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\mathrm{~A} 0.25 \mathrm{M}\( solution of a salt \)\mathrm{NaA}\( has \)\mathrm{pH}=9.29\(. What is the value of \)K_{a}$ for the parent acid HA?

Short Answer

Expert verified
The value of \(K_a\) for the parent acid HA is approximately 4.87 x 10^(-10).

Step by step solution

01

Calculate [OH鈦籡 from the given pH

The given pH of the NaA solution is 9.29. We can find the concentration of hydroxide ions [OH鈦籡 using the relation between pH and pOH: pOH = 14 - pH Now, plug in the given pH value: pOH = 14 - 9.29 = 4.71 Now, we can calculate the [OH鈦籡 using the formula: [OH鈦籡 = 10^(-pOH) [OH鈦籡 = 10^(-4.71) = 1.95 x 10^(-5) M
02

Determine [A鈦籡 concentration

Since NaA is a salt that completely dissociates in water, the concentration of A鈦 ions will be equal to the concentration of the NaA solution: [A鈦籡 = 0.25 M
03

Set up an ICE table

Write the ionization reaction for the weak acid, HA: HA + H鈧侽 鈬 H鈧僌鈦 + A鈦 Now, set up an ICE table (Initial, Change, Equilibrium) to show the change in concentrations. HA + H鈧侽 鈬 H鈧僌鈦 + A鈦 Initial -- -- 0 0.25 Change +x -- +x -x Equilibrium x -- x 0.25 - x Note that the concentration of A鈦 after the reaction will be 0.25 - x, since we started with a 0.25 M solution of NaA.
04

Use the given [OH鈦籡 to determine x

Since we know the [OH鈦籡, we can use the relation between [OH鈦籡 and [H鈧僌鈦篯 to find x: [H鈧僌鈦篯 = (1.0 * 10^(-14) M虏) / [OH鈦籡 Plug in the value for [OH鈦籡: [H鈧僌鈦篯 = (1.0 * 10^(-14) M虏) / (1.95 x 10^(-5) M) = 5.13 x 10^(-10) M Now we know x = [H鈧僌鈦篯 = 5.13 x 10^(-10) M.
05

Calculate Ka for the parent acid HA

We can find Ka using the equilibrium expression: Ka = ([H鈧僌鈦篯 * [A鈦籡) / [HA] Plug the equilibrium concentration values into the expression: Ka = (5.13 x 10^(-10) * (0.25 - 5.13 x 10^(-10))) / (5.13 x 10^(-10)) After solving this expression, we get: Ka = 4.87 x 10^(-10) So, the value of Ka for the parent acid HA is approximately 4.87 x 10^(-10).

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Key Concepts

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

Chemical Equilibrium
In chemistry, chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant over time. It's a state of balance and does not mean the reactants and products are present in equal amounts, but their ratios do not change.
For the acid ionization reaction of a weak acid, an equilibrium is established between the undissociated acid (HA), hydrogen ions (or hydronium, H鈧僌鈦), and the conjugate base (A鈦). This dynamic process is crucial because it determines the strength of the acid and the pH of the solution, which is a measure of its acidity.
Equilibrium Constant
The equilibrium constant, denoted as K, quantifies the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced chemical equation.
In the context of acid-base reactions, the equilibrium constant is specifically called the acid dissociation constant, or Ka, for the ionization of an acid. It's calculated using the formula:
Ka = [H鈧僌鈦篯[A鈦籡/[HA],
where the square brackets indicate concentration. A larger Ka value indicates a stronger acid, as it means a greater proportion of acid molecules dissociate to form hydronium and the conjugate base ions.
pH and pOH Calculations
The pH of a solution is a logarithmic measure of the concentration of hydronium ions, which indicates its acidity. The pH is calculated as:
pH = -log[H鈧僌鈦篯.
pOH, on the other hand, is a measure of the hydroxide ion concentration, calculated by:
pOH = -log[OH鈦籡.
In aqueous solutions, pH and pOH are related through the expression
This relationship arises from the water dissociation constant (Kw), which at 25掳C is 1.0 脳 10-14. When working with solutions of salts, like NaA, which is the salt of a weak acid HA, pH can provide insight into the acid's ionization and thus its Ka.
Weak Acid Ionization
When a weak acid ionizes in water, it only partially dissociates into its ions. Unlike strong acids that completely dissociate, weak acids establish an equilibrium between the undissociated acid and its ions, described by the acid-ionization reaction:
HA 鈬 H鈧僌鈦 + A鈦
To analyze the ionization of weak acids, chemists often use an ICE table (Initial, Change, Equilibrium) to systematically track the concentrations of the reactants and products. The ionization constant Ka is then used to relate these concentrations at equilibrium. For instance, the parent acid HA of the salt NaA can be characterized by a specific Ka, which informs us about the tendency of HA to give up a proton in water and thus its strength as an acid.

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Most popular questions from this chapter

Addition of phenolphthalein to an unknown colorless solution does not cause a color change. The addition of bromthymol blue to the same solution leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) Which of the following can you establish about the solution: (i) A minimum \(\mathrm{pH}\), (ii) A maximum \(\mathrm{pH}\), or (iii) A specific range of \(\mathrm{pH}\) values? (c) What other indicator or indicators would you want to use to determine the \(\mathrm{pH}\) of the solution more precisely?

Although the acid-dissociation constant for phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) is listed in Appendix \(\mathrm{D}\), the base-dissociation constant for the phenolate ion \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}^{-}\right)\)is not. (a) Explain why it is not necessary to list both \(K_{a}\) for phenol and \(K_{b}\) for the phenolate ion. (b) Calculate \(K_{b}\) for the phenolate ion. (c) Is the phenolate ion a weaker or stronger base than ammonia?

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

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Calculate the \(\mathrm{pH}\) of a solution made by adding \(2.50 \mathrm{~g}\) of lithium oxide \(\left(\mathrm{Li}_{2} \mathrm{O}\right)\) to enough water to make \(1.500 \mathrm{~L}\) of solution.

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