# In this acid base reaction: NaOH+HCl->NaCl+H_2O, if we measured out 10.0 mL of a 2.0 M solution of HCl, how many grams of sodium hydroxide(NaOH; MW=40 g/mol) would we need to neutralize the HCl solution?

Sep 27, 2016

$0.8 \cdot g$ of sodium hydroxide are required.

#### Explanation:

You have the stoichiometric equation. There is clearly 1:1 equivalence. Thus $\text{moles of hydrochloric acid}$ $=$ $\text{moles of sodium hydroxide}$.

$\text{Moles of hydrochloric acid}$ $=$ $10.0 \times {10}^{-} 3 L \times 2.0 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.020$ $m o l$ $H C l$.

$\text{Mass of sodium hydroxide req'd}$ $=$ $0.020 \cdot m o l \times 40.00 \cdot g \cdot m o {l}^{-} 1$ $=$ $0.80 \cdot g$.

It would be hard to measure such a quantity of sodium hydroxide as it is deliquescent, and once removed from the tin (or if the tin is poorly sealed), it would absorb moisture from the atmosphere.