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The following observations are made about a diprotic acid \(\mathrm{H}_{2} \mathrm{~A}\) : (i) \(\mathrm{A} 0.10 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{~A}\) has \(\mathrm{pH}=3.30\). (ii) \(\mathrm{A} 0.10 \mathrm{M}\) solution of the salt NaHA is acidic. Which of the following could be the value of \(\mathrm{p} K_{a 2}\) for \(\mathrm{H}_{2} \mathrm{~A}\) : (i) \(3.22\), (ii) \(5.30\), (iii) 7.47, or (iv) \(9.82 ?\)

Short Answer

Expert verified
Any of the provided pKa鈧 values could be the correct one for H鈧侫: (i) 3.22 (ii) 5.30 (iii) 7.47 (iv) 9.82

Step by step solution

01

Determine the concentration of H鈦 ions in the H鈧侫 solution

From the first piece of information, we know that the pH of the 0.10 M H鈧侫 solution is 3.30. We can use the pH formula to determine the concentration of H鈦 ions: pH = -log[H鈦篯 Rearranging the formula, we get: [H鈦篯 = 10^(-pH) Plugging in the given pH value: [H鈦篯 = 10^(-3.3) = 5.0 x 10^(-4) M
02

Calculate the Ka鈧 value from the H鈧侫 solution

Let's denote the change in concentration of H鈦 ions due to the dissociation of H鈧侫 as x. As H鈧侫 is a strong acid, we can assume that [H鈦篯=[A鈦籡=[x]. Since the initial concentration of H鈧侫 is 0.1 M, the equilibrium expression for Ka鈧 can be written as: Ka鈧 = [H鈦篯[A鈦籡/[H鈧侫] = (x)(x)/(0.1 - x) Since [H鈦篯 = 5.0 x 10^(-4) = x, we can simplify the equation: Ka鈧 = (5.0 x 10^(-4))^2 / (0.1 - 5.0 x 10^(-4)) = 2.50 x 10^(-4) Taking the negative log of Ka鈧, we can find the pKa鈧 value: pKa鈧 = -log(Ka鈧) = -log(2.50 x 10^(-4)) = 3.60
03

Determine the Ka鈧 value from the given pKa鈧 values

Calculate the pKa鈧 values for each given pKa鈧 value using the relationship: pKa鈧 + pKa鈧 = pKw (value of pKw is 14) Option (i): pKa鈧 = 14 - 3.22 = 10.78 Option (ii): pKa鈧 = 14 - 5.30 = 8.70 Option (iii): pKa鈧 = 14 - 7.47 = 6.53 Option (iv): pKa鈧 = 14 - 9.82 = 4.18
04

Check which pKa鈧 value correctly describes the acidity of the NaHA solution

A 0.10 M solution of the salt NaHA is acidic. An acidic solution occurs when the pKa鈧 value for the given pKa鈧 is larger than the actual pKa鈧 of H鈧侫. We calculated the pKa鈧 of H鈧侫 to be 3.60 in step 2. Now, let's compare this value to the pKa鈧 values calculated for each given pKa鈧: Option (i): pKa鈧 = 10.78 (> 3.60) Option (ii): pKa鈧 = 8.70 (> 3.60) Option (iii): pKa鈧 = 6.53 (> 3.60) Option (iv): pKa鈧 = 4.18 (> 3.60) Since all given pKa鈧 values result in pKa鈧 values that are larger than the actual pKa鈧 of H鈧侫, each of them would produce an acidic NaHA solution, as required by the information. Therefore, any of the provided pKa鈧 values could be the correct one for H鈧侫: (i) 3.22 (ii) 5.30 (iii) 7.47 (iv) 9.82

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Key Concepts

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

pH Calculation
Understanding the pH of a solution is crucial because it indicates how acidic or basic that solution is. The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
In the context of a diprotic acid \( H_{2}A \), determining the pH involves calculating the concentration of hydrogen ions \( [H^+] \) in the solution. From the exercise, we can derive that for a 0.10 M \( H_{2}A \) solution with a pH of 3.30, the concentration of hydrogen ions is \( 5.0 \times 10^{-4} \) M. This is found by taking the inverse log of the pH value:
\[ [H^+] = 10^{-pH} \]
This process is essential for understanding the solution's acidity and is a foundation for analyzing acid dissociation and the related constants.
Acid Dissociation Constant
The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution. It expresses the equilibrium between the acid (HA) and its ions (H鈦 and A鈦). For a diprotic acid like \( H_{2}A \) with two dissociable protons, there are two constants: Ka鈧 and Ka鈧 for the first and second dissociation steps, respectively. The formula to calculate Ka is:
\[ Ka = \frac{[H^+][A^-]}{[HA]} \]
During the first dissociation of a diprotic acid, Ka鈧 can be determined by the concentration of hydrogen ions squared, divided by the initial concentration of the acid minus the ion concentration, as indicated in the exercise solution. These constants help chemists understand how the acid will behave in different environments.
Ka and pKa Relationship
The relationship between Ka and pKa is another fundamental aspect of acid-base chemistry. The pKa value is simply the negative logarithm of the Ka value, which provides a more convenient way to express acid strength. The formula to find pKa is:
\[ pKa = -\log(Ka) \]
As seen in the exercise, once the Ka鈧 value is calculated, the corresponding pKa鈧 can be determined. As the magnitude of Ka decreases, indicating a weaker acid, the pKa value increases. Knowing the pKa helps predict the extent of dissociation of an acid at various pH levels, which is vital for controlling reactions in a laboratory or industrial setting.
Acidity of Salt Solutions
The acidity of salt solutions is determined by the nature of the ions in the solution. Salts derived from a weak acid and a strong base, like NaHA, can often form solutions with acidic pH because the conjugate base (A鈦) from the weak acid has the ability to accept a proton from water and generate additional hydrogen ions. This effect is related to the hydrolysis of the anion.

In the provided problem, we infer that since solutions of NaHA are acidic, Ka鈧 must be significantly smaller than Kw, the ion product constant for water, and the related pKa鈧 must be greater than 7 to result in a pH less than 7. By extension, since pKa鈧 is also fairly small, the acid's second dissociation step must be weaker than the first, as the given pKa鈧 options also imply.

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