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

Step-by-step solution

Number of moles of ( HCl = 0.192 36.5 =0.0053 ) Volume (=0.5 ~L ) Molarity of ( HCl = 0.0053 0.5 =1.06 10⁻² M ) ( [ H ⁺ ] )in HCl solution (=1.06 10⁻² 95 100 1 10⁻² ) ( pH =- [ H ⁺ ]=- (1 10⁻² )=2 )