Question

25mL of N / 10 NaOH solution exactly neutralises 20mL of an acid solution. what is the normality of the acid solution?​

Answer 1

Answer:

Avoid using an equation that looks like this: V1M1 = V2M2 or VaMa = VbMb. Such an equation works great for doing dilution problems, although the unit-factor method will just as well. But the aforementioned equation fails miserably for certain acids and bases.

Instead, use the unit-factor method just as you would for any stoichiometry problem.

HCl(aq) + NaOH(aq) → NaCl(aq) + HOH(l)

20.0 mL… 40.0mL

0.100M ……..?M

0.0200L x (0.100 mol HCl / 1L) x (1 mol NaOH / 1 mol HCl) / 0.0400L = 0.0500M NaOH

Of course, this problem is simple enough to do in your head. You don’t really need a mathematical “setup”. Since the ratio of acid to base is 1: 1 and the volume of base is twice the concentration of the acid, the base must be half has concentrated as the acid. Hence, the base is 0.05M.

Answer 2

N₁V₁ = N₂V 2 (NaOH) (Acid)

1 10 x 25 = N₂ x 20

N₂ = 25 10 × 20 = 0.125

Strength = Normality X Eq.mass Eq. mass of the acid 7.875 0.125 = 63.00.

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