Question

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?

Answer 1

`0.8*g` of sodium hydroxide are required.

Explanation:

You have the stoichiometric equation. There is clearly 1:1 equivalence. Thus `"moles of hydrochloric acid"` `=` `"moles of sodium hydroxide"`.

`"Moles of hydrochloric acid"` `=` `10.0xx10^-3Lxx2.0*mol*L^-1` `=` `0.020` `mol` `HCl`.

`"Mass of sodium hydroxide req'd"` `=` `0.020*molxx40.00*g*mol^-1` `=` `0.80*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.