Answer:
6x10^-6 M
Explanation:
The relationship between pOH and [OH-] is
pOH = -log[OH-]
Solving this equation for [OH-] is
[OH-] = 10^-pOH
Plugging in the pOH makes
[OH-] = 10^-5.2
= 6.31x10^-6
Rounding to 1 sig fig since pOH and pH sig figs are only the number of decimals means to round to 1 sig fig or
6 x 10^-6 M
