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

Concentration of
`H_3PO_4` in sample is 0.25 M.

Step-by-step explanation:

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

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

`n=0.11* (24.83)/(1000)\ mol=2.73* 10^(-3)\ mol.`

Therefore,
`2.73* 10^(-3)` mol of NaOH reacts with
`3* 2.73* 10^(-3)`
`H_3PO_4`.

So, concentration of
`H_3PO_4`
`=(no\ of \ moles)/(volume\ in\ liter)=(3* 2.73* 10^(-3))/((32)/(1000))=(3* 2.73)/(32)=0.25\ M.`

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

Hence, this is the required solution.