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 $16.23\times 10^{-5}g$.

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = $380\times 10^{-3}$      $(1ml=10^{-3}L)$

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

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

$pOH=-log[OH^-]$

$5.06=-log[OH^-]$

$[OH^-]=0.00000871=8.71\times 10^{-6}mole/L$

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

Formula used : $Moles=Concentration\times Volume$

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

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

$\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles$

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 $H_2SO_4$

$3309.8\times 10^{-9}$ moles of $H^+$ ion is given by $\frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9}$ 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 = $(1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g$

Therefore, the mass of sulfuric acid needed is $16.23\times 10^{-5}g$.