What is the concentration of an HCl solution if 23.1 mL of HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2

Question

I cannot get the right answer

(Macic)(23.1mL)=(0.101M)(35.0mL)
Macid = .153

Answer 1

The problem is that the simplified formula you are using works ONLY for acids and bases that react 1:1. For example, in the titration of HCl and NaOH you have HCl + NaOH ==> NaCl + H2O and the formula you have shown works very well. In the case of HCl and Ba(OH)2; however, the reaction is
2HCl + Ba(OH)2 ==> BaCl2 + 2H2O
The best way to work these is to calulate moles of the base you have. That is M x L = 0.101 x 0.035 = 0.003535. Then using the coefficients to convert from moles Ba(OH)2 to moles HCl. It takes 2HCl to equal 1 Ba(OH)2 so moles HCl must be 2*0.0335 = 0.00707. Then M HCl = moles HCl/L HCl = 0.00707/0.0231 = ?
Many students don't like to use so many zeros (when using volumes in L --and I don't either) so I use millimoles.
millimoles = mL x M = ?
Then convert mmoles Ba(OH)2 to mmoles HCl. Finally M HC = mmoles/mL.
(Remember M = moles/L but if we multiply top and bottom by 1000, we get mmoles on top and mL on bottom)