Respuesta :
Explanation:
Reaction equations for the given species is as follows.
[tex]H_{3}PO_{4} + 2NaOH \righleftharpoons Na_{2}HPO_{4} + 2H_{2}O[/tex]
[tex]NaH_{2}PO_{4} + NaOH \rightleftharpoons Na_{2}HPO_{4} + H_{2}O[/tex]
At the first equivalence point we need 2 × 10 mmol NaOH.
At the second equivalence point we need 5 mmol of NaOH.
Hence, total moles of NaOH required is as follows.
(20 + 5) mmol = 25 mmol
We assume that volume of NaOH required is V.
[tex]25 mmol NaOH \times \frac{10^{-3} mol NaOH}{1 mmol NaOH} \times \frac{1000 ml V}{0.10 \text{mol NaOH}}[/tex]
= 250 ml V
Thus, we can conclude that 250 ml of 0.10 M NaOH must be added to reach second equivalence point of the titration of the [tex]H_{3}PO_{4}[/tex] and NaOH.
In the given case, 250 ml of 0.10 M NaOH must be added to attain the second equivalence point.
Calculation based on equivalence point:
The reactions taking place in the given case are:
- H₃PO₄ + 2NaOH ⇔ Na₂HPO₄ + 2H₂O
- NaH₂PO₄ + NaOH ⇔ Na₂HPO₄ + H₂O
Based on the given information, a solution comprises 10 mmol of H₃PO₄ and 5 mmol of NaH₂PO₄.
For the first equivalence point, there is a need of 2 × 10 mmol NaOH.
For the second equivalence point, there is a need of 5 mmol NaOH.
Now the total moles of NaOH needed is,
= 20 + 5 mmol
= 25 mmol
Now let the volume of NaOH required be V. Now putting the values we get,
[tex]= 25 mmol NaOH * \frac{10^{-3} mole NaOH}{1 mmol NaOH} * \frac{1000 mLV}{0.10 mol NaOH} \\= 250 ml[/tex]
Thus, 250 ml of 0.10 M NaOH is need to be added to attain the second equivalence point of the titration.
Find out more information about the calculations based on equivalence point here:
https://brainly.com/question/14014457
