Degree of dissociation of an acid HCl is 95% 0.192 g of the acid is present in 0.5 L of solution.The pH of the solution is

Question

Degree of dissociation of an acid HCl is 95% 0.192 g of the acid is present in 0.5 L of solution.The pH of the solution is (A) 2 (B) 1 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

No. of moles of HCl=(0.192/36.5)=0.0053 Volume = 0.5 It Molarity of HCl =(0.0053/0.5)=1.06×10-2 M [H+] in Hcl solution =1.06×10-2×(95/100) =100.7×10-4=1×10-2 Thus pH=-log[H]+ =-log(1×10-2)=2

Q. Degree of dissociation of an acid HCl is 95% 0.192 g of the acid is present in 0.5 L of solution.The pH of the solution is

2

1

3

4

No. of moles of HCl=36.50.192​=0.0053 Volume =0.5 It Molarity of HCl=0.50.0053​=1.06×10−2M [H+] in Hcl solution =1.06×10−2×10095​ =100.7×10−4=1×10−2 Thus pH=−log[H]+ =−log(1×10−2)=2