"In the titration of a solid acid, an endpoint is reached after 22.0mL of 0.120 M $\ce{NaOH}$ has been added. Assuming that each acid particle contains two acidic hydrogens and both are neutralized, what is the molar mass of the unknown acid?"

I really don't know how to solve this. I've figured out how to calculate the moles of acid were in the sample but don't see how to find the molar mass from that...any ideas?

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    $\begingroup$ From the data above, all you'll get is the moles. For molar mass calculation you need more data, like the weight of acid added, or the density of the solution. $\endgroup$ – udiboy1209 Aug 7 '13 at 15:21
  • $\begingroup$ If the acid is solid, then the final solution has the same density as pure water due to neutralization, so you basically know the volume and the total mass contained, and you know the mass of $\ce{NaOH}$ added; I think the book may be right... $\endgroup$ – TheVal Jan 20 '14 at 11:47

Your data is not enough to calculate molecular weight. You may have missed the mass of the solid acid you have weighed. Let's assume you have weighed $a$ gram of acid. Then: $Molecular Weight = \frac{Weight}{Moles_{acid}} = \frac{Weight}{2 \times Moles_{NaOH}} = \frac{Weight}{2 \times Molarity_{NaOH} \times Volume_{NaOH}} = \frac{a}{2 \times 0.120 \times \frac{22.0}{1000}}$

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  • $\begingroup$ That's what I thought, must be an error in the book. $\endgroup$ – jaykirby Aug 8 '13 at 4:12

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