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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) \(0.182 \mathrm{M} \mathrm{KOH},\) (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 \(0.015 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\).

Short Answer

Expert verified
(a) [OH鈦籡 = 0.182 M; pH = 13.26 (b) [OH鈦籡 = 0.1128 M; pH = 13.053 (c) [OH鈦籡 = 4.2 脳 10鈦烩伌 M; pH = 10.62 (d) [OH鈦籡 = 1.5467 脳 10鈦宦� M; pH = 12.19

Step by step solution

01

Calculate the moles of OH鈦� ions

KOH is a strong base that dissociates completely in water according to the equation: KOH 鈫� K鈦� + OH鈦�. Therefore, 1 mole of KOH produces 1 mole of OH鈦�. The concentration of OH鈦� ions is the same as that of KOH: 0.182 M.
02

Calculate pH

Now we'll use the OH鈦� concentration to calculate the pH using the formula: pOH = -log[OH鈦籡 and pH = 14 - pOH. pOH = -log(0.182) = 0.74 pH = 14 - 0.74 = 13.26 (a) [OH鈦籡 = 0.182 M; pH = 13.26 (b) 3.165 g of KOH in 500.0 mL of solution
03

Calculate the moles of OH鈦� ions

First, we'll calculate the moles of KOH by dividing the mass by the molar mass of KOH (56.11 g/mol). moles of KOH = 3.165 g / 56.11 g/mol = 0.0564 mol Since 1 mole KOH produces 1 mole OH鈦�, moles of OH鈦� = moles of KOH = 0.0564 mol
04

Calculate the concentration of OH鈦� ions

To find the concentration of OH鈦� ions, divide the moles of OH鈦� by the volume of the solution in liters. [OH鈦籡 = 0.0564 mol / 0.5 L = 0.1128 M
05

Calculate pH

pOH = -log(0.1128) = 0.947 pH = 14 - 0.947 = 13.053 (b) [OH鈦籡 = 0.1128 M; pH = 13.053 (c) 10.0 mL of 0.0105 M 颁补(翱贬)鈧� diluted to 500.0 mL
06

Calculate the moles of OH鈦� ions

The dissociation of 颁补(翱贬)鈧� in water: 颁补(翱贬)鈧� 鈫� Ca虏鈦� + 2OH鈦�. Thus, 1 mole of 颁补(翱贬)鈧� produces 2 moles of OH鈦� ions. moles of 颁补(翱贬)鈧� = 0.0105 M 脳 0.01 L = 1.05 脳 10鈦烩伌 moles of OH鈦� = 1.05 脳 10鈦烩伌 脳 2 = 2.1 脳 10鈦烩伌
07

Calculate the concentration of OH鈦� ions

[OH鈦籡 = 2.1 脳 10鈦烩伌 mol / 0.5 L = 4.2 脳 10鈦烩伌 M
08

Calculate pH

pOH = -log(4.2 脳 10鈦烩伌) = 3.38 pH = 14 - 3.38 = 10.62 (c) [OH鈦籡 = 4.2 脳 10鈦烩伌 M; pH = 10.62 (d) A solution formed by mixing 20.0 mL of 0.015 M 叠补(翱贬)鈧� with 40.0 mL of 8.2 脳 10鈦宦� M NaOH.
09

Calculate the moles of OH鈦� ions

The dissociation of 叠补(翱贬)鈧�: 叠补(翱贬)鈧� 鈫� Ba虏鈦� + 2OH鈦�. Thus, 1 mole of 叠补(翱贬)鈧� produces 2 moles of OH鈦� ions. moles of OH鈦� from 叠补(翱贬)鈧� = 0.015 M 脳 0.02 L 脳 2 = 6 脳 10鈦烩伌 mol moles of OH鈦� from NaOH = 8.2 脳 10鈦宦� M 脳 0.04 L = 3.28 脳 10鈦烩伌 Total moles of OH鈦� = 6 脳 10鈦烩伌 + 3.28 脳 10鈦烩伌 = 9.28 脳 10鈦烩伌 mol
10

Calculate the concentration of OH鈦� ions

Total volume = 20 mL + 40 mL = 60 mL = 0.06 L [OH鈦籡 = 9.28 脳 10鈦烩伌 mol / 0.06 L = 1.5467 脳 10鈦宦� M
11

Calculate pH

pOH = -log(1.5467 脳 10鈦宦�) = 1.81 pH = 14 - 1.81 = 12.19 (d) [OH鈦籡 = 1.5467 脳 10鈦宦� M; pH = 12.19

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Key Concepts

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

KOH
Potassium hydroxide (KOH) is a common example of a strong base. When KOH dissolves in water, it fully dissociates into potassium ions (K鈦�) and hydroxide ions (OH鈦�). This complete dissociation is a characteristic feature of strong bases.
  • You can easily calculate the concentration of OH鈦� ions; it is the same as the initial concentration of KOH.
  • For example, if you have a solution with 0.182 M KOH, the concentration of OH鈦� ions is also 0.182 M.
Consequently, calculating pH also becomes straightforward because it requires converting OH鈦� concentration to pOH and then to pH.
颁补(翱贬)鈧�
Calcium hydroxide ( 颁补(翱贬)鈧� ) is another potent base, frequently utilized in various applications such as construction. Unlike KOH, 颁补(翱贬)鈧� dissociates in a two-step process:
  • Each molecule of 颁补(翱贬)鈧� produces two hydroxide ions ( OH鈦� ) when it dissolves in water.
For instance, in a solution of 0.0105 M 颁补(翱贬)鈧�, each mole of 颁补(翱贬)鈧� releases two moles of OH鈦� ions, leading to 0.021 M OH鈦� . Understanding this multiplication factor is essential when calculating the net OH鈦� concentration for pH calculations.
叠补(翱贬)鈧�
Barium hydroxide, 叠补(翱贬)鈧�, shares similarities with 颁补(翱贬)鈧� in that it also dissociates to form more than one hydroxide ion. Specifically, 叠补(翱贬)鈧� dissociates as follows:
  • One mole of 叠补(翱贬)鈧� produces two moles of OH-).
This is a key point for calculating pH, because it significantly affects the concentration of hydroxide ions in the solution.
Considering the total dissociation in the solution, you multiply the concentration of 叠补(翱贬)鈧� by two to find the total OH鈦� concentration.
Strong Base Dissociation
Strong bases, like KOH, 颁补(翱贬)鈧�, and 叠补(翱贬)鈧�, are unique in their full dissociation into ions in aqueous solutions. This complete breakdown underlies their ability to significantly increase OH鈦� concentration if they are present.
  • For KOH, this means that each molecule turns into one hydroxide ion.
  • For 颁补(翱贬)鈧� and 叠补(翱贬)鈧�, each molecule results in two hydroxide ions.
Understanding this full dissociation principle is pivotal because it directly influences the pH, positioning these solutions as among the most basic when dissolved in water.
pH and pOH Relationship
The relationship between pH and pOH is a fundamental concept for understanding acid-base chemistry. Both values measure different aspects of hydrogen and hydroxide ion concentrations in solutions:
  • pOH is calculated using the negative logarithm of the hydroxide ion concentration, [OH鈦籡.
  • Then, calculate pH by subtracting pOH from 14 (i.e., pH = 14 - pOH).
This relationship is vital in assessing the acidity or basicity of a solution. The higher the concentration of hydroxide ions, the lower the pOH, and consequently, the higher the pH value, indicating a more basic (or alkaline) solution. This interplay provides insight into the chemical nature of strong bases.

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Most popular questions from this chapter

Identify the Lewis acid and Lewis base in each of the following reactions: (a) \(\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\) (c) \(\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\) (d) \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\)

Caproic acid \(\left(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{COOH}\right)\) is found in small amounts in coconut and palm oils and is used in making artificial flavors. A saturated solution of the acid contains \(11 \mathrm{~g} / \mathrm{L}\) and has a \(\mathrm{pH}\) of 2.94. Calculate \(K_{a}\) for the acid.

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right),\) a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). What is the \(\mathrm{pH}\) of a solution obtained by dissolving two extra-strength aspirin tablets, containing \(500 \mathrm{mg}\) of acetylsalicylic acid each, in \(250 \mathrm{~mL}\) of water?

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side: $$ \text { (a) } \mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q) $$ (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) $$ \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) $$

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 \(\mathrm{BrO}_{2}^{-},(\mathbf{c})\) \(\mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)
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