Question

358ml of an H3PO4 solution is completely titrated by 876mL of 0.0102M Ba(OH)2 solution?

a) Write the balanced neutralization reaction
b) Calculate the molarity of H3PO4
c) Will the pH at the quivalence point be 7, above 7 or below 7

Answer 1

`H_3PO_4(aq) + Ba(OH)_2(aq) rarr BaHPO_4(aq) + 2H_2O(l)`

Explanation:

You will have to check with your teacher and text on this. But it is a fact that phosphoric acid behaves as a `"DIPROTIC"` acid in water.

Given this we can answer the questions according to the reaction:

`H_3PO_4(aq) + Ba(OH)_2(aq) rarr BaHPO_4(aq) + 2H_2O(l)`

Moles of `Ba(OH)_2` `=` `876xx10^-3*Lxx0.0102*mol*L^-1` `=` `8.94xx10^-3mol`.

Given the 1:1 equation, the concentration of the starting acid solution was:

`(8.94xx10^-3*mol)/(0.358*L)` `~=` `0.0250*mol*L^-1` with respect to `H_3PO_4`.

At the equivalence point, given the basicity of biphosphate, `pH>7`.

Note that I assume you are an undergrad not an A-level student, and at this level you should know that phosphoric acid is a diacid and that aqueous titration with base contains just 2 equivalence points.. See here for further details.