Question

If 114mL of 0.008 04M NaOH completely titrates 118mL of H3PO4 solution, what is the molarity of the H3PO4 solution?

Answer 1

Question #60268, Chemistry / Other

If $114\mathrm{mL}$ of 0.008 04M NaOH completely titrates $118\mathrm{mL}$ of H3PO4 solution, what is the molarity of the H3PO4 solution?

**Answer:**

The amount of used NaOH upon titration is determined as follows:

$$v(\mathrm{NaOH}) = C(\mathrm{NaOH}) \times V(\mathrm{NaOH}), \text{ where } C - \text{the concentration of NaOH and } V - \text{the volume of NaOH}.$$ $$v(\mathrm{NaOH}) = 0.00804 \, \mathrm{M} \times 114 \, \mathrm{ml} = 0.91656 \, \mathrm{mmol}$$

Taking into the account that sodium hydroxide reacts with $\mathrm{H}_3\mathrm{PO}_4$ according to the equation:

$$3 \mathrm{NaOH} + \mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{Na}_3\mathrm{PO}_4 + 3 \mathrm{H}_2\mathrm{O}$$

it is clear that 3 molecules of NaOH neutralize 1 molecule of $\mathrm{H}_3\mathrm{PO}_4$.

Therefore the amount of titrated acid is:

$$v(\mathrm{H}_3\mathrm{PO}_4) = v(\mathrm{NaOH}) / 3 = 0.91656 \, \mathrm{mmol} / 3 = 0.30552 \, \mathrm{mmol}$$

Then the concentration of $\mathrm{H}_3\mathrm{PO}_4$ solution is defined:

$$C(\mathrm{H}_3\mathrm{PO}_4) = v(\mathrm{H}_3\mathrm{PO}_4) / V(\mathrm{H}_3\mathrm{PO}_4) = 0.30552 \, \mathrm{mmol} / 118 \, \mathrm{ml} = 0.00259 \, \mathrm{M}$$

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