Respuesta :

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

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