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

Short Answer

Expert verified
The pKa for bromcresol green is 4.68, which is calculated using the Henderson-Hasselbalch equation: \(pH = pKa + log\frac{[A^-]}{[HA]}\). Given that the acid and base forms of the indicator are present in equal concentrations at pH 4.68, the ratio \(\frac{[A^-]}{[HA]}\) is equal to 1, simplifying the equation to \(4.68 = pKa\).

Step by step solution

01

Understand the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH, pKa, and the ratio of conjugate base concentration ([A-]) to weak acid concentration ([HA]): pH = pKa + log\(\frac{[A^-]}{[HA]}\) In this problem, we know the pH and the concentrations of the conjugate base and weak acid forms are equal, which makes the ratio \(\frac{[A^-]}{[HA]}\) equal to 1.
02

Insert the given values into the equation

Now, substitute the given pH value and the equal concentration ratio into the Henderson-Hasselbalch equation: \(4.68 = pKa + log\frac{[A^-]}{[HA]}\) Since \(\frac{[A^-]}{[HA]}\) = 1, we can write the equation as: \(4.68 = pKa + log(1)\)
03

Calculate the pKa value

We know that log(1) = 0, so our equation simplifies to: \(4.68 = pKa\) The pKa for bromcresol green is 4.68.

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Most popular questions from this chapter

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is 0.010\(M\) in \(\mathrm{Ba}^{2+}(a q)\) and 0.010\(M\) in \(\mathrm{Sr}^{2+}(a q) .\) (a) What concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. \(\mathrm{BaSO}_{4} : K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}\): \(K_{s p}=3.2 \times 10^{-7} . )\) (b) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) when the second cation begins to precipitate?

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