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

Answer:

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

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 = $\frac{0.2*17.5}{1000}$

= 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

= $0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}$

= 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 =  $\frac{mole \ \ of \ soulte }{ Volume \ of \ solution }$

Molar Concentration = $\frac{0.00166 \ mole \ of \ H_3PO_4 }{10}*1000$

Molar Concentration = 0.166 M

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