Eo aa bh, 4 (] 0.63 g of a@ibasic acid)was dissolved in water to make 100 ml of solution. 20 ml of this solution required 10 ml of N/5 NaOH solution for complete neutralization. Find out the equivalent weight and molecular weight of the acid. Ea

1. Normality and the concept of an equivalent

• Normality, $N$ tells us "how many equivalents are present per litre".
• For acids an equivalent is the amount that can furnish one mole of $H^+$ ions.
• For bases (NaOH) an equivalent is the amount that can furnish one mole of $OH^-$.
  Because NaOH has only one $OH^-$ per formula unit, molarity = normality for NaOH.

2. The titration relation

At the end-point (neutralisation):

$N_{\text{acid}}\,V_{\text{acid}} = N_{\text{base}}\,V_{\text{base}}$

This is simply a restatement of the fact that the number of equivalents of acid = number of equivalents of base.

3. Connecting normality with mass of solute

If you have a solution prepared by dissolving w grams of a substance having equivalent weight $E$ in a final volume $V$ (in litres), then

$N = \frac{w}{E\;V}$

Re-arranging gives

$E = \frac{w}{N V}$

4. Molecular weight vs. equivalent weight for a poly-basic acid

For an acid that can release $n$ hydrogen ions (called basicity),

$\text{Molecular weight} = n \times \text{Equivalent weight}$

So for a dibasic acid ($n = 2$),

$M = 2E$

That is the final bridge from equivalent weight to molecular weight.

What’s happening here?

Imagine you have a lemon juice (the unknown acid) and you want to know how strong it is.
You mix a little of this juice with water and then slowly add a known amount of soap water (NaOH) that cancels out the sourness.
By carefully measuring how much soap water you needed, you can figure out how strong (how many sour particles) your lemon juice had.

Key idea in kid-speak

1. Weigh the lemon juice powder → $0.63\,\text{g}$
2. Make a big glass (100 mL) of its drink.
3. Take a small sip (20 mL) from that glass.
4. Count drops of soap water (10 mL of a known strength) needed to make that sip no longer sour.
5. Use the counting to tell how strong the whole glass was.
6. From that strength, work backwards to tell how heavy one "sour-packet" (equivalent) is and then how heavy one full molecule is (because this acid can give 2 sour packets).

That’s all the fancy chemistry boiled down!

Step-by-Step Solution

1. Data extraction

   • Mass of acid used, $w = 0.63\,\text{g}$
   • Total volume of solution, $V_{\text{total}} = 100\,\text{mL} = 0.1\,\text{L}$
   • Aliquot taken for titration, $V_{\text{acid}} = 20\,\text{mL}$
   • NaOH used, $V_{\text{base}} = 10\,\text{mL}$
   • Normality of NaOH, $N_{\text{base}} = \dfrac{1}{5}\,\text{N} = 0.2\,\text{N}$

2. Find normality of the acid solution using the titration formula:

$N_{\text{acid}} V_{\text{acid}} = N_{\text{base}} V_{\text{base}}$

Plugging in values (volumes in mL work because both sides share the same unit):

$N_{\text{acid}} \times 20 = 0.2 \times 10$ $N_{\text{acid}} \times 20 = 2$ $N_{\text{acid}} = \frac{2}{20} = 0.1\,\text{N}$

3. Relate normality to the mass present in the entire 100 mL solution
   For the whole 100 mL (0.1 L):

$N_{\text{acid}} = \frac{w}{E \; V_{\text{total}}}$

Substitute $N_{\text{acid}} = 0.1$, $w = 0.63\,\text{g}$, $V_{\text{total}} = 0.1\,\text{L}$:

$0.1 = \frac{0.63}{E \times 0.1}$

Multiply both sides by $E \times 0.1$:

$0.1 \times E \times 0.1 = 0.63$

$0.01 E = 0.63$

$E = \frac{0.63}{0.01} = 63$

Equivalent weight of the acid, $E = 63$ g eq$^{-1}$

4. Calculate molecular weight (acid is dibasic, $n = 2$):

$M = n \times E = 2 \times 63 = 126$

Molecular weight of the acid, $M = 126$ g mol$^{-1}$

Final answers:

• Equivalent weight = 63 g per equivalent
• Molecular weight = 126 g per mole