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`        This is a question of chapter equilibrium. .. and i just want to know the formula used in it`
2 months ago

Arun
22989 Points
```							Ka for acetic acid is 1.8 x 10-5.Acetic acid, CH3COOH is a weak acid which ionizes partially and sodium acetate is a strong electrolyte which ionizes completely in the solution. Let x be the number of moles of acetic acid ionized.Then, the concentrations of various species are:CH3COOH (aq) + H2O [H3O+] = 3.6 x 10-6 mol L-1[H3O+] = 1.8 x 10-5 x 0.02 / 0.11.8 x 10-5 = [H3O+] x 0.1 / 0.02Ka for acetic acid is 1.8 x 10-5 (given)[CH3COOH]Ka = [H3O+] [CH3COO–]Ionization constant, Ka is:[CH3COOH] = 0.02 – x = 0.02 mol L-1[CH3COO–] = 0.1 + x = 0.1 mol L-1Now, x is very small in comparison to 0.1 so thatConcentration of acetic acid left unionized = 0.02 – xTherefore, total [CH3COO–] = 0.1 + xThe concentration of acetate ion from sodium acetate = 0.1 MSince sodium acetate is completely ionized,Concentration of acetate ions from acetic acid = xCH3COONa (aq) —————-> CH3COO– (aq) + Na+(aq) CH3COO– (aq) + H3O+ (aq)
```
2 months ago
Vikas TU
9476 Points
```							Dear student There is no special formula used in this , The formula used gere is only , the Product side concentration / Reactant side concentration . Hope this clears your doubt . Good Luck
```
2 months ago
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### Course Features

• 54 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions