Question

what is the ratio of pH of 0.1 M${{H}_{2}}S{{O}_{4}}$and 0.1 N${{H}_{2}}S{{O}_{4}}$?

Answer 1

Hint: pH of any solution is the negative log of hydrogen ion concentration. The molarity (M) and normality (N) are units in which concentration of any solution is expressed. Molarity is the number of moles of solute in per liter of solution, while normality is the gram equivalent of solute in per liter of solution.

Complete answer:
pH of any solution tells us about the hydrogen ion concentration or the potency of hydrogen ions or hydronium ions in that solution. We have been given the concentration of sulfuric acid${{H}_{2}}S{{O}_{4}}$ in terms of molarity and normality, for which we have to find the ratio of pH of 0.1 M and 0.1 N.
The dissociation of sulfuric acid, gives us the hydronium ion, whose concentration tells us the pH of the solution, the equation is ${{H}_{2}}S{{O}_{4}}(\ell )+2{{H}_{2}}O(\ell )\to 2{{H}_{3}}{{O}^{+}}+S{{O}_{4}}^{2-}$
As molarity is the number of moles of solute per liter solvent, so, the concentration of 0.1 mol/L will be 0.2 mol/L due to 2 moles of ${{H}_{3}}{{O}^{+}}$ ions, so pH will be
$pH=-{{\log }_{10}}(0.2)$
pH = 0.7
While, normality is the number of gram equivalent of solute upon liter of solution, so sulfuric acid being diprotic, so its normality will be divided by its n factor 2, so, $\dfrac{0.1}{2}=0.05mol/L$ which multiplied to 2 moles will give, 0.1 mol/L, so the pH will be,
$pH=-{{\log }_{10}}(0.1)$
pH = 1
So, the ratio will become, 0.7 : 1 which when rounded off will be, 7 : 10.
Hence, the ratio of pH of 0.1 M ${{H}_{2}}S{{O}_{4}}$ and 0.1 N${{H}_{2}}S{{O}_{4}}$is 7 : 10 ( 0.7 : 1).

Note:
The relationship between molarity and normality is molarity $\times $ valence factor = normality. Valence factor may be the number of liberating protons. It is also called “n factor” that is the number of reacting protons that can be liberated into a solution, so sulfuric acid is diprotic with n factor or valence factor as 2.