[johnnyjohn993]

What volume of 0.1292 M of Ba(OH)2 would neutralize 50.00 mL of HCl solution standardized in the sample problem above?

my solution:

Ba(OH)₂ + 2HCl  BaCl₂ +H₂O
Let x be the volume of  Ba(OH)₂

moles of Ba(OH)₂ : x L sol'n  *  0.1292 mol Ba(OH)₂
                                                           __________________________
                                                                         1 L sol'n
= 0.1293 (x) mol  Ba(OH)₂

Molarity of HCl :   0.1293 (x) mol  Ba(OH)₂ *  2mol HCl
                                                                    _____________
                                                                              1mol Ba(OH)₂

      0.2584 (x) mol HCl 
=   __________________ =  0.1292  M HCl
         0.05 L

x = 0.025 L of 0.1292 M  Ba(OH)₂


QUESTIONs:
1) what molarity of HCl should I use, coz i used the same molarity of Ba(OH)₂ to calculate the volume, is that wrong ?
2) what does it mean to neutralize HCl is there a specific molarity of neutralize HCl ?

[Borek]

HCl solution standardized in the sample problem above?

There is no problem above.

0.1292 mol Ba(OH)₂

Where you got it from? I am not saying it is wrong, I am not saying it is right, I am saying - it came from nowhere.