In a certain titration, 34.5 mL of 0.20 M #NaOH# is required to neutralize 15.3 mL of #HCl#. What is the concentration of the acid (in normality and molarity)?

Question

2574 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-22T06:16:13

#NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)#

The equation demonstrates 1:1 stoichiometry: moles of sodium hydroxide are equivalent to moles of hydrochloric acid.

#"Moles of NaOH"#

#=# #34.5xx10^-3Lxx0.20*mol*L^-1=0.0069*mol#.

Given the stoichiometry, there were an equivalent quantity of moles in the given volume of #HCl#.

And thus #[HCl]=(0.0069*mol)/(15.3xx10^-3L)# #~=# #0.50*mol*L^-1#.

For a monoprotic acid such as hydrochloric acid,

#"molarity "=" normality"#.

Capisce?