What is the ph of a 0.001 62 m naoh solution?

The answer

first of all, we should know that NaOH is a strong base. For such a product, the conentration of the OH - is equivalent to the concentration of the NaOH itself.

that means:

[ OH - ] = [ NaOH] = 0.001 62

and for a strong basis, pH can be calculated as pH = 14 + log [ OH - ]

first we compute log [ OH - ]:

log [ OH - ] = log (0.001 62) = - 2.79

finally pH = 14 - 2.79 = 11.20