What is #[HCl]# if a #17.04*mL# volume of #0.468*mol*L^-1# concentration of #[Ca(OH)_2]# was titrated with #13.7*mL# of said acid?

1 Answer
May 25, 2016

Answer:

Approx. #1* mol*L^-1#

Explanation:

We need (i) the balanced chemical equation:

#2HCl(aq) + Ca(OH)_2(aq) rarr CaCl_2(aq) + 2H_2O(l)#

And (ii) the amount in moles of calcium hydroxide.

#"Moles of calcium hydroxide"# #=# #17.04xx10^-3Lxx0.468*mol*L^-1# #=# #7.98xx10^-3*mol#

Twice this molar quantity of #HCl# was present in the #13.7*mL# #HCl(aq)#.

Thus #[HCl]=(2xx7.98xx10^-3*mol)/(13.7xx10^-3*L)# #=# #??mol*L^-1#

Note that the concentration of #Ca(OH)_2# is unreasonable....