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Chapter 26: Problem 644

A small amount of Methyl Orange is added to a solution containing equimolar amounts of acetic acid and sodium acetate. Is this solution yellow or red?

Short Answer

Expert verified
The solution containing equimolar amounts of acetic acid and sodium acetate has a pH of approximately 4.74. Since Methyl Orange is red in acidic solutions (pH < 3.1) and yellow in basic solutions (pH > 4.4), and the calculated pH is higher than 4.4, the Methyl Orange indicator in this solution will be yellow.

Step by step solution

01

Write the equilibrium expression for acetic acid

We first need to write down the equilibrium expression for the weak acid (acetic acid), to have an idea of the ions present in the solution. Acetic acid (CH3COOH) dissociates into acetate ions (CH3COO-) and hydrogen ions (H+) as follows: \[CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}\]
02

Find the acid dissociation constant, Ka, for acetic acid

We now need to find the acid dissociation constant, Ka, for acetic acid. Ka is a measure of the strength of an acid in solution. The higher the Ka value, the stronger the acid. For acetic acid, the Ka is approximately 1.8 x 10^{-5}.
03

Use the Henderson-Hasselbalch equation to determine the pH of the solution

The Henderson-Hasselbalch equation relates the pH of a solution to the ratio of the concentrations of the undissociated acid ([HA]) and its conjugate base ([A-]), and the acid dissociation constant, Ka. The equation is given as: \[pH = pKa + log\left(\frac{[A^{-}]}{[HA]}\right)\] We are given that the solution contains equimolar amounts of acetic acid and sodium acetate, meaning the concentrations of CH3COOH and CH3COO- are equal. Thus the ratio of their concentrations is 1, which makes the logarithm term equal to zero. Plugging in the values, we have: \[pH = -log(1.8 \times 10^{-5})\]
04

Calculate the pH of the solution

Now we can calculate the pH of the solution: \[pH = -log(1.8 \times 10^{-5}) \approx 4.74\]
05

Determine the color of the Methyl Orange

With the pH of the solution calculated in Step 4, we can determine the color of the Methyl Orange indicator. Since Methyl Orange is red in acidic solutions (pH < 3.1) and yellow in basic solutions (pH > 4.4), and our calculated pH is approximately 4.74, which is higher than 4.4, the Methyl Orange indicator will be yellow in this solution.

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Most popular questions from this chapter

A compound \(\mathrm{A}\) has the molecular formula \(\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{BrClO}\). Treatment of A With diazomethane gave B, which upon heating with silver oxide and water was converted to another compound, C, \(\mathrm{C}_{8} \mathrm{H}_{7} \mathrm{BrO}_{2}\). The nmr spectrum of \(\mathrm{C}\) is as follows: \(\delta 2.7\) singlet, \(\mathrm{A}=2 ; \delta 7.0 \mathrm{multiplet}, \mathrm{A}=4 ; \delta 12.0\) singlet, \(\mathrm{A}=1 .\) What are the structures of compounds \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) ?

N, N, 4- Trimethylaniline has \(\mathrm{K}_{\mathrm{B}}\) of \(3 \times 10^{-9}\), quinuclidine has \(\mathrm{K}_{\mathrm{B}}=4 \times 10^{-4}\) and benzoquinuclidine has \(\mathrm{K}_{\mathrm{B}}=6 \times 10^{-7}\). What conclusions may be drawn from these results as to the cause(s) of the reduced base strength of aromatic amines relative to saturated aliphatic or alicyclic amines? Explain.

Suggest the syntheses of (a) \(\beta\) -naphthol from naphthalene (b) Orange II from \(\beta\) -naphthol and sulfanilic acid.

Diazotization of \(2,4-\) dinitroaniline in aqueous solution is accompanied by some conversion to phenols in which a nitro group is replaced by a hydroxy group. Give a reasonable mechanism for this reaction.

(a) Besides destabilizing the anilinium ion, how else might a nitro group affect basicity? (b) Why does the nitro group exert a larger base-weakening effect from the para position than from the nearer meta position?
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