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Chapter 16: Problem 84

Using data from Appendix D, calculate [OH \(^{-}\) ] and pH for each of the following solutions: (a) \(0.105 \mathrm{M}\) NaF, (b) \(0.035 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S},(\mathrm{c})\) a mixture that is \(0.045 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(0.055 \mathrm{M}\) in \(\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \mathrm{Ba} .\)

Short Answer

Expert verified
For the given solutions, the [OH鈦籡 and pH values are: (a) 0.105 M NaF: [OH鈦籡 = \( 1.23 \times 10^{-6} \) M; pH 鈮 8.09 (b) 0.035 M Na鈧係: [OH鈦籡 = \( 2.19 \times 10^{-4} \) M; pH 鈮 10.34 (c) 0.045 M CH鈧僀OONa and 0.055 M (CH鈧僀OO)鈧 Ba: [OH鈦籡 = \( 1.95 \times 10^{-5} \) M; pH 鈮 9.29

Step by step solution

01

(a) Calculate [OH鈦籡 and pH for 0.105 M NaF solution

1. Finding the Kb of F鈦: The anion of NaF is F鈦, which acts as a weak base. Its conjugate acid is HF with a known Ka. First, we need to find the Kb of F鈦 using the relation between Ka and Kb: Ka * Kb = Kw. Here, Kw is the ion product of water, 1x 10^{-14}. Ka of HF = 6.8 x 10^{-4} Kb = Kw / Ka = \( \frac{1 \times 10^{-14}}{6.8 \times 10^{-4}} \) = \( 1.47 \times 10^{-11} \) 2. Set up an ICE table for the hydrolysis reaction of F鈦: F鈦 + H鈧侽 鈬 HF + OH鈦 Initial: 0.105 M - - 0 Change: -x +x +x Equilibrium: 0.105-x x x 3. Solve for x ([OH鈦籡): Kb = [HF][OH鈦籡/[F鈦籡 = \( \frac{x \times x}{0.105 - x} \) = 1.47 x 10^{-11} Since the concentration of F鈦 is much greater than Kb, we can assume x is much smaller than 0.105 and ignore it in the denominator: x squared 鈮 0.105 x (1.47 x 10^{-11}) x 鈮 [OH鈦籡= 1.23 x 10^{-6} M 4. Calculate the pOH and pH: pOH = -log[OH鈦籡 = -log(1.23 x 10^{-6}) 鈮 5.91 pH = 14 - pOH = 14 - 5.91 鈮 8.09
02

(b) Calculate [OH鈦籡 and pH for 0.035 M Na鈧係 solution

1. Finding the Kb of S虏鈦: S虏鈦 is the anion in Na鈧係, which acts as a weak base. Its conjugate acid is HS鈦, with a known Ka. First, we need to find the Kb of S虏鈦 using the relation between Ka and Kb: Ka * Kb = Kw. Here, Kw is the ion product of water, 1x 10^{-14}. Ka of HS鈦 = 1.3 x 10^{-7} Kb = Kw / Ka = \( \frac{1 \times 10^{-14}}{1.3 \times 10^{-7}} \) = \( 7.69 \times 10^{-8} \) 2. Set up an ICE table for the hydrolysis reaction of S虏鈦: S虏鈦 + H鈧侽 鈬 HS鈦 + OH鈦 Initial: 0.035 M - - 0 Change: -x +x +x Equilibrium: 0.035-x x x 3. Solve for x ([OH鈦籡): Kb = [HS鈦籡[OH鈦籡/[S虏鈦籡 = \( \frac{x \times x}{0.035 - x} \) = 7.69 x 10^{-8} Since the concentration of S虏鈦 is much greater than Kb, we can assume x is much smaller than 0.035 and ignore it in the denominator: x squared 鈮 0.035 x (7.69 x 10^{-8}) x 鈮 [OH鈦籡 = 2.19 x 10^{-4} M 4. Calculate the pOH and pH: pOH = -log[OH鈦籡 = -log(2.19 x 10^{-4}) 鈮 3.66 pH = 14 - pOH = 14 - 3.66 鈮 10.34
03

(c) Calculate [OH鈦籡 and pH for a mixture that is 0.045 M CH鈧僀OONa and 0.055 M (CH鈧僀OO)鈧侭a solution

1. Calculate the total concentration of CH鈧僀OO鈦: Since both salts share the same anion, CH鈧僀OO鈦, we can find the total concentration of CH鈧僀OO鈦: [CH鈧僀OO鈦籡 = 0.045 + 2 * 0.055 = 0.155 M 2. Finding the Ka of CH鈧僀OOH and Kb of CH鈧僀OO鈦: Ka of CH鈧僀OOH = 1.8 x 10^{-5} Kb of CH鈧僀OO鈦 = Kw / Ka = \( \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} \) = \( 5.56 \times 10^{-10} \) 3. Set up an ICE table for the hydrolysis reaction of CH鈧僀OO鈦: CH鈧僀OO鈦 + H鈧侽 鈬 CH鈧僀OOH + OH鈦 Initial: 0.155 M - - 0 Change: -x +x +x Equilibrium: 0.155-x x x 4. Solve for x ([OH鈦籡): Kb = [CH鈧僀OOH][OH鈦籡/[CH鈧僀OO鈦籡 = \( \frac{x \times x}{0.155 - x} \) = 5.56 x 10^{-10} Since the concentration of CH鈧僀OO鈦 is much greater than Kb, we can assume that x is much smaller than 0.155 and ignore it in the denominator: x squared 鈮 0.155 x (5.56 x 10^{-10}) x 鈮 [OH鈦籡 = 1.95 x 10^{-5} M 5. Calculate the pOH and pH: pOH = -log[OH鈦籡 = -log(1.95 x 10^{-5}) 鈮 4.71 pH = 14 - pOH = 14 - 4.71 鈮 9.29

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Key Concepts

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

Hydrolysis Reaction
In the realm of chemistry, a hydrolysis reaction is a pivotal process which typically occurs when a salt, made from a weak acid or base, dissolves in water. In simpler terms, hydrolysis involves a salt reacting with water to form an acid or a base. For instance, when sodium fluoride (NaF) is dissolved in water, the fluoride ion (F鈭) undergoes hydrolysis with water to yield hydroxide ions (OH鈭) and hydrofluoric acid (HF).

This reaction can be represented as:
F鈭 + H2O 鈬 HF + OH鈭.

Understanding the behavior of these ions in water is crucial for calculating pH. The generated OH鈭 increases the solution's basicity, thus affecting the overall pH level.
Ionic Product of Water (Kw)
The ionic product of water, or Kw, is an essential constant in the field of acid-base chemistry. It signifies the product of the molar concentrations of H+ and OH鈭 ions in pure water at a given temperature. Generally, at 25掳C, Kw equals 1 脳 10鈭14.

The equation can be expressed as:
Kw = [H+][OH鈭].

This constant is foundational because it connects the concentrations of hydrogen and hydroxide ions, which are pivotal for understanding the pH and pOH of solutions. Additionally, it provides a relation between the acid dissociation constant (Ka) and base dissociation constant (Kb) for conjugate acid-base pairs: Ka 脳 Kb = Kw.
ICE Table Method
The ICE Table Method is an organized way to solve for concentrations of reactants and products in an equilibrium reaction. ICE stands for Initial, Change, and Equilibrium. You can depict the concentrations at each of these stages for all entities in the reaction.

For instance, consider the fluoride ion hydrolysis described earlier. The ICE table would be:

Initial: [F鈭] = 0.105 M, [HF] = 0, [OH鈭] = 0
Change: [F鈭] = 鈭抶, [HF] = +x, [OH鈭] = +x
Equilibrium: [F鈭] = (0.105 - x) M, [HF] = x, [OH鈭] = x

In this method, 'x' represents the change in concentration. This method significantly simplifies the process of finding equilibrium concentrations in acid-base reactions and thus assists in pH calculations.
Acid-Base Equilibrium
Acid-base equilibrium pertains to the balance that exists in a solution between its acid (H+) and base (OH鈭) components. It is crucial for the determination of pH, as it touches on the extents to which acids donate hydrogen ions or bases accept them. The equilibrium lies at the core of the calculations involving the ICE table method.

When solving for pH, one calculates the equilibrium concentration of OH鈭 by setting up an equation with the base dissociation constant (Kb), which can be derived from the ionic product of water (Kw) and the acid dissociation constant (Ka). In summary, acid-base equilibrium is indispensable for accurately determining the acidity or basicity of a solution.

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Most popular questions from this chapter

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}, \mathrm{O}_{4}\right)\), 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?

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), (b) \(\mathrm{HCO}_{3}^{-}\)

A \(0.100 \mathrm{M}\) solution of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) is \(11.0 \%\) ionized. Using this information, calculate \(\left.\left[\mathrm{ClCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{ClCH}_{2} \mathrm{COOH}\right)\right]\), and \(K_{a}\) for chloroacetic acid.

Predict which member of each pair produces the more acidic aqueous solution: (a) \(\mathrm{K}^{4}\) or \(\mathrm{Cu}^{2+}\), (b) \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\), (c) \(\mathrm{Al}^{3+}\) or \(\mathrm{Ga}^{3+}\). Explain.

If a substance is a Lewis acid, is it necessarily a Br酶nsted-Lowry acid? Is it necessarily an Arrhenius acid? Explain.
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