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Chapter 4: Problem 128

A 6.50-g sample of a diprotic acid requires \(137.5 \mathrm{~mL}\) of a \(0.750\) \(M\) NaOH solution for complete neutralization. Determine the molar mass of the acid.

Short Answer

Expert verified
The molar mass of the diprotic acid is 126.02 \(\mathrm{g/mol}\).

Step by step solution

01

Write the Balanced Equation for the Reaction

Let's call our diprotic acid H2A. Then the balanced equation for the reaction between NaOH and the diprotic acid is: H2A + 2NaOH -> 2H2O + Na2A Note that a diprotic acid has two acidic protons, so the stoichiometry is 1 mol of the acid reacts with 2 mol of NaOH.
02

Calculate the Amount of NaOH in Moles

We need to find out how many moles of NaOH were used in the reaction. We are given the volume and concentration of the NaOH solution, so we can use the formula: Moles of NaOH = Volume x Concentration Moles of NaOH = \(137.5\mathrm{~mL}\) × 0.750 M Note that we need to convert the volume of NaOH into liters so that the units are matched: Moles of NaOH = \(0.1375\mathrm{~L}\) × 0.750 M = 0.103125 mol
03

Calculate the Amount of the Acid in Moles

Next, we need to convert the moles of NaOH to moles of the diprotic acid using the stoichiometric coefficients in the balanced equation. We can use the ratio of moles of H2A to moles of NaOH: Moles of H2A = Moles of NaOH × \(\frac{1 \ \mathrm{mol\ H2A}}{2 \ \mathrm{mol\ NaOH}}\) Moles of H2A = \(0.103125\mathrm{~mol}\) × \(\frac{1 \ \mathrm{mol\ H2A}}{2 \ \mathrm{mol\ NaOH}}\) = 0.0515625 mol
04

Calculate the Molar Mass of the Acid

Now that we have the moles of H2A, we can use the given mass of the sample to find the molar mass, using the formula: Molar Mass = \(\frac{\mathrm{Mass\ of\ a\ Sample}}{\mathrm{Moles}}\) Molar Mass = \(\frac{6.50\mathrm{~g}}{0.0515625 \mathrm{~mol}}\) = 126.02 \(\mathrm{g/mol}\) So, the molar mass of the diprotic acid is 126.02 \(\mathrm{g/mol}\).

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