NEETChemistryIonic Equilibrium
The ionization constant of HF is 3.2 × 10 -4 . Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H₃O⁺, F⁻, and HF) in the solution and its pH.
Options
- ADegree of dissociation = 0.017, [H₃O⁺] = 0.037 M, [F⁻] = 0.017 M, [HF] = 1.963 M, pH = 0.71
- BDegree of dissociation = 0.032, [H₃O⁺] = 0.066 M, [F⁻] = 0.032 M, [HF] = 1.968 M, pH = 2.62
- CDegree of dissociation = 0.020, [H₃O⁺] = 0.040 M, [F⁻] = 0.020 M, [HF] = 1.980 M, pH = 0.69
- DDegree of dissociation = 0.025, [H₃O⁺] = 0.049 M, [F⁻] = 0.025 M, [HF] = 1.975 M, pH = 0.70
Correct answer
C. Degree of dissociation = 0.020, [H₃O⁺] = 0.040 M, [F⁻] = 0.020 M, [HF] = 1.980 M, pH = 0.69
Step-by-step solution
Correct Option is : (C) Degree of dissociation = 0.020, [H₃O⁺] = 0.040 M, [F⁻] = 0.020 M, [HF] = 1.980 M, pH = 0.69