A 30 ml volume of HCl is titrated with 23 mL of 0.20 M NaOH. How would you calculate the molarity of HCl in this solution?

1 Answer
Dec 3, 2015

Correctly! And according to the equation:

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

Explanation:

Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of #HCl# was dissolved in a #30xx10^-3L# volume.

Volume (#L#) #xx# concentration (#mol*L^-1#) gives an answer in #mol#. Note that #1# #mL# #=# #1xx10^-3*L#

So #(23xx10^-3cancelLxx0.20*mol*cancel(L^-1))/(30xx10^-3L)# #=# #???mol*L^-1#