Answer
57.6 mL
Work Step by Step
First, we must examine the reactants of the reaction. We see that there is a reaction between an acid and a base. In order to solve this problem, we need to write a balanced reaction. We know that the products of the reaction would be $Ba_3(PO_4)_2$ and water. Next, we write the skeleton of the reaction to be $Ba(OH)_2 + H_3PO_4 => Ba_3(PO_4)_2 + H_2O$
We must balance this equation in order to follow the law of conservation of mass, meaning that there needs to be the same number of atoms of each element on both sides of the equation.
The balanced equation is : $3Ba(OH)_2 + 2 H_3PO4 => Ba_3(PO_4)_2 + 6H_2O$
Next, we calculate the number of moles of acid we have which is $0.141\times 0.0142$, or 0.0020022 moles of acid. From the balanced equation, we need 3 moles of the base for every 2 moles of acid. Hence, we need $0.0020022\times 3/2$ or 0.0030033 moles of base. By knowing the concentration of the base, we can calculate the number of liters of the base we need by dividing the moles we need by the concentration. Doing $0.0030033\div.0521$, we obtain that we need 0.0576 L. Converting this to mL, we get 57.6 mL.
