Asolution of naoh has a concentration of 5.51 x 10-3 m. what is the ph of this solution? ph = 2.26 ph = 5.51 ph = 11.74 ph = 3.00

ph = 11.74

Explanation:

Given the dissociation of NaOH in water is NaOH Imported Asset Na+ + OH-.

Hence, if the concentration of NaOH is 5.51 x 10-3 M,

the concentration of OH- must also be 5.51 x 10-3 M,

Given that a mole to mole ratio of 1:1.

To find the pH of the solution, one will then use pOH = -log[OH-] and subtracting from 14, which is = pH+pOH = 14 for water at 298K

Hence, pOH = −log105.51×10−3

=2.26

Therefore, pH = 14−2.26 = 11.74