How many milliliters of a 0.1500-M solution of KOH will be required to titrate 40.00 mL of a 0.0656-M solution of H₃PO₄?

\({{\rm{H}}_3}{\rm{P}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2KO}}{{\rm{H}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ }}{{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2}}{{\rm{H}}_2}{{\rm{O}}_{({\rm{l}})}}\)

According to the question, it is given that

The concentration of H₃PO₄ is 0.0656 Mand the given volume is 40.00 ml. Thus, for determining the moles of H₃PO₄, we can use the relation between volume, concentration, and moles which is given by:

\(\begin{array}{l}{\rm{Moles = Concentration }} \times {\rm{ Volume}}\\{\rm{Moles = 0}}{\rm{.0656 }} \times {\rm{ 40 = 2}}{\rm{.624 g}}\end{array}\)

Now, when 1 mole of \({{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}\) reacts with potassium hydroxide it requires 2 moles of KOH, thus the number of moles of potassium hydroxide will be:

\({\rm{2 }} \times {\rm{ 2}}{\rm{.624 = 5}}{\rm{.248 moles}}\)

As the number of moles of potassium hydroxide is 5.248 and the concentration is 0.1500 M. Thus, the volume of potassium hydroxide will be:

\(\begin{array}{l}{\rm{Volume = }}\frac{{{\rm{Moles}}}}{{{\rm{Concentration}}}}\\{\rm{Volume = }}\frac{{5.248}}{{0.1500}}{\rm{ = 34}}{\rm{.98 mL}}\end{array}\)