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Chapter 17: Problem 34

Solution (a) is \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) and solution (b) is \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\). A few drops of thymol blue indicator are added to each solution. What is the color of each solution? What is the color of the solution obtained when these two solutions are mixed?

Short Answer

Expert verified
Solution (a) will turn the thymol blue red. Solution (b) will turn the thymol blue blue. When the two solutions are mixed, the thymol blue will turn yellow/green.

Step by step solution

01

Identify the nature of solutions

First identify the nature of solutions. Here, solution (a) is hydrochloric acid (HCl), therefore it's acidic. Solution (b) is sodium acetate (NaCH3COO), a salt of a strong base and weak acid, so it is slightly basic.
02

Determine the color of the indicator in each solution

Thymol blue changes color based on the pH of the solution it is in. In acidic solutions (pH < 2.8), it is red. In neutral solutions (pH = 7), it is yellow/green and in basic solutions (pH > 8.2) it is blue. So, solution (a) will turn the thymol blue red because it's acidic. Solution (b), being slightly basic, will turn the thymol blue blue.
03

Determine the color of indicator when solutions are mixed

When these two solutions are mixed, they will neutralize each other as an acid reacts with a base to make a salt and water. The pH of the resulting solution will be about 7, which is neutral. So, upon mixing, the thymol blue will become yellow/green.

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Most popular questions from this chapter

Sulfuric acid is a diprotic acid, strong in the first ionization step and weak in the second \(\left(K_{\mathrm{a}_{2}}=1.1 \times 10^{-2}\right)\) By using appropriate calculations, determine whether it is feasible to titrate \(10.00 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) to two distinct equivalence points with \(0.100 \mathrm{M} \mathrm{NaOH}\)

The neutralization of \(\mathrm{NaOH}\) by \(\mathrm{HCl}\) is represented in equation (1), and the neutralization of \(\mathrm{NH}_{3}\) by HCl in equation (2). 1. \(\mathrm{OH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O} \quad K=?\) 2\. \(\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \quad K=?\) (a) Determine the equilibrium constant \(K\) for each reaction. (b) Explain why each neutralization reaction can be considered to go to completion.

Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH},\) is found in sour milk. A solution containing \(1.00 \mathrm{g} \mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) in 100.0 \(\mathrm{mL}\) of \(0.0500 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) has a \(\mathrm{pH}=4.11 .\) What is \(K_{\mathrm{a}}\) of lactic acid?

Calculate the change in pH that results from adding (a) \(0.100 \mathrm{mol} \mathrm{NaNO}_{2}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{2}(\mathrm{aq})\) (b) \(0.100 \mathrm{mol} \mathrm{NaNO}_{3}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}(\mathrm{aq})\) Why are the changes not the same? Explain.

The ionization constants of ortho-phthalic acid are \(K_{\mathrm{a}_{1}}=1.1 \times 10^{-3}\) and \(K_{\mathrm{a}_{2}}=3.9 \times 10^{-6}\) 1. \(\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{COOH})_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}\) 2\. \(\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{COO}^{-}\right)_{2}\) What are the pH values of the following aqueous solutions: (a) 0.350 M potassium hydrogen orthophthalate; (b) a solution containing 36.35 g potassium ortho-phthalate per liter? (f) \(0.68 \mathrm{M} \mathrm{KCl}, 0.42 \mathrm{M} \mathrm{KNO}_{3}, 1.2 \mathrm{M} \mathrm{NaCl},\) and \(0.55 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) with \(\mathrm{pH}=6.4\)
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