If pH of the solution is zero, what would be the normality of $250 \, \mathrm{ml}$ $\ce{H2SO4}$ solution?

My work: $\mathrm{pH} = 0$, hence the concentration of $\ce{H+}$ ions is $\pu{1 mol L^-1}$, which implies that the molarity of the solution is $\pu{1M}$ hence the normality of the solution must be $2~\mathrm{N}$. Is this correct?


$\ce{H2SO4}$ has 2 $\mathrm pK_\mathrm a$s, one that is strong whereby $\ce{H2SO4}$ ionizes to $\ce{HSO4-}$ and $\ce{H+}$ and another that is weak whereby $\ce{HSO4-}$ ionizies to $\ce{SO4^{2-}}$ and $\ce{H+}$

At $\mathrm p[\ce{H+}] = 0, [\ce{H+}] = \pu{1M}$

At $\mathrm {pH} = 0$, sulfuric acid is substantially $\ce{HSO4-}$ and $\ce{H+}$, with little $\ce{H2SO4}$ or $\ce{SO4^{2-}}$.

Therefore at $\mathrm p[\ce{H+}] = 0$, sulfuric acid concentration is $\pu{1M}$, because each mole of sulfuric acid has released one mole of $\ce{H+}$.

Normality of an acid is the concentration of the acid times the number of acidic protons per molecule. Sulfuric acid has two acidic protons per molecule.

Normality $= \pu{1M} \times 2 = \pu{2N}$


Here the $\mathrm{pH}$ is $0$, it means it has concentration of $\pu{1 M}$, i.e. $[\ce{H+}]$ is $1$. $$[\ce{H+}] = 10^{-\mathrm{pH}} = 10^{-0} = \pu{1 M}$$ Hence the molarity will be $1$. So its normality will be, $$\text{normality} = \text{molarity}\times\text{n-factor} = 1\times 1 = \pu{1 N}.$$ Here we have taken $\text{n-factor}$ for $\ce{H2SO4}$ will be $1$, because $\mathrm{pH}$ is given as $0$, so it will be a strong acid and in strong condition the $\text{n-factor}$ or valency factor becomes $1$ and so it forms to ions $\ce{H+}$ and $\ce{HSO4-}$, i.e. $\text{n-factor} = \text{no. of $\ce{H+}$ replaceable ions is 1}$, hence normality will be $\pu{1N}$.

  • $\begingroup$ You might want to clean up this answer a bit. See the guidelines in the Help Center to assist you in formatting your answer properly. $\endgroup$ – Todd Minehardt Jun 9 '17 at 15:33
  • 1
    $\begingroup$ No of substitutable H in H2SO4 is 2 right? $\endgroup$ – Mockingbird Jun 9 '17 at 15:34

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