The image you've sent describes a quantitative analysis experiment (titration), not qualitative analysis. Assuming you're asking for the formulas relevant to the experiment shown in the image, here

The image you've sent describes a quantitative analysis experiment (titration), not qualitative analysis. Assuming you're asking for the formulas relevant to the experiment shown in the image, here they are:

Formulas for the Quantitative Analysis Experiment

1. Preparation of Standard Potassium Iodate (KIO₃) Solution (Procedure 1):

• Molar Mass of KIO₃: To calculate the molar mass, you'd sum the atomic masses of its constituent elements: $$M_{\text{KIO}_3} = M_{\text{K}} + M_{\text{I}} + (3 \times M_{\text{O}})$$ Using standard atomic masses: $$M_{\text{KIO}_3} = 39.10 \text{ g/mol} + 126.90 \text{ g/mol} + (3 \times 16.00 \text{ g/mol}) = 214.00 \text{ g/mol}$$
• Moles of KIO₃: $$\text{Moles of KIO}_3 = \frac{\text{Mass of KIO}_3 \text{ (from Table 1)}}{\text{Molar Mass of KIO}_3}$$
• Concentration of KIO₃ solution: The solution is made up to 250 cm³. $$C_{\text{KIO}_3} = \frac{\text{Moles of KIO}_3}{\text{Volume of solution (in dm}^3)}$$ $$\text{Volume in dm}^3 = \frac{250 \text{ cm}^3}{1000 \text{ cm}^3/\text{dm}^3} = 0.250 \text{ dm}^3$$

2. Standardization of Sodium Thiosulphate (Na₂S₂O₃) Solution (Procedure 2):

This involves two sequential redox reactions:

• Reaction 1: Iodate reacts with Iodide in acidic medium to produce Iodine. $$\text{IO}_3^{-}\text{(aq)} + 5\text{I}^{-}\text{(aq)} + 6\text{H}^{+}\text{(aq)} \rightarrow 3\text{I}_2\text{(aq)} + 3\text{H}_2\text{O(l)}$$
• Reaction 2: Iodine is then titrated with Thiosulphate. $$\text{I}_2\text{(aq)} + 2\text{S}_2\text{O}_3^{2-}\text{(aq)} \rightarrow 2\text{I}^{-}\text{(aq)} + \text{S}_4\text{O}_6^{2-}\text{(aq)}$$
• Stoichiometry: From Reaction 1, 1 mole of IO₃⁻ produces 3 moles of I₂. From Reaction 2, 1 mole of I₂ reacts with 2 moles of S₂O₃²⁻. Therefore, 1 mole of IO₃⁻ reacts with $3 \times 2 = 6$ moles of S₂O₃²⁻. $$\frac{\text{Moles of IO}_3^{-}}{1} = \frac{\text{Moles of S}_2\text{O}_3^{2-}}{6}$$
• Calculation of Concentration of Na₂S₂O₃: $$\text{Moles of S}_2\text{O}_3^{2-} = 6 \times \text{Moles of IO}_3^{-}$$ $$C_{\text{Na}_2\text{S}_2\text{O}_3} \times V_{\text{Na}_2\text{S}_2\text{O}_3} = 6 \times C_{\text{KIO}_3} \times V_{\text{KIO}_3}$$ Where $V_{\text{KIO}_3}$ is the volume of KIO₃ solution used (25 cm³ = 0.025 dm³) and $V_{\text{Na}_2\text{S}_2\text{O}_3}$ is the mean titre from Table 2. $$C_{\text{Na}_2\text{S}_2\text{O}_3} = \frac{6 \times C_{\text{KIO}_3} \times V_{\text{KIO}_3}}{V_{\text{Na}_2\text{S}_2\text{O}_3}}$$

3. Determination of Concentration of Copper (II) Sulphate (CuSO₄) Solution (Procedure 3):

This also involves two sequential redox reactions:

• Reaction 1: Copper(II) ions react with Iodide to produce Iodine. $$2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^{-}\text{(aq)} \rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)}$$
• Reaction 2: Iodine is then titrated with Thiosulphate. $$\text{I}_2\text{(aq)} + 2\text{S}_2\text{O}_3^{2-}\text{(aq)} \rightarrow 2\text{I}^{-}\text{(aq)} + \text{S}_4\text{O}_6^{2-}\text{(aq)}$$
• Stoichiometry: From Reaction 1, 2 moles of Cu²⁺ produce 1 mole of I₂. From Reaction 2, 1 mole of I₂ reacts with 2 moles of S₂O₃²⁻. Therefore, 2 moles of Cu²⁺ react with 2 moles of S₂O₃²⁻, which simplifies to a 1:1 molar ratio. $$$$