50 ml decimolar H2SO4 when mixed with 50 ml decimolar NaOH then normality of resultant solution is?

0.05 N

Explanation:

You can calculate the normality of an acid by multiplying the molarity by the number of acid hydrogens. So, the normality for H₂SO₄ is its molarity multiplied by 2.

For bases, the normality is the product of the molarity and the number of OH⁻ ions. So, for NaOH the normality is its same molarity.

Also remember the definitions and formulae:

Molarity = number of moles of solute / volume in liters of the solution Normality = number of equivalents of solute / volume in liters of solution Normality = molarity * number of hydrogens or OH⁻.

The chemical equation for the reaction of H₂SO₄ with NaOH is:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

As you see, the mole ratio is 1 mol H₂SO₄ : 2 mol NaOH meaning that every mol of H₂SO₄ neutralizes 2 moles of NaOH.

Since both reactants have the same molarity (decimolar = 0.1M) and the same volume, there are the same number of moles of each, but they have different normalities, which means different number of equivalents.

At the same molarity, the number of equivalents of H₂SO₄ is double than the number of equivalents of NaOH.

Then, after reaction half the number of equivalents of H₂SO₄ will remain in solution. The calculations are:

Before reacting:

Number of moles of each compound before reacting:

H₂SO₄: 0.050 liter * 0.1 M = 0.005 mol

NaOH: 0.050 liter * 0.1 M = 0.005 mol

After reacting:

Since ther reaction is 1 mol acid : 2 moles base, the base is the limiting reactant (it will be fully consumed).

Only 0.005 / 2 mol of acid will react and 0.005 / 2 mol will remain in solution. That is 0.0025 mol.

The volume of the solution will be 50 ml + 50 ml = 100 ml = 0.10 liter

So, after reaction you have:

Molarity of acid = 0.0025 mol / 0.10 liter = 0.025 M

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Normality of acid = 0.025 * 2 = 0.05 N ← answer