Someone posted this but i still dont understand why you cant use the M1xV1=M2xV2 formula :(
Question #1
If 23.67 mL of a solution of Ba(OH)2 requires 15.05 mL of a 1.082 M solution of HNO3 for complete titration, what is the molarity of Ba(OH)2 solution? answer in units of M
"First, write the balanced equation:
2 HNO3 + Ba(OH)2 → Ba+2aq + NO3-aq + 2 H2O
Figure moles of nitric acid used: 15.05 mL HNO3 x (1 L / 1000 mL) x (1.082 mole/L) = 0.016 moles HNO3
Use the stoichiometric ratio in the eqn: 0.016 moles HNO3 x ( 1 mole Ba(OH)2/ 2 moles HNO3) =
8.142 x 10-3 moles Ba(OH)2
Figure the concentration of the Ba(OH)2: (8.142 x 10-3 moles Ba(OH)2 / 23.67 mL) x (1000 mL/L) = 0.344 M Ba(OH)2
check: 23.67 mL x (1 L / 1000 mL) x (0.344 mole/L) = 8.143 x 10-3 moles
So, the answer is that the concentration of the Ba(OH)2 soln is 0.344 M"