Question

A 32.00 mL sample of an unknown H3PO4 solution is titrated with a 0.110 M NaOH solution. The equivalence point is reached when 24.83 mL of NaOH solution is added.What is the concentration of the unknown H3PO4 solution? The neutralization reaction is
H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)

Answer 1

Answer:

Concentration of $H_3PO_4$ in sample is 0.25 M.

Explanation:

From the reaction, one mole of $H_3PO_4$ reacts with 3 moles of NaOH.

Now, number of moles of NaOH, n = $molarity \times volume(in \ liters).$

$n=0.11\times \dfrac{24.83}{1000}\ mol=2.73\times 10^{-3}\ mol.$

Therefore, $2.73\times 10^{-3}$ mol of NaOH reacts with $3\times 2.73\times 10^{-3}$ $H_3PO_4$.

So, concentration of $H_3PO_4$ $=\dfrac{no\ of \ moles}{volume\ in\ liter}=\dfrac{3\times 2.73\times 10^{-3}}{\dfrac{32}{1000}}=\dfrac{3\times 2.73}{32}=0.25\ M.$

Therefore, concentration of $H_3PO_4$ in sample is 0.25 M.

Hence, this is the required solution.