3.34 g of an unknown acid is dissolved in water to make 250 mL solution. While performing the titration, 46.43 mL of the acid is needed to neutralize 7.41 mL of 0.50 M NaOH is a titration. What is the molar mass of the dry acid?
(Assume the acid is monoprotic)

Question

3.34 g of an unknown acid is dissolved in water to make 250 mL solution. While performing the titration, 46.43 mL of the acid is needed to neutralize 7.41 mL of 0.50 M NaOH is a titration. What is the molar mass of the dry acid?
(Assume the acid is monoprotic)

aaronq · Answer

You first need to find the molarity of the acid. Knowing the moles you can trace back to the molar mass.

You need these two equations:

$M=\dfrac{n_{solute}}{L_{solution}}$ and  $n=\dfrac{m}{MW}$

anonymous · Answer

Is the answer 167 g/mol? @aaronq

aaronq · Answer

i dont know, i havent worked it out. you're welcome to post your work if you'd like me to check it.

anonymous · Answer

Well using your hint. And some of the things that my teach taught me. I did:
$$M_{a}=?$$$$V_{a}=46.43 mL$$$$M_{b}=0.50M$$$$V_{b}=7.41mL$$
$$M_{a}*(46.43mL)=(0.50M)*(7.41mL)$$
$$M_{a}=0.080 M$$
$$\frac{ 0.080mol }{ 1.00L }=\frac{ 3.34g }{ 0.25L }* Xmol/g$$$$Xg/mol=\frac{ 3.34g*1.00L }{ 0.25L*0.080mol }$$
The molar mass of the unknown acid is 167 g/mol.
@aaronq

anonymous · Answer

teacher*

aaronq · Answer

$\checkmark$ that is correct, nice formatting too.

anonymous · Answer

:D Thank you for the help @aaronq

aaronq · Answer

no problem dude!