Question

How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94

Answer 1

Answer : The mass of sulfuric acid needed is
`16.23* 10^(-5)g`.

Solution : Given,

pH = 8.94

Volume of solution = 380 ml =
`380* 10^(-3)`
`(1ml=10^(-3)L)`

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

`pH+pOH=14\\pOH=14-8.94=5.06`

`pOH=-log[OH^-]`

`5.06=-log[OH^-]`

`[OH^-]=0.00000871=8.71* 10^(-6)mole/L`

Now we have to calculate the moles of
`OH^-`.

Formula used :
`Moles=Concentration* Volume`

`\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]* Volume\\\text{ Moles of }[OH^-]=(8.71* 10^(-6)mole/L)* (380* 10^(-3)L)=3309.8* 10^(-9)moles`

For neutralization, equal number of moles of
`H^+` ions will neutralize same number of
`OH^-` ions.

`\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8* 10^(-9)moles`

As,
`H_2SO_4\rightarrow 2H^++SO^(2-)_4`

From this reaction, we conclude that

2 moles of
`H^+` ion is given by the 1 mole of
`H_2SO_4`

`3309.8* 10^(-9)` moles of
`H^+` ion is given by
`(3309.8* 10^(-9))/(2)=1654.9* 10^(-9)` moles of
`H_2SO_4`

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of
`H_2SO_4` × Molar mass of sulfuric acid

Mass of sulfuric acid =
`(1654.9* 10^(-9)moles)* (98.079g/mole)=162310.94* 10^(-9)=16.23* 10^(-5)g`

Therefore, the mass of sulfuric acid needed is
`16.23* 10^(-5)g`.