25 ml of a solution of barrium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of b rium hydroxide solution was

The reaction involved in the titration is:

Ba(OH)2 + 2HCl -------> BaCl2 + 2H2O

The equation that is used to find the molarity of unknown solutions in titrations is:

M1V1/n1 = M2V2/n2

Where

M1 and M2 are, respectively, the molarities of Ba(OH)2 & HCl solutions.

V1 and V2 are, respectively, the volumes of Ba(OH)2 & HCl solutions.

n1 = Stoichiometric coefficient of Ba(OH)2 = 1

n2 = Stoichiometric coefficient of HCl = 2

Therefore:

molarity of barium hydroxide solution, M1 = (M2V2/n2) x (n1/V1) = (0.1 x 35 / 2) x (1/25) = 0.07 M

Number of moles in 1000 ml of 0.1 M solution of HCl = 0.1 moles Volume of Ba(OH)2 solution = 25 mlNumber of moles in 35 ml of 0.1 M solution of HCl = (0.1/1000) x 35 = 0.0035 moles According to the following equation...... 1 mole of Ba(OH)2 reacts with 2 moles of HCl Number of moles of Ba(OH)2 required to react with 0.0035 moles of HCl = 0.0035/ 2 =0.00175 moles Molarity = Number of moles of Ba(OH)2 / volume of solution in L 0.00175/0.025 = 0.7 M.