calculate molarity if H3PO4 from phenolphthalein titre (21ml) assuming it to be dibasic

H3PO4 + 2NaOH ---> Na2HPO4 + 2H2O

To calculate the molarity of H3PO4 (phosphoric acid) from the phenolphthalein titre, we first need to determine the amount of NaOH (sodium hydroxide) used in the reaction.

From the balanced chemical equation:
H3PO4 + 2NaOH → Na2HPO4 + 2H2O

We can see that 1 mole of H3PO4 reacts with 2 moles of NaOH.

Given that the titre of NaOH used is 21 mL, we need to convert this volume to moles. To do this, we'll use the molarity of NaOH, which should be provided in the question.

Let's assume the molarity of NaOH is represented by "C" moles per liter (C M).

The volume of NaOH used is 21 mL, which is equivalent to 0.021 L.

According to the equation, 1 mole of H3PO4 reacts with 2 moles of NaOH. So, if x moles of NaOH had reacted with the H3PO4, the molar ratio would be:

1 mole H3PO4 : 2 moles NaOH
1 mole H3PO4 : x moles NaOH

Using the equation for the molar ratio:

Moles of NaOH used = (C M) × (0.021 L)

Since the H3PO4 in the reaction is assumed to be dibasic, it means that there are 2 moles of H3PO4 for every mole that reacts with NaOH.

Therefore, the moles of H3PO4 would be half of the moles of NaOH used in the reaction:

Moles of H3PO4 = 0.5 × moles of NaOH used

Finally, to calculate the molarity of H3PO4 (M H3PO4), divide the moles of H3PO4 by the total volume (in liters) used for the reaction:

M H3PO4 = (0.5 × moles of NaOH used) / total volume (L)

Make sure to substitute the actual molarity value for NaOH into the equation to calculate the molarity of H3PO4.