# In a titration, 16.02 mL of 0.100M NaOH was required to titrate 0.2011 g of an unknown acid, HX. What is the molar mass of the acid?

Nov 8, 2015

We assume the acid to be monoprotic.

#### Explanation:

Rxn: $H X \left(a q\right) + N a O H \left(a q\right) \rightarrow {H}_{2} O \left(l\right) + N a X \left(a q\right)$

Therefore, moles of NaOH used $\equiv$ moles of HX, the unknown acid.

Moles of NaOH:$16.02 \times {10}^{- 3} L \times 0.100 \cdot m o l \cdot {L}^{- 1} = 16.02 \times {10}^{- 4} \cdot m o l$.

Thus your sample mass represents $16.02 \times {10}^{- 4} \cdot m o l$.

The molar mass is simply the quotient, $\frac{0.2011 \cdot g}{16.02 \times {10}^{- 4} \cdot m o l}$, about 120*g*mol^(-1)? At least I know it is right dimensionally, which is why I pfaff about with the units.