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Question: The pH of 200 mL of 0.1N \({{H}_{2}}S{{O}_{4}}\) is: [A] 2 [B] 1 [C] log 40 [D] No option...

The pH of 200 mL of 0.1N H2SO4{{H}_{2}}S{{O}_{4}} is:
[A] 2
[B] 1
[C] log 40
[D] No option

Explanation

Solution

HINT: To solve this firstly convert the normality of the given solution to find the molarity and use it to find the number of moles. From the number of moles of hydrogen ions after dissociation of sulphuric acid you can find the concentration of them in 200mL solution. Use the concentration to find the pH of the solution. pH of a solution is given as- pH=−log⁡[H+]pH=-\log [{{H}^{+}}].

COMPLETE STEP BY STEP SOLUTION: We know that the pH of a solution gives us an idea about the strength of the solution. We use the term pH to measure the acidity of a solution in terms of the concentration of hydrogen ions in the solution.
Now let us define pH-
pH is the measure of acidity of the solution. It is equal to the negative log the molar concentration of hydronium ions released in the solution. We can write pH as-
pH=−log⁡[H+]pH=-\log [{{H}^{+}}]
We can see from the above relation that to find the pH of a solution we have to find the concentration of the hydronium ions in it.
In the given question we have 200 mL of 0.1N solution of H2SO4{{H}_{2}}S{{O}_{4}}. Firstly, let us find the number of moles of sulphuric acid.
For that we have to convert normality into molarity and then use it to find the number of moles.
Now for that we can use the formula, Normality = Molarity ×\times valency factor.
We know that sulphuric acid is dibasic. Therefore, the valency factor of sulphuric acid is 2.
Therefore, 0.1 N = molarity × 2\times \text{ 2}
Or, molarity = 0.05 mol.
Now, from the definition of molarity we know that,
Molarity = number of moles of solutevolume of solvent(L)\dfrac{number\text{ }of\text{ }moles\text{ }of\text{ }solute}{\text{volume }of\text{ }solvent(L)} or, number of moles = molarity×vol. of solute (L)molarity\times vol.\text{ of solute (L)}
The volume is given to us as 200mL. We can write it as 0.2 L.
Therefore, the number of moles of sulphuric acid = 0.05mol ×\times 0.2L = 0.01 mol/L.
Therefore,
H2SO4→2H++SO42−{{H}_{2}}S{{O}_{4}}\to 2{{H}^{+}}+S{{O}_{4}}^{2-}
We can write that 0.01 moles of H2SO4{{H}_{2}}S{{O}_{4}} dissociated to give us 0.02 moles of H+{{H}^{+}}.
Therefore, concentration of the hydrogen ions = 0.02200×1000=0.1\dfrac{0.02}{200}\times 1000=0.1
Now that we know the concentration of hydrogen ions we can find the pH-
pH=−log⁡[0.1]=1pH=-\log [0.1]=1
We can understand from the above discussions that pH of 200 mL of 0.1N H2SO4{{H}_{2}}S{{O}_{4}} is 1.

Therefore, the correct answer is option [B] 1

NOTE: pH 1 means the solution is acidic and pH=7 means the solution is completely neutralised. A neutral solution is formed when the number of hydroxyl ions and hydronium ions are equal. An acidic solution has a pH less than 7 and pH of an alkaline solution is greater than 7 and pH = 7 means that the solution is neutral with no excess ions.