Respuesta :
Answer : The mass of sulfuric acid needed is [tex]16.23\times 10^{-5}g[/tex].
Solution : Given,
pH = 8.94
Volume of solution = 380 ml = [tex]380\times 10^{-3}[/tex] [tex](1ml=10^{-3}L)[/tex]
Molar mass of sulfuric acid = 98.079 g/mole
As we know,
[tex]pH+pOH=14\\pOH=14-8.94=5.06[/tex]
[tex]pOH=-log[OH^-][/tex]
[tex]5.06=-log[OH^-][/tex]
[tex][OH^-]=0.00000871=8.71\times 10^{-6}mole/L[/tex]
Now we have to calculate the moles of [tex]OH^-[/tex].
Formula used : [tex]Moles=Concentration\times Volume[/tex]
[tex]\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles[/tex]
For neutralization, equal number of moles of [tex]H^+[/tex] ions will neutralize same number of [tex]OH^-[/tex] ions.
[tex]\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles[/tex]
As, [tex]H_2SO_4\rightarrow 2H^++SO^{2-}_4[/tex]
From this reaction, we conclude that
2 moles of [tex]H^+[/tex] ion is given by the 1 mole of [tex]H_2SO_4[/tex]
[tex]3309.8\times 10^{-9}[/tex] moles of [tex]H^+[/tex] ion is given by [tex]\frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9}[/tex] moles of [tex]H_2SO_4[/tex]
Now we have to calculate the mass of sulfuric acid.
Mass of sulfuric acid = Moles of [tex]H_2SO_4[/tex] × Molar mass of sulfuric acid
Mass of sulfuric acid = [tex](1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g[/tex]
Therefore, the mass of sulfuric acid needed is [tex]16.23\times 10^{-5}g[/tex].
