Question

• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL volumetric flask and diluting to 25.00 mL prior to removing the 10.00 mL sample used above. What was the molar concentration of phosphoric acid in the original stock solution?

Answer 1

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c) 0.166 M

Step-by-step explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

`H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O`

1 mole 3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

`H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O`

10 ml 17.50 ml

(x) M 0.200 M

Molarity =
`(0.2*17.5)/(1000)`

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

=
`0.0035 \ mole \ of NaOH* (1 mole of H_3PO_4)/(3 \ mole \ of \ NaOH)`

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =
`(mole \ \ of \ soulte )/( Volume \ of \ solution )`

Molar Concentration =
`(0.00166 \ mole \ of \ H_3PO_4 )/(10)*1000`

Molar Concentration = 0.166 M

∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M