# Question 02722

May 21, 2017

f.wt. = 125 g/mole

#### Explanation:

Assume the following acid/base rxn

HA + NaOH => NaA + HOH
Rxn is 1:1 mole ratio of HA:HOH which means that moles of acid will be completely neutralized by an equal number of moles of base. That is ...

moles acid neutralized = moles of base used

moles = mass(g)/formula wt and in solution, moles = Molarity x Volume in Liters

Given:
$\text{moles" of HA = 0.25 "grams" / "f.wt. of acid}$
moles of NaOH = $\left(\frac{0.20 m o l}{\mathrm{dL}} ^ 3\right) \left(0.010 L i t e r s\right)$ = $\left(\frac{0.20 m o l}{L}\right) \left(0.010 L\right)$ = $0.002 \text{mole}$

using moles acid = moles base

$\frac{0.25 g r a m s}{f . w t a c i d}$ = $\left(0.002 \text{mole}\right)$

f.wt acid = (0.25 g)/((0.002"mole")# = $125 \frac{g}{\text{mol}}$