If the ionization constant of hypochlorous acid (HOCl) is 2.5 10⁻⁵ , the pH of 1.0 M of its solution is ( 5=0.7)

Step-by-step solution

aligned & Given, ionisation constant (K_a )=2.5 10⁻⁵ & and molarity (c)=1.0 M & K_a=C ^2 & aligned & = K_a C = 2.5 10⁻⁵ 1 & =0.5 10⁻³ or & =5 10⁻³ aligned aligned or Also, for dissociation of HOCl , the required relation is as follows ( C=1 ~m ) aligned & [ H ⁺ ]= =5 10⁻⁴ & and pH =- [ H ⁺ ] & =- [5 10⁻³ ] & =- 5+3 10 & =-0.7+3 & pH =2.3 & aligned