Question

The pH value of 0.1M solution of anilinium chloride is:(Ka of C6H5NH3+=10−6)

Answer 1

Explanation:
Given:The concentration of the solution of aniline chloride =0.1 MDissociation constant, Ka of C6H5NH3+=10−6We have to find the pH value of the solution.Dissociation of C6H5NH3+ is given as:C6H5NH3+⇋C6H5NH2++H+Considering Ostwald's dilution law, dissociation constant, Ka is given as:ka=Cα2α=KaCα=3.16×10−3Now, [H+] is given as:[H+]=C×α.....(i)On substituting the values in equation (i), we get[H+]=0.1×3.16×10−3[H+]=3.16×10−4Also, [H+] is related to pH as:pH=−log⁡[H+]...(ii)pH=−log⁡(3.16×10−4)pH=−log⁡(3.16)+4log⁡10pH=−0.4997+4(As,log⁡10=1)pH=3.5003PH≃3.5So, the correct answer is 3.50First Alternative SolutionpH of weak acid is given as:pH=12(pKa−log⁡C)or,pH=12(−log⁡Ka−log⁡C)On substituting the values of Ka, we getpH=12(−log⁡10−6−log⁡0.1)pH=12(6log⁡10−log⁡0.1)pH=12(6+1)pH=3.5Hence, the correct answer is 3.5.