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Chapter 20: Problem 462

(a) How many equivalents of base would be neutralized by one mole of phthalic acid? What is the neutralization equivalent of phthalic acid? (b) What is the relation between neutralization equivalent and the number of acidic hydrogens per molecule of acid? (c) What is the neutralization equivalent of \(1,3,5\) -benzenetricarboxylic acid? Of mellitic acid, \(\mathrm{C}_{6}(\mathrm{COOH})_{6}\) ?

Short Answer

Expert verified
(a) One mole of phthalic acid would be neutralized by two equivalents of base. The neutralization equivalent of phthalic acid is 83.07 g/mol. (b) The relation between neutralization equivalent and the number of acidic hydrogens per molecule of acid is: Neutralization equivalent = Molar mass / Number of acidic hydrogens. (c) The neutralization equivalent of 1,3,5-benzenetricarboxylic acid is 70.05 g/mol, and of mellitic acid (C6(COOH)6) is 43.03 g/mol.

Step by step solution

01

Identify the number of acidic hydrogens in phthalic acid

Phthalic acid has the molecular formula \( C_8H_6O_4 \), which can be written as \( C_6H_4(COOH)_2 \). It can be seen that phthalic acid has two acidic hydrogens (from the two carboxyl groups (COOH).
02

Calculate the equivalents of base required for neutralization of phthalic acid

Since there are two acidic hydrogens in phthalic acid, it will require two equivalents of base for complete neutralization.
03

Calculate the neutralization equivalent of phthalic acid

The neutralization equivalent is the mass of the acid that will react with one equivalent of the base. So, firstly we need to find the molar mass of phthalic acid: Molar mass of \( C_8H_6O_4 \) = (8 × 12.01) + (6 × 1.01) + (4 × 16.00) = 166.14 g/mol Since it takes two equivalents of base to neutralize one mole of phthalic acid, the neutralization equivalent is half of its molar mass: Neutralization equivalent of phthalic acid = Molar mass / Number of acidic hydrogens = 166.14 g/mol / 2 = 83.07 g/mol
04

Establish the relationship between neutralization equivalent and acidic hydrogens

From the above analysis, we can see that the neutralization equivalent is related to the molar mass of the acid and the number of acidic hydrogens it contains. Thus, the relationship can be given by: Neutralization equivalent = Molar mass / Number of acidic hydrogens
05

Calculate the neutralization equivalent of 1,3,5-benzenetricarboxylic acid

1,3,5-benzenetricarboxylic acid has the molecular formula \( C_9H_6O_6 \). It has three acidic hydrogens (from the three carboxyl groups, COOH). Molar mass of \( C_9H_6O_6 \) = (9 × 12.01) + (6 × 1.01) + (6 × 16.00) = 210.15 g/mol Using the relationship we established in Step 4: Neutralization equivalent of 1,3,5-benzenetricarboxylic acid = Molar mass / Number of acidic hydrogens = 210.15 g/mol / 3 = 70.05 g/mol
06

Calculate the neutralization equivalent of mellitic acid (C6(COOH)6)

Mellitic acid has the molecular formula \( C_6(COOH)_6 \). It has six acidic hydrogens (from the six carboxyl groups, COOH). Molar mass of \( C_6(COOH)_6 \) = (6 × 12.01) + (6 × 6 × 1.01) + (6 × 6 × 16.00) = 258.16 g/mol Using the relationship we established in Step 4: Neutralization equivalent of mellitic acid = Molar mass / Number of acidic hydrogens = 258.16 g/mol / 6 = 43.03 g/mol

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