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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 pH is \(4.68 .\) What is the \(p K_{a}\) for bromcresol green?

Short Answer

Expert verified
The \( pK_a \) for bromcresol green is 4.68.

Step by step solution

01

Understanding the Problem

We are given that the yellow acid and blue base forms of bromcresol green are present in equal concentrations at a pH of 4.68. This means that this is equivalent to the pH at which the indicator changes color.
02

Using Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \( pH = pK_a + \log\left( \frac{[Base]}{[Acid]} \right) \). Since the concentrations of the base and acid forms are equal, \( \log\left( \frac{[Base]}{[Acid]} \right) = 0 \).
03

Simplifying the Equation

Given the condition \( \log(1) = 0 \), the Henderson-Hasselbalch equation simplifies to \( pH = pK_a \).
04

Finding \( pK_a \)

Since we established that \( pH = pK_a \) at this point, the \( pK_a \) of bromcresol green is equal to 4.68.

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Key Concepts

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

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a vital tool in chemistry. It describes the relationship between the pH of a solution and the pKa of the acid in the solution. This equation is especially useful when dealing with buffer solutions and acid-base indicators, such as bromcresol green. The equation is given by:

\[pH = pK_a + \log\left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]

This formula allows us to calculate the pH of a solution when we know both the pKa and the ratio of concentrations of the base and acid forms. It’s particularly helpful for understanding how indicators work — substances that change color at a specific pH threshold. In the case of bromcresol green, knowing that the concentrations of the yellow acid and blue base forms are identical when pH equals 4.68, we are easily able to use this equation to find out more about the indicator's properties.
pKa
The pKa is a measure of the strength of an acid in solution. It stands for the negative base-10 logarithm of the acid dissociation constant (Ka). The pKa value can tell you how readily a compound will give up its proton (H+ ion) to the solution.

- **Lower pKa:** Stronger acid, easily donates proton. - **Higher pKa:** Weaker acid, less eager to donate proton.

    In the problem with bromcresol green, we find that its pKa is 4.68 because the solution's pH is precisely where the acid and base forms of the indicator are equal. At this point, bromcresol green partially exists in each form, acting as a perfect indicator for changes around this pH value. Therefore, understanding pKa helps predict the behavior of acids in different environments, which is crucial for utilizing indicators effectively.
Bromcresol Green
Bromcresol green is a popular acid-base indicator used in various scientific fields, particularly in titrations and experiments that measure pH. It is known for its clear color change from yellow to blue, allowing it to easily indicate the acidic or basic nature of a solution. Here's how it works:

- **In Acidic Solutions:** Bromcresol green appears yellow. - **In Neutral to Basic Solutions:** It shifts towards blue.
This property is due to the molecular structure of bromcresol green and the balance between its protonated (acidic) and deprotonated (basic) forms. It shines in applications where precise pH determination is essential, such as in medical labs or during chemical education. Remember that the pKa of bromcresol green is 4.68, meaning it changes from yellow to blue as the pH transitions near this value. Thus, its color change points are highly valuable for experiments that require exact pH monitoring. Mastering its use can help improve the accuracy of many pH-related procedures.

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Most popular questions from this chapter

A solution containing several metal ions is treated with dilute HCl; no precipitate forms. The pH is adjusted to about \(1,\) and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8 . Again, \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). No precipitate forms. Which of these metal cations are either possibly present or definitely absent: \(\mathrm{Al}^{3+}, \mathrm{Na}^{+}, \mathrm{Ag}^{+}, \mathrm{Mg}^{2+} ?\)

Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{BaCrO}_{4}, \mathrm{CuS}, \mathrm{PbCl}_{2}\) and \(\mathrm{LaF}_{3}\).

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{~A}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and \(\mathrm{a}\) \(1.0 \mathrm{M}\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH}\) \(6.50 ?\) (Ignore any volume change.)

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

What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(111.5 \mathrm{kPa}\) ? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}-\mathrm{kPa} .\)

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