Question

calculate ph of 0.004M of NaOH

Answer 1

PH = 14 - pOH
pOH = -Log[OH-]
        =- Log[0.004] = -Log[4×10^(-3)]
        = -Log4 - Log[10^(-3)] = -Log[2²] - Log[10^(-3)] 
        = -2 Log2 +3
Log2 = 0.301
pOH = -2×0.301 +3 = -0.602 +3 = 2.398
now, pH = 14 - 2.398 = 11.602
Note that, pH of base should be greater than 7 but less than 14.