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Chapter 17: Problem 86

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 \(\mathrm{p} K_{a}\) for bromcresol green?

Short Answer

Expert verified
The \(\mathrm{p} K_{a}\) for bromcresol green is 4.68, as determined by using the Henderson-Hasselbalch equation with the given pH and equal concentrations of the acid and base forms of the indicator.

Step by step solution

01

Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is as follows: \(\mathrm{pH} = \mathrm{p} K_{a} + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) where \(\mathrm{pH}\) is the pH of the solution, \(\mathrm{p} K_{a}\) is the pKa of the weak acid, \([\mathrm{A}^{-}]\) is the concentration of the conjugate base, and \([\mathrm{HA}]\) is the concentration of the weak acid. In this exercise, the acid and base forms of bromcresol green are present in equal concentrations, so \([\mathrm{A}^{-}] = [\mathrm{HA}]\).
02

Set up the equation with the given values

Now, let's plug in the given values into the Henderson-Hasselbalch equation: \(\mathrm{pH} = \mathrm{p} K_{a} + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) \(4.68 = \mathrm{p} K_{a} + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) Since \([\mathrm{A}^{-}] = [\mathrm{HA}]\), the fraction in the equation becomes: \(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]} = \frac{[\mathrm{HA}]}{[\mathrm{HA}]} = 1\)
03

Calculate the pKa of bromcresol green

Now the equation becomes: \(4.68 = \mathrm{p} K_{a} + \log_{10}(1)\) Since \(\log_{10}(1) = 0\), we have: \(4.68 = \mathrm{p} K_{a}\) So the \(\mathrm{p} K_{a}\) for bromcresol green is 4.68.

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Most popular questions from this chapter

Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\) ) reacts with acids in foods to form carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right),\) which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(g)\) forms bubbles and causes the cake to rise. \((\mathbf{a})\) A rule of thumb in baking is that \(1 / 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\). Write the chemical equation for this neutralization reaction. (b) The density of baking soda is \(2.16 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of mol/L. (One cup \(=236.6 \mathrm{~mL}=48\) teaspoons \() .(\mathbf{c})\) If \(1 / 2\) teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at a pressure of \(101.3 \mathrm{kPa}\), in an oven set to \(177^{\circ} \mathrm{C}\).

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{+}\), (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Mn}^{2+} .\) Suggest how each mixture might be separated.

Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathbf{a})\) a solution formed by adding \(30.0 \mathrm{~g}\) of furoic acid and \(25.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.300 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) $\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\( and \)30.0 \mathrm{~mL}\( of \)0.250 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$ and diluting the total volume to \(125 \mathrm{~mL},(\mathbf{c})\) a solution prepared by adding $25.0 \mathrm{~mL}\( of \)1.00 \mathrm{M} \mathrm{NaOH}\( solution to \)100.0 \mathrm{~mL}\( of \)0.100 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$

A 1.00-L solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{PbI}_{2}\). Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\).

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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