# Calculation of Titration problems - Should be Easy?

## Homework Statement

1. Calculate the volume (mL) of 0.100 M H3PO4 required to neutralize 30.0 mL of 0.050 M

2. If 0.2178 g of a diprotic solid acid is neutralized by 44.81 mL of 0.0953 M NaOH, calculate the molar mass of the solid acid.

## Homework Equations

Molar Mass = Mass/Moles
Concentration = Moles/Volume(mL)
Neutralization --> Moles of H+ = Moles of OH-

## The Attempt at a Solution

First Problem
For this problem, I've already done work to arrive to my answer, I just require someone else's opinion on whether it is right or not. I also felt lazy to type out the work for this one, since this problem is not that important and is easier compared to the second one (which is why I showed work for the second problem and not this one, hopefully you guys understand).

The volume I derived from the problem was that you needed 10 mL of H3PO4; is this correct?

Second Problem

For the second problem, I will show work:

44.81 mL x 0.0953 M = 4.27 moles of NaOH = 4.27 moles of OH-

4.27 moles of OH- imply that you need 4.27 moles of H+ for neutralization

Because the solid acid is diprotic, that means there are two acidic hydogens, which means the chemical formula for the acid is H2X, X being the remaining part of the acid's formula.

To get 4.27 moles of H+, then you would need (4.27/2), or 2.14 moles of H2X?

Since there was 0.2178 g of the diprotic solid acid, then the molar mass would be mass/moles, or 0.2178 g/ 2.14 moles = 0.101 g/mol

At this point, I got confused because the molar mass was so small; can anyone confirm whether or not this is truly the right answer?

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Borek
Mentor
1. Calculate the volume (mL) of 0.100 M H3PO4 required to neutralize 30.0 mL of 0.050 M
Of what? Your answer can be right, can be wrong, impossible to tell.

44.81 mL x 0.0953 M = 4.27 moles of NaOH = 4.27 moles of OH-
Your approach is correct in general, but 44.81 mL of this solution doesn't contain 4.27 moles.

Oh I see, I completely missed the point where molarity is moles/volume(L) and not mL. This gave me, for the first problem, 10 mL still and for the second problem, 102 g/mol.