Question

A 25.0 mL of 0.175M H3PO4 reacts with 25.0 mL of 0.205 M KOH. Write a balanced chemical equation to show this reaction. Calculate the concentrations of H3PO4 and KOH that remain in solution, as well as the concentration of the salt that is formed during the reaction.

Answer 1

A 25.0 mL of 0.175M H3PO4 reacts with 25.0 mL of 0.205 M KOH

moles of H3PO4 = 25 x 0.175 / 1000

                           = 0.004375

moles of KOH = 25 x 0.205 / 1000

                       = 0.005125

H3PO4 +    3 KOH ------------------------> K3PO4 + 3 H2O

1                    3

0.004375    0.005125

here limiting reagent is KOH so total KOH consumed in the reaction

1 mole H3PO4 ---------------> 3 mole KOH

howmany moles needed ------------> 0.005125 moles KOH

H3PO4 needed = 0.005125 /3

                           = 0.001708 moles

remaining moles of H3PO4 = 0.004375 - 0.001708

                                             = 0.002667

concentration of H3PO4 remaining = moles / total volume

                                                        = 0.002667 / (25 x 10^-3 + 25 x 10^-3)

                                                        =0.0533 M

concentration of H3PO4 remaining = 0.0533 M ------------------------------> answer

concentration of KOH remaining    = 0 M ------------------------------> answer

next salt foramtion :

3 moles KOH --------------------> 1 mole salt

0.005125 moles KOH gives ------------------> ?? moles salt

salt moles = 0.005125 / 3 = 0.001708 moles

salt concentration = moles / total volume

                           = 0.001708 / (50 x 10^-3)

                           = 0.0342 M

salt concentration = 0.0342 M ------------------------------> answer