Question

What is the molarity of a solution of nitric acid if 0.216 of barium hydroxide is required to neutralize 20.00mL of nitric acid?

Answer 1

I assume that it is `0.216` `g` of barium hydroxide, and `20.00` `mL` nitric acid.

Explanation:

`0.216` `g` `Ba(OH)_2` `=` `(0.216*g)/(171.34*g*mol^-1)` `=` `1.26xx10^(-3)` `mol` `Ba(OH)_2`.

Now, of course I need a chemical equation:

`Ba(OH)_2(aq) + 2HNO_3 rarr Ba(NO_3)_2(aq) + 2H_2O(l)`.

Given the stoichiometry of the equation above (what do I mean here?), in `20.00` `mL` solution, there were `2.52xx10^(-3)` moles of nitric acid present.

So concentration of the `HNO_3(aq)` `=` `n/V` `=` `(2.52xx10^(-3) mol)/(20.00xx10^(-3) L)` `~=` `0.1*mol*L^-1`. You'll have to work out the actual figure; my calculator is never handy.