Calculate the ph of a solution that is 0.111 m hcl. ph=?

HCl(aq) > H⁺(aq)  + Cl(aq)  
     ↓                    ↓            ↓
i :   0.111 M          0               0
v: - 0.111 M     +0.111 M   +0.111 M

eq:    0            +0.111 M    +0.111 M 

HCl is a strong acid, so it ionizes 100% in solution :

PH = - log [ H+]

PH = - log [ 0.111]

PH = 0.9546

hope this helps!