titration - chemistry

Question

To titrate 1.18 g of a dicarboxylic acid COOH-(CH2)n-COOH to the second equivalence point, 25 ml of a 0.80mol/L NaOH solution is needed. What is the value of n?

answer is 2

Answer 1

To find n, we first need to find the MOLES of COOH-(CH2)n-COOH. Once we find the moles, and knowing the mass (1.18g) we can determine the molar mass and hence, n.

moles COOH-(CH2)n-COOH:

COOH-(CH2)n-COOH + 2 NaOH ==> COONa-(CH2)n-COONa + 2H2O

mols NaOH = 25 ml x 1 L/1000 ml x 0.80 mol/L = 0.02 mols NaOH

mols COOH-(CH2)n-COOH = 0.02 mols NaOH x 1 mol COOH-(CH2)n-COOH / 2 mol NaOH = 0.01 mols

molar mass of COOH-(CH2)n-COOH = 1.18 g / 0.01 mols = 118 g/mol

How many -CH2 units do we need to add to 2 COOH units to get a molar mass of 118?

Each -COOH unit = 45 g for a total of 90 g

118 - 90 = 28 g

each -CH2 unit = 14

# of -CH2 units needed = 2 (2x14 = 28)

So, n = 2

COOH-(CH2)₂-COOH molar mass = 118