50 ml of 0.001M phosphoric acid is exactly neutralised by addition of 0.005N NaOH.the molarity of resulting salt solution is

To determine the molarity of the resulting salt solution after neutralizing phosphoric acid with sodium hydroxide, we first need to understand the neutralization reaction and how to calculate the molarity of the resulting salt solution. Let's break this down step by step.

Understanding the Neutralization Reaction

Phosphoric acid (H₃PO₄) is a triprotic acid, meaning it can donate three protons (H⁺ ions). Sodium hydroxide (NaOH) is a strong base that will react with phosphoric acid to form sodium phosphate and water. The balanced chemical equation for the reaction can be represented as:

• H₃PO₄ + 3 NaOH → Na₃PO₄ + 3 H₂O

In this case, we are using 0.001 M phosphoric acid and 0.005 N sodium hydroxide. The normality of NaOH is equal to its molarity since it is a strong base that provides one hydroxide ion (OH⁻) per molecule. Therefore, 0.005 N NaOH is also 0.005 M.

Calculating Moles of Reactants

Next, we need to calculate the number of moles of phosphoric acid and sodium hydroxide involved in the reaction.

• Moles of H₃PO₄:

  Volume = 50 ml = 0.050 L

  Molarity = 0.001 M

  Moles = Molarity × Volume = 0.001 mol/L × 0.050 L = 0.00005 moles

• Moles of NaOH:

  To find the moles of NaOH, we need to know how much was added. Since we know the normality and the volume of NaOH used, we can calculate:

  Assuming we added enough NaOH to completely neutralize the acid, we can use the stoichiometry of the reaction. For every mole of H₃PO₄, we need 3 moles of NaOH.

  Thus, moles of NaOH required = 3 × moles of H₃PO₄ = 3 × 0.00005 = 0.00015 moles.

Volume of NaOH Solution Used

Now, we can find out how much volume of NaOH solution was used to provide 0.00015 moles:

• Volume of NaOH:

  Using the formula: Moles = Molarity × Volume, we can rearrange it to find Volume:

  Volume = Moles / Molarity = 0.00015 moles / 0.005 M = 0.03 L = 30 ml.

Calculating the Molarity of the Resulting Salt Solution

After the reaction, we will have a solution of sodium phosphate (Na₃PO₄) in water. The total volume of the resulting solution will be the sum of the volumes of the phosphoric acid and sodium hydroxide solutions:

• Total Volume:

  Volume of H₃PO₄ = 50 ml

  Volume of NaOH = 30 ml

  Total Volume = 50 ml + 30 ml = 80 ml = 0.080 L.

Now, we can calculate the molarity of the resulting sodium phosphate solution:

• Molarity of Na₃PO₄:

  Moles of Na₃PO₄ produced = moles of H₃PO₄ = 0.00005 moles (since 1 mole of H₃PO₄ produces 1 mole of Na₃PO₄).

  Molarity = Moles / Volume = 0.00005 moles / 0.080 L = 0.000625 M.

Final Result

The molarity of the resulting salt solution, sodium phosphate, after the complete neutralization of 50 ml of 0.001 M phosphoric acid with 0.005 N NaOH is 0.000625 M.