base, releasing two OH鈦� ions per molecule.When preparing solutions with these bases:
  • KOH's molarity equals \([\text{OH}^-]\) in the solution.
  • Ca(OH)\(_2\)'s \([\text{OH}^-]\) is double its molarity due to two hydroxide ions per formula unit.
These properties make such calculations in solutions straightforward, but you must apply them accurately to evaluate their pH effectively.

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Most popular questions from this chapter

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-},(\mathbf{b}) \mathrm{PO}_{4}^{3-}\) or \(\mathrm{AsO}_{4}^{3-},\) (c) \(\mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}\).

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14}\). Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature.

The volume of an adult's stomach ranges from about 50 \(\mathrm{mL}\) when empty to \(1 \mathrm{~L}\) when full. If the stomach volume is \(400 \mathrm{~mL}\) and its contents have a \(\mathrm{pH}\) of 2 , how many moles of \(\mathrm{H}^{+}\) does the stomach contain? Assuming that all the \(\mathrm{H}^{+}\) comes from HCl, how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?

The amino acid glycine \(\left(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)\) can participate in the following equilibria in water: \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{\mathrm{a}}=4.3 \times 10^{-3} $$$$ \begin{aligned} \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \\ &{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_{\mathrm{b}}=6.0 \times 10^{-5} \end{aligned} $$ (a) Use the values of \(K_{a}\) and \(K_{b}\) to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-} $$ (b) What is the pH of a \(0.050 \mathrm{M}\) aqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with \(\mathrm{pH} 13\) ? With \(\mathrm{pH} 1\) ?

An unknown salt is either \(\mathrm{KBr}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{KCN},\) or \(\mathrm{K}_{2} \mathrm{CO}_{3} .\) If a \(0.100 M\) solution of the salt is neutral, what is the identity of the salt?
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