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Chapter 16: Problem 47

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, \((c) 100 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},(\mathrm{~d})\) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

Short Answer

Expert verified
(a) The hydroxide ion concentration [OH鈦籡 for the 1.5 x 10鈦宦� M Sr(OH)鈧� solution is 3.0 x 10鈦宦� M, and the pH is approximately 11.48. (b) The [OH鈦籡 for the 2.250 g of LiOH in 250.0 mL of solution is 0.375 M, and the pH is approximately 13.575. (c) The [OH鈦籡 for the 100 mL of 0.175 M NaOH diluted to 2.00 L is 8.75 x 10鈦宦� M, and the pH is approximately 11.942. (d) The [OH鈦籡 for the solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 10鈦宦� M Ca(OH)鈧� is approximately 0.16875 M, and the pH is approximately 13.227.

Step by step solution

01

a) Calculate [OH鈦籡 and pH for 1.5 x 10鈦宦� M Sr(OH)鈧� solution

To calculate the [OH鈦籡 in Sr(OH)鈧� solution, remember that one molecule of Sr(OH)鈧� produces 2 OH鈦� ions. So we multiply the molarity of Sr(OH)鈧� by 2: \[ [\mathrm{OH}^{-}] = 1.5\times 10^{-3} \mathrm{M} \cdot 2 = 3.0 \times 10^{-3} \mathrm{M} \] pOH can be calculated using the formula: \[ \mathrm{pOH} = -\log [\mathrm{OH}^{-}] \] Plug in the value for [OH鈦籡: \[ \mathrm{pOH} = -\log (3.0 \times 10^{-3}) \approx 2.52 \] Since pH + pOH = 14: \[ \mathrm{pH} = 14 - 2.52 \approx 11.48 \] So, the [OH鈦籡 is 3.0 x 10鈦宦� M, and the pH is approximately 11.48.
02

b) Calculate [OH鈦籡 and pH for 2.250 g of LiOH in 250.0 mL of solution

First, we determine the moles of LiOH from its mass: \[ \mathrm{moles} = \frac{\mathrm{mass}}{\mathrm{molar~mass}} =\frac{2.250 \mathrm{~g}}{23.95 \mathrm{~g/mol}}\approx 0.0939 \mathrm{~mol}\] Total volume of the solution is 250 mL. Therefore, the molarity of LiOH is: \[ \mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{volume (L)}}=\frac{0.0939 \mathrm{~mol}}{0.250 \mathrm{~L}}\approx 0.375 \mathrm{M} \] Since one molecule of LiOH produces one OH鈦� ion, the concentration of OH鈦� ions is equal to that of LiOH. \[ [\mathrm{OH}^{-}] = 0.375 \mathrm{M} \] Calculating pOH: \[ \mathrm{pOH} = -\log (0.375) \approx 0.425 \] And the pH: \[ \mathrm{pH} = 14 - 0.425\approx 13.575 \] So, the [OH鈦籡 is 0.375 M and the pH is approximately 13.575.
03

c) Calculate [OH鈦籡 and pH for 100 mL of 0.175 M NaOH diluted to 2.00 L

First, find the moles of NaOH in 100 mL of 0.175 M solution: \[ \mathrm{moles} = \mathrm{volume (L)} \times \mathrm{Molarity} = 0.1 \mathrm{~L} \times 0.175 \mathrm{M} = 0.0175 \mathrm{~mol} \] After dilution, the total volume of the solution is 2.00 L. To find the new molarity of NaOH after dilution, we use: \[ \mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{volume (L)}}=\frac{0.0175 \mathrm{~mol}}{2 \mathrm{~L}}= 8.75 \times 10^{-3} \mathrm{M} \] Since one molecule of NaOH produces one OH鈦� ion, the concentration of OH鈦� ions is equal to that of NaOH. \[ [\mathrm{OH}^{-}] = 8.75 \times 10^{-3} \mathrm{M} \] Calculating pOH: \[ \mathrm{pOH} = -\log (8.75 \times 10^{-3})\approx 2.058\] And the pH: \[ \mathrm{pH} = 14 - 2.058\approx 11.942 \] So, the [OH鈦籡 is 8.75 x 10鈦宦� M and the pH is approximately 11.942.
04

d) Calculate [OH鈦籡 and pH for a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 10鈦宦� M Ca(OH)鈧�

First, we need to find the total moles of OH鈦� ions from each base in the mixed solution. For KOH: \[ \mathrm{moles} = \mathrm{volume (L)} \times \mathrm{Molarity} = 0.005 \mathrm{~L} \times 0.105 \mathrm{M} = 5.25 \times 10^{-4} \mathrm{~mol} \] For Ca(OH)鈧�, remember one molecule of Ca(OH)鈧� produces 2 OH鈦� ions: \[ \mathrm{moles} = \mathrm{volume (L)} \times \mathrm{Molarity} \times 2 = 0.015 \mathrm{~L} \times 9.5 \times 10^{-2} \mathrm{M} \times 2 \approx 0.00285 \mathrm{~mol} \] Now, add the moles of OH鈦� ions from both sources: \[ \mathrm{Total~moles~of~OH^{-}} = 5.25 \times 10^{-4} + 0.00285 鈮� 0.003375\mathrm{~mol} \] Total volume of mixed solution is 20.0 mL or 0.020 L. Calculate the molarity of OH鈦� ions in the mixed solution: \[ [\mathrm{OH}^{-}] = \frac{\mathrm{moles}}{\mathrm{volume (L)}} = \frac{0.003375 \mathrm{~mol}}{0.020 \mathrm{~L}} \approx 0.16875 \mathrm{M} \] Calculating pOH: \[ \mathrm{pOH} = -\log (0.16875) \approx 0.773 \] And the pH: \[ \mathrm{pH} = 14 - 0.773 = 13.227 \] So, the [OH鈦籡 is approximately 0.16875 M and the pH is approximately 13.227.

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Key Concepts

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

Strong Bases
Strong bases such as Sr(OH)鈧�, NaOH, and KOH are compounds that fully dissociate in water to produce hydroxide ions (OH鈦�). This complete dissociation means that they efficiently increase the hydroxide concentration in a solution.
For example, Sr(OH)鈧� dissociates into Sr虏鈦� and 2 OH鈦� ions, effectively doubling the concentration of hydroxide ions compared to its initial concentration.
  • Full dissociation: When added to water, strong bases like these dissociate completely.
  • High hydroxide ion concentration: This means they are highly effective in raising the OH鈦� concentration in solutions.
Recognizing whether a base is strong or weak is important when predicting and calculating solution behavior, such as in pH determination.
Molarity
Molarity (M) is a measure of concentration that expresses the number of moles of solute per liter of solution. It's a crucial concept in chemistry because it quantifies how much of a substance is present in a given volume.
For instance, if we have 0.375 mol of LiOH in 1 liter of solution, the molarity is 0.375 M.
  • Understanding moles: It's essential to convert grams to moles using the compound's molar mass for accurate calculations.
  • Volume in liters: Remember, the solution's volume must be in liters for molarity calculations.
By calculating molarity, we can easily determine how changes, such as dilution, will affect a solution's concentration.
Dilution
Dilution involves adding more solvent to a solution to decrease the concentration of solutes, like OH鈦� ions, without adding more solute. The total amount of solute remains constant.
For instance, diluting 100 mL of a 0.175 M NaOH solution to 2.00 L reduces the concentration significantly.
  • Volume increase: By increasing the solution's volume, the molarity decreases proportionally.
  • Constant moles: The total moles of solute remain unchanged during dilution.
This technique is commonly used to prepare solutions with precise concentrations from more concentrated stock solutions, simplifying many laboratory procedures.
Solution Concentration
Solution concentration defines how much solute is dissolved in a given amount of solvent, typically expressed in units like molarity (M). Keeping track of concentration changes is essential when mixing or diluting solutions.
In the exercise, calculating the combined concentration of multiple solutions like KOH and Ca(OH)鈧� helps in determining the pH accurately.
  • Combining solutions: When solutions are mixed, calculate the total number of moles of solute for accurate concentration determination.
  • Mole balance: Ensure that you consider the volume and concentration of each component in the mixture.
Proficiently understanding solution concentration enables precise control over chemical reactions and solution properties, simplifying complex tasks such as titrations and pH adjustments.

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Most popular questions from this chapter

Designate the Bronsted-Lowry acid and the BronstedLowry base on the left side of each equation, and also designate the conjugate acid and conjugate base on the right side. (a) \(\mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q)\) (b) \(\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons\) \(\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{BrO}^{-}\) or \(\mathrm{ClO}^{-}\), (b) \(\mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-},(\mathrm{c}) \mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

The acid-dissociation constant for hypochlorous acid (HClO) is \(3.0 \times 10^{-8}\). Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{ClO}^{-}\), and \(\mathrm{HClO}\) at equilibrium if the initial concentration of \(\mathrm{HClO}\) is \(0.0090 \mathrm{M}\).

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\). Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 \mathrm{M}\); (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;(\mathrm{c})\) solution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\).
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