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?

1 Answer
Nov 9, 2015

Answer:

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.