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If Kb for #NX_3# is #7.5x10^-6# , what is the pOH of a .175 M aqueous solution of NX3?

1 Answer

Feb 22, 2017

#sf(2.94)#

Explanation:

#sf(pOH=1/2[pK_b-logb])#

#sf(pK_b=-logK_b=-log(7.5xx10^(-6))=5.12)#

#sf(b=[NX_3]=0.175color(white)(x)"mol/l")#

#:.##sf(pOH=1/2[5.12-log(0.175)])#

#sf(pOH=2.94)#

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