# Diprotic Acid Titration: Determining Molar Mass

Chapter 14, Problem 98P

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QUESTION:

A 0.105-g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1288 M NaOH. The equivalence point is reached after adding 15.2 mL of base. What is the molar mass of the unknown acid?

QUESTION:

A 0.105-g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1288 M NaOH. The equivalence point is reached after adding 15.2 mL of base. What is the molar mass of the unknown acid?

Step 1 of 3

Calculate the moles of NaOH used in the titration:

Given Molarity of NaOH = 0.1288 M

$$\text { Volume of } \mathrm{NaOH}=15.2 \mathrm{mL}=15.2 / 1000 \mathrm{L}=0.0152 \mathrm{L}$$

$$\begin{array}{l} \text { Moles of } \mathrm{NaOH}=\text { Molarity } \times \text { Volume } \\ =0.1288 \mathrm{mol} / \mathrm{L} \times 0.0152 \mathrm{L} \\ =0.00195616 \mathrm{mol} \end{array}$$

##### Diprotic Acid Titration: Determining Molar Mass

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Discover how to find the molar mass of an unknown diprotic acid through a titration experiment with 0.1288 M NaOH. Learn step by step in this chemistry tutorial.