Question

1. Titration reveals that 11.60 mL of 3.0 M HCl are required to neutralize a 25.00 mL of Ba(OH)₂ What is the molarity of the Ba(OH)₂ solution?

1. 3 M
2. 2.784 M
3. 1.392 M
4. 0.696 M

Answer 1

Titration reveals that 11.60 mL of 3.om Hal are required to neutralize a 25.00 mL of Ba(OH) What is the rolanity of the BacoHla solution A 3M 3 2.784 M 11. 392M 0.696 M Answer D = 0.696M - Explanation given data Volume of Hcl = 11.60 mL (Y) Molarity of Hel = 3.0 M CMOJ volume of Bacolt), = 25.00 mL (V2) nolarity of Bacol), = ? (M2) Dilution principle mov, = M2Y₂ M2 = Mg = 3.0 x 11.60 25.00 34.8 25.00 M2= 1.392 MI Molarity of Bacot), = 1.392 M.

21 dl + BaCOH)- Back + 2 H20 2 moles I more Mole ratio of Hel and Bacolti as 2:1 Then. Go G96 moles of BaLOH)2 requied for neutralization of 3.0 moles of Hal, Then finally the molarity of BaCOH)2 = 0.696