The pH of a \(0.02 \mathrm{M}\) solution of an acid was measured at 4.6. a. What is the \(\left[\mathrm{H}^{+}\right]\) in this solution? b. Calculate the acid dissociation constant \(K_{\mathrm{a}}\) and \(\mathrm{p} K_{\mathrm{a}}\) for this acid.

Short Answer

Expert verified
a. The hydrogen ion concentration [H+] in the solution is \(2.51 \times 10^{-5}\) M. b. The acid dissociation constant Ka is \(1.26 \times 10^{-9}\) and the pKa is 8.90.

Step by step solution

01

Calculate [H+] from pH

From the definition of pH = -log[H+], rearrange for H+ to find H+ = 10^-pH. Substitute the given pH value (4.6) in this equation to calculate the hydrogen ion concentration.
02

Calculate Ka

The Ka for a weak acid dissociation can be calculated using [H+] and the initial concentration of the acid. We know that Ka = [H+]^2/[HA]. Given that the initial concentration of the acid (HA) is 0.02 M and we have calculated [H+] in the previous step, we can substitute both values into the equation to calculate Ka.
03

Calculate pKa

The pKa is calculated as the negative log base 10 of the Ka value. Therefore, once Ka is calculated, use pKa = -log(Ka) to calculate the pKa of the acid.

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Key Concepts

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

pH Calculation
Understanding how to calculate the pH of a solution is essential for numerous scientific disciplines, particularly chemistry and biology. pH is a measure of the acidity or basicity of an aqueous solution, which can impact reactions, living organisms, and environmental factors. It is expressed on a logarithmic scale ranging from 0 to 14; a pH less than 7 indicates an acidic solution, a pH of 7 is neutral, and a pH greater than 7 indicates a basic solution.

To calculate pH, we use the formula:
\r\[\text{pH} = -\log\left(\left[\mathrm{H}^+\right]\right)\]
where \(\left[\mathrm{H}^+\right]\) represents the hydrogen ion concentration in moles per liter. In our exercise, we took the provided pH value of 4.6 and reversed the calculation to find the hydrogen ion concentration. This foundational step allows us to delve further into understanding the properties of the acidic solution in question.
Hydrogen Ion Concentration
The concentration of hydrogen ions \(\left[\mathrm{H}^+\right]\) in a solution is a critical parameter determining the solution's pH level. These hydrogen ions are what make a solution acidic. The higher the concentration of hydrogen ions, the lower the pH and the more acidic the solution is.

In our textbook example, we used the pH value to find the hydrogen ion concentration by transforming the pH equation to:\[\left[\mathrm{H}^+\right] = 10^{-\text{pH}}\]
By inserting the pH value of 4.6 into this formula, we determine the concentration of the hydrogen ions in the solution. This value is crucial as it is also used to compute the acid dissociation constant for the acid in the solution.
pKa Calculation
The calculation of the pKa value is a pivotal concept in acid-base chemistry, as it provides insights into the strength of an acid. It is defined as the negative logarithm of the acid dissociation constant (Ka):\[\text{pKa} = -\log(K_a)\]
This value is essential because it helps ascertain how much an acid dissociates in a solution. A low pKa value suggests a strong acid that readily gives off hydrogen ions, while a high pKa indicates a weaker acid.

In the provided exercise, we first had to determine the Ka using the calculated \(\left[\mathrm{H}^+\right]\) and the initial concentration of the acid. With Ka calculated, we could then move on to find the pKa by simply taking the negative logarithm of the Ka value, giving us an understanding of the acid's dissociation characteristics in solution.
Acid-Base Equilibrium
The concept of acid-base equilibrium is related to the reversible reaction of acids donating hydrogen ions to bases. At equilibrium, the rate at which the acid dissociates to form hydrogen ions is equal to the rate at which these ions recombine to form the acid. The position of the equilibrium can be quantified by the Ka, which is the acid dissociation constant.

\rThe generic equilibrium for a weak acid HA in water is:
\r\[\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-\]
The equilibrium expression for the dissociation of a weak acid is given by:
\r\[K_a = \frac{\left[\mathrm{H}^+\right]\times\left[\mathrm{A}^-\right]}{\left[\mathrm{HA}\right]}\]
In the problem provided, assuming the concentration of \(A^-\) is approximately equal to \(\left[\mathrm{H}^+\right]\) due to the 1:1 dissociation, we can then conclude that the Ka calculated reflects the extent of dissociation for this particular acid at the given concentration.

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Most popular questions from this chapter

The \(K_{\mathrm{a}}\) for formic acid is \(1.78 \times 10^{-4} M.\) a. What is the pH of a \(0.1 \mathrm{M}\) solution of formic acid? b. \(150 \mathrm{mL}\) of \(0.1 \mathrm{MNaOH}\) is added to \(200 \mathrm{mL}\) of \(0.1 \mathrm{M}\) formic acid, and water is added to give a final volume of 1 L. What is the pH of the final solution?

Tris-hydroxymethyl aminomethane (TRIS) is widely used for the preparation of buffers in biochemical research. Shown here is the structure of TRIS in its protonated form: Its acid dissociation constant, \(K_{\mathrm{a}},\) is \(8.32 \times 10^{-9} M .\) You have available at your lab bench a \(0.1 \mathrm{M}\) solution of TRIS in its protonated form, 0.1 \(M\) solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), and ample distilled water. Describe the preparation of a 1 L solution of 0.02 M TRIS buffer, pH 7.8.

When a \(0.1 \mathrm{M}\) solution of a weak acid was titrated with base, the following results were obtained: $$\begin{array}{cc}\begin{array}{c}\text { Equivalents of } \\\\\text { base added }\end{array} & \text { pH observed } \\\\\hline 0.05 & 3.4 \\\0.15 & 3.9 \\\0.25 & 4.2 \\\0.40 & 4.5 \\\0.60 & 4.9 \\\0.75 & 5.2 \\\0.85 & 5.4 \\\0.95 & 6.0\end{array}$$ Plot the results of this titration and determine the \(\mathrm{p} K_{\mathrm{a}}\) of the weak acid from your graph.

a. Draw the titration curve for Bicine, assuming the \(\mathrm{p} K_{\mathrm{a}}\) for its free COOH group is 2.3 and the \(\mathrm{p} K_{\mathrm{a}}\) for its tertiary amino group is 8.3 b. Draw the structure of the fully deprotonated form (completely dissociated form) of bicine. c. You have available a \(0.1 ~ M\) solution of Bicine at its isoelectric point \(\left(\mathrm{pH}_{\mathrm{I}}\right), 0.1 \mathrm{M}\) solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\), and ample distilled \(\mathrm{H}_{2} \mathrm{O}\) Describe the preparation of 1 L of 0.04 M Bicine buffer, pH 7.5 d. What is the concentration of fully protonated form of Bicine in your final buffer solution?

Given \(0.1 M\) solutions of acetic acid and sodium acetate, describe the preparation of \(1 \mathrm{L}\) of \(0.1 \mathrm{M}\) acetate buffer at a pH of 5.4.

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