THis is the formula I used: M2=V2/M1V1

While this is technically correct, "using" "special formula" leads nowhere in this case - as it doesn't reflect stoichiometry of the reaction, thus it won't work for 2

^{nd} and 3

^{rd} question. But all these questions are identical and easy - they are just stoichiometry problems that you solve using balanced reaction equations.

Let's see how it works for the Ba(OH)

_{2} and KHP. First - balanced reaction equation:

Ba(OH)

_{2} + 2KHC

_{8}H

_{4}O

_{4} -> BaC

_{8}H

_{4}O

_{4} + 2H

_{2}O + K

_{2}C

_{8}H

_{4}O

_{4}For us now the only important conclusion of this equation is that barium hydroxide reacts with hydrogen phtalate 1:2 - 1 mole of hydroxide with 2 moles of hydrogen phtalate. You were told there was 1.190g of potassium hydrogen phtalate - convert it to moles (it should give you 9.304x10

^{-3} mole). As they react 1:2 hydrogen phtalate reacts with half the amount of barium hydroxide - so there was 9.304x10

^{-3}/2 = 4.653x10

^{-3} mole of barium hydroxide. Now - you know number of moles and you know volume, as it was given to you - you have everything needed to calculate concentration, just use definition.

"Formula" you have tried to use won't work here, as it is a shortcut that can be used only for reactions where the stoichiometry is 1:1.