What is #[HCl]# if a #25*mL# volume of #[NaOH]# of #0.250*mol*L^-1# concentration reaches an endpoint with a #22.5*mL# volume of the acid?

1 Answer
Jul 8, 2017

Answer:

You mean to start with a solution of #NaOH(aq)#

Explanation:

#NaOH(aq)# reacts with #HCl(aq)# with 1:1 stoichiometry.......

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

Given that #"Concentration"="Moles of titrant"/"Volume of titrant"# we can work out the molar equivalence......

#"Moles of NaOH"# #=# #0.250*mol*L^-1xx25xx10^-3*L=6.25xx10^-3*mol#, and given the 1:1 stoichiometry, this was also the number of moles of base neutralized.

This molar quantity was equivalent to the amount of moles contained in #22.5*cm^3# #HCl# titrant..........so..........

#"Concentration"# #=# #(6.25xx10^-3*mol)/(22.5xx10^-3*L)#

#[HCl]=0.278*mol*L^-1#.........

Note that #1*L=1000*cm^3=1000*mL#...........