Respuesta :
Answer : The final hydrogen ion concentration is [tex]1.58\times 10^{-13}M[/tex]
Explanation :
The chemical reaction equation will be:
[tex]H_3PO_4+3NaOH\rightarrow Na_3PO_4+3H_2O[/tex]
In this reaction, 1 mole of [tex]H_3PO_4[/tex] reacts with 3 mole NaOH.
So, the number of moles of [tex]H_3PO_4[/tex] present in 150 ml of 0.1 M solution is calculated as follows.
Number of moles = Concentration × Volume
Number of moles = 0.1 M ×0.150 L = 0.015 mol
As it reacts with 3 moles of NaOH.
Number of moles of NaOH = 3 × 0.015 mol = 0.045 mol
So, moles of NaOH in 400 mL of 0.2 M NaOH is as follows.
Number of moles = 0.2 M × 0.4 L = 0.080 mol
Number of moles remained after the reaction = (0.080 - 0.045) mol = 0.035 mol NaOH in 550 ml (400 ml + 150 ml)
As molarity is the number of moles present in liter of solution. Hence, molarity of NaOH is as follows.
[tex]\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution}}[/tex]
[tex]\text{Molarity}=\frac{0.035mol}{0.550L}=0.0636M[/tex]
Now we have to determine the hydroxide ion concentration.
As, [tex][OH^-][/tex] = 0.0636 M
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log 0.0636[/tex]
[tex]pOH=1.20[/tex]
Now we have to determine the pH.
As, pH + pOH = 14
pH = 14 - pOH
pH = 14 - 1.20
pH = 12.8
Now we have to determine the hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
[tex]12.8=-\log [H^+][/tex]
[tex][H^+]=1.58\times 10^{-13}M[/tex]
Therefore, the final hydrogen ion concentration is [tex]1.58\times 10^{-13}M[/tex]
