1.

0.002 M kOH calculate pH

Answer»


(d) 0.002 M KOH

KOH ↔ K+ + OH-

[OH-] = [KOH]

[OH-] = 0.002

pOH = - LOG [OH-] =-log[0.002] = 2.69

pH= 14-2.69 = 11.70



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