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Question: A sulphuric acid solution has pH = 3. Its normality is...

A sulphuric acid solution has pH = 3. Its normality is

Explanation

Solution

Normality (N) is equal to molarity (M) multiplied by the n factor. So first we have to find the molarity with the help of pH and we should also know the n factor for sulphuric acid.
Formula used:
N=M×n−factorN = M \times n - factor
Here, NN = Normality
MM = Molarity
n−factorn - factor = number of replaceable hydrogen ion

Complete step by step answer:
The pH of a reaction is given as the negative logarithm (base 10) of its hydrogen ion concentration.
We can represent mathematically as:
pH=−log⁡10[H+]pH = - {\log _{10}}\left[ {{H^ + }} \right]
We have given pH=3 in the question:
3=−log⁡10[H+]3 = - {\log _{10}}\left[ {{H^ + }} \right]
Now by taking the negative sign on the left side of the equation we get:
−3=log⁡10[H+]- 3 = {\log _{10}}\left[ {{H^ + }} \right]
Now by taking the log on the left side we get antilog:
AL(−3)=[H+]AL\left( { - 3} \right) = \left[ {{H^ + }} \right]
So we get the concentration of hydrogen ion as:
[H+]=10−3\left[ {{H^ + }} \right] = {10^{ - 3}} (equation 1)
Now let us see the dissociation of sulphuric acid:
H2SO4→2H++SO4−2{H_2}S{O_4} \to 2{H^ + } + S{O_4}^{ - 2}
Now let the initial concentration of sulphuric acid be C i.e. [H2SO4]=C\left[ {{H_2}S{O_4}} \right] = C
So at equilibrium, the concentration of hydrogen ion becomes 2C i.e. [H+]=2C\left[ {{H^ + }} \right] = 2C
Now comparing the concentration of hydrogen ion with equation 1:
2C=10−32C = {10^{ - 3}}
So the value of C will come:
C=10−32C = \dfrac{{{{10}^{ - 3}}}}{2}
But we know that the initial concentration of sulphuric acid is denoted by C. So we can write:
[H2SO4]=C=10−32\left[ {{H_2}S{O_4}} \right] = C = \dfrac{{{{10}^{ - 3}}}}{2}
Now, we know that in sulphuric acid we have two replaceable hydrogen atoms. So, we can write:
n−factor=2n - factor = 2
So normality can be given as:
N=M×n−factorN = M \times n - factor
By substituting the value we get:
N=10−32×2N = \dfrac{{{{10}^{ - 3}}}}{2} \times 2
By canceling the 2 we get:
N=10−3NormalN = {10^{ - 3}}Normal

Therefore we can conclude that the answer to this question is 10−3N{10^{ - 3}}N.

Note:
To find the n-factor for sulphuric acid, focus on finding the concentration from the given pH as normality depends on both concentration and n-factor of the substance.