What is the pH of a solution that has a [OH-] = 0.0033 M?

1 Answer
Apr 19, 2018

pH=11.52

Explanation:

Here [OH^-]=33xx 10^(-4)M

rArr pOH=-log_10[OH^-]

rArr pOH=-log_10[33xx10^(-4)]

rArr pOH=-log_10(33)-log_10(10^(-4))

rArr pOH=-1.52+4.00

rArr pOH=2.48

We know that pH+pOH=pK_w............{pK_w=14}

rArr pH+2.48=14

:.pH=11.52