The acid-base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the \(\mathrm{pH}\) is \(4.68 .\) What is the \(\mathrm{p} K_{a}\) for bromcresol green?

Short Answer

Expert verified
The pKa of bromcresol green can be found using the Henderson-Hasselbalch equation: \(pH = pKa + \log \frac{[A^-]}{[HA]}\). Since the concentrations of the acid and its conjugate base are equal, the equation simplifies to \(4.68 = pKa + 0\). Therefore, the pKa of bromcresol green is \(4.68\).

Step by step solution

01

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \( pH = pKa + \log \frac{[A^-]}{[HA]} \) where: - pH is the acidity of the solution - pKa is the weak acid dissociation constant - [A^-] is the concentration of the conjugate base - [HA] is the concentration of the weak acid
02

Identify given values from the problem

We are given that the pH of the solution is 4.68 when the concentrations of the acid ([HA]) and its conjugate base ([A^-]) are equal. Therefore, we can set [HA] = [A^-] and substitute the pH value in the equation.
03

Substitute known values and solve for pKa

Plug the given values into the Henderson-Hasselbalch equation: \( 4.68 = pKa + \log \frac{[A^-]}{[HA]} \) Since [HA] = [A^-], the equation becomes: \( 4.68 = pKa + \log \frac{[A^-]}{[A^-]} \) Now, we simplify the equation. Since \(\log 1 = 0\), \( 4.68 = pKa + 0 \)
04

Determine pKa

Now we can find the pKa value: \( pKa = 4.68 \) The pKa of bromcresol green is 4.68.

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Key Concepts

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

Understanding Acid-Base Indicators
Acid-base indicators are compounds that change color in response to changes in pH, making them valuable tools in chemistry for determining the acidity or basicity of a solution. Indicators are themselves weak acids or bases which exist in two forms: one with a color that shows in acidic environments and another with a different color that shows in basic conditions.

For instance, bromcresol green, the indicator mentioned in the exercise, is yellow when it forms its acid (HA) and blue when it forms its conjugate base (A-). When the pH of the solution equals the pKa of the indicator, both forms are present in equal concentrations, which results in a distinct intermediate color. This is the principle used to deduce the pKa value for an acid-base indicator.
Deciphering pKa (Acid Dissociation Constant)
The pKa of a substance is a quantitative measurement that reflects the strength of an acid in solution. It is derived from the acid dissociation constant (Ka), which measures the ability of an acid to donate a proton (H+). The pKa is simply the negative base-10 logarithm of this constant: \( pKa = -\text{log}_{10}(Ka) \)

A lower pKa value indicates a stronger acid because it suggests a greater tendency to lose the proton. As seen in our example with bromcresol green, when the concentrations of an acid and its conjugate base are equal, the pH of the solution is equal to the pKa. This relationship provides an essential way to understand how different chemical species will behave under varying pH conditions.
Calculating pH
To calculate the pH of a solution, which is a measure of its acidity or basicity, we often use the Henderson-Hasselbalch equation as a tool for solutions involving weak acids and their salts (which form a conjugate base). The equation \( pH = pKa + \text{log} \frac{[A^-]}{[HA]} \) helps to determine the pH knowing the concentration of the acid and its conjugate base. In solutions where these concentrations are equal, as in a buffer system at its pKa, the log term becomes zero, and the pH is equal to the pKa. The exercise provided demonstrates a clear situation where this application is useful for determining the pKa of a weak acid; by knowing the pH when the acid and base forms are in equilibrium, the pKa value is readily obtained.
Properties of Weak Acids
Weak acids are characterized by their partial dissociation in solution, which contrasts with strong acids that dissociate completely. This dissociation behavior can be quantified by the acid dissociation constant (Ka) and consequently the pKa. A key property of weak acids is that they establish an equilibrium between the un-dissociated acid (HA) and its dissociate form (A- and H+).

Bromcresol green functions as a typical weak acid in the given textbook exercise. It demonstrates a fundamental behavior of weak acids—a balance between their acid and base forms at a specific pH level, which is the acid's pKa. It also exemplifies how weak acids create buffer solutions that can resist changes in pH upon the addition of small amounts of acids or bases, due to the presence of both the acid and its conjugate base.

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Most popular questions from this chapter

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2 .(\mathrm{a})\) Write the equilibrium equation and equilibrium-constant expression for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\). (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and \(B\) ? (c) What is the value of \(K_{b}\) for B?

Aspirin has the structural formula At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{a}\) for aspirin equals \(3 \times 10^{-5}\). If two aspirin tablets, each having a mass of \(325 \mathrm{mg},\) are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is \(2,\) what percent of the aspirin is in the form of neutral molecules?

(a) Calculate the percent ionization of \(0.0075 \mathrm{M}\) butanoic acid \(\left(K_{a}=1.5 \times 10^{-5}\right) .\) (b) Calculate the percent ionization of \(0.0075 \mathrm{M}\) butanoic acid in a solution containing \(0.085 \mathrm{M}\) sodium butanoate.

(a) A 0.1044 -g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid. Do both the molecular weight and \(K_{a}\) value agree with your choice?

Calculate the pH at the equivalence point for titrating \(0.200 \mathrm{M}\) solutions of each of the following bases with \(0.200 \mathrm{M} \mathrm{HBr}\) : (a) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\).

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