9. During a titration, 50.0 ml of 0.2M NaOH were required to neutralize 50.0ml of H_{3}*P * O_{4} What's the concentration of the H_{3}*P * O_{4} solution?

0.067M H3PO4

Explanation:

H3PO4 reacts with NaOH as follows:

H3PO4 + 3NaOH → 3H2O + Na3PO4

Where 1 mole of H3PO4 reacts with 3 moles of NaOH

To solve trhis question we need to find the moles of NaOH required. With the chemical equation we can find the moles of H3PO4 and its concentration as follows:

Moles NaOH:

50.0mL = 0.0500L * (0.20moles /L) = 0.0100 moles NaOH

Moles H3PO4:

0.0100 moles NaOH * (1mol H3PO4 / 3mol NaOH) = 0.00333 moles H3PO4

Concentration:

0.00333 moles H3PO4 / 0.0500L = 0.067M H3PO4

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.