Question

2. (5 pts) You perform an acid-base titration to standardize an HCl solution by placing 50.00 mL of 0.100M HCI in a flask with a few drops of indicator solution. You put 0.1292 M Ba(OH)2 into the buret, how many mL of the Ba(OH)2 are necessary to titrate the HCI?

Answer 1

Consider a reaction, 2 HCl + Ba(OH) ₂ BaCl ₂+ 2 H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base = 2/1=2

We have correlation, M _acid$\times$ V _acid = M _base$\times$ V _base$\times$ stoichiometric ratio.

Therefore, 0.100 M  $\times$ 50.0 ml = 0.1292 M $\times$ V _base$\times$ 2

V _base= 0.100 M  $\times$ 50.0 ml / 0.1292 M $\times$ 2

V _base = 19.3 ml

ANSWER: Volume of Ba(OH) ₂ solution required for titration with 50.00 ml 0.100 M HCl  is 19.3 ml