1.

Calculate [OH–] if pOH = 8.3.

Answer»

pOH = -log10[OH]; 8.3 

= -log10[OH]

Taking antilog on both the sides

[OH] = antilog (-8.3) 

= [OH] = antilog (-9 - 8.3 + 9) 

= antilog [(0.7) × 10-9]

= 5.012 × 10-9 mol/dm3



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