Deduce the normality equation, N1V1 = N2V2 . Experimental data obtained by titrating decinormal solution of oxalic acid with potassium permanganate solution are given below. [2] Expt. No. Volume of…

Derivation of the Normality Equation

Normality (N) is defined as the number of equivalents of solute present per litre of solution.

$$N=\frac{\text{equivalents of solute}}{\text{volume of solution (L)}}$$

For a titration, the equivalents of the titrant that react are equal to the equivalents of the analyte (titrand) at the equivalence point:

$${\text{equiv}}_1={\text{equiv}}_2$$

Using the definition of normality for each solution:

$$N_1{V}_1=N_2{V}_2$$

where

• $N_1,V_1$ – normality and volume of the titrant (solution 1)
• $N_2,V_2$ – normality and volume of the analyte (solution 2)

Thus the normality equation $N_1{V}_1=N_2{V}_2$ is obtained.

---

Part I – Oxalic acid – Potassium permanganate titration

a) Name of the titration  [1]

Redox (oxidation‑reduction) titration – oxalic acid (reducing agent) is titrated with acidic potassium permanganate (oxidizing agent).

Oxalic acidOxalic acid

b) Equivalent weight of $\mathrm{KMnO}{}_4$  [1]

In acidic medium the half‑reaction of permanganate is

2MnO₄^-+5C₂O₄^2-+16H⁺2Mn²⁺+10CO₂+8H₂O

$$\mathrm{MnO}{}_4{}^{-}+8\mathrm{H}{}^{+}+5\mathrm{e}{}^{-}\to \mathrm{Mn}{}^{2+}+4\mathrm{H}{}_2\mathrm{O}$$

Thus 5 equivalents of electrons are transferred per mole of $\mathrm{KMnO}{}_4$.

$$\text{Equivalent weight}=\frac{\text{Molar mass}}{n_{\text{e}}}$$

Molar mass of $\mathrm{KMnO}{}_4$ = 39.10 (K) + 54.94 (Mn) + 4 × 16.00 (O) = 158.04 g mol⁻¹

$$\boxed{\text{E.W. of }\mathrm{KMnO}{}_4=\frac{158.04}{5}=31.61{\text{g eq}}^{-1}}$$

c) Normality of the $\mathrm{KMnO}{}_4$ solution  [2]

The oxalic acid solution is decinormal ($0.10\text{N}$).
Moles of oxalic acid used per experiment:

$$\text{Equivalents of oxalic acid}=N_{\text{ox}}\times V_{\text{ox}}(\text{L})=0.10\text{N}\times 0.010\text{L}=1.0\times 10^{-3}\text{eq}$$

At the equivalence point the equivalents of $\mathrm{KMnO}{}_4$ equal those of oxalic acid:

$$N_{\text{KMnO4}}\times V_{\text{KMnO4}}(\text{L})=1.0\times 10^{-3}\text{eq}$$

Calculate $N_{\text{KMnO4}}$ for each experiment and take the average.

Average normality

$$\overline{N}=\frac{0.08696+0.09091+0.09091}{3}=0.0896\mathbf{N}$$

d) Reason for adding dilute $\mathrm{H}{}_2\mathrm{SO}{}_4$  [1]

Dilute sulfuric acid provides the strongly acidic medium required for the permanganate reduction:

Sulphuric acidSulphuric acid

$$\mathrm{MnO}{}_4{}^{-}+8\mathrm{H}{}^{+}+5\mathrm{e}{}^{-}\to \mathrm{Mn}{}^{2+}+4\mathrm{H}{}_2\mathrm{O}$$

The acid (i) prevents precipitation of $\mathrm{MnO}{}_2$, (ii) supplies the $\mathrm{H}{}^{+}$ ions needed for the half‑reaction, and (iii) keeps the reaction fast and complete.

e) Identify titrant and titrand  [1]

---

Part II – Alternative questions

a) Applications of the solubility‑product principle & common‑ion effect in qualitative analysis  [4]

b) Difference between Brønsted‑Lowry and Lewis concepts of bases  [2]

c) Degree of ionization of $\mathrm{HCN}$ (0.01 M)  [2]

For the weak acid dissociation:

$$\mathrm{HCN}\rightleftharpoons \mathrm{H}{}^{+}+\mathrm{CN}{}^{-},K_a=4.8\times 10^{-10}$$

Let $x$ = degree of ionization (mol L⁻¹ of $\mathrm{H}{}^{+}$ formed).

$$K_a=\frac{x^2}{c-x}\approx \frac{x^2}{c}(\text{since }x\ll c)$$

$$x=\sqrt{K_a\times c}=\sqrt{(4.8\times 10^{-10})(0.01)}=\sqrt{4.8\times 10^{-12}}=6.93\times 10^{-6}\text{M}$$

Degree of ionization

$$\alpha =\frac{x}{c}=\frac{6.93\times 10^{-6}}{0.01}=6.93\times 10^{-4}=6.9\times 10^{-4}(0.069\%)$$

$[\mathrm{H}{}^{+}]$ concentration

$$[\mathrm{H}{}^{+}]=x=6.9\times 10^{-6}\text{M}$$

pH

$$\text{pH}=-{\log }_{10}[\mathrm{H}{}^{+}]=-{\log }_{10}(6.9\times 10^{-6})=5.16$$

$$\boxed{\text{pH}=5.16}$$